Simple (?) probability problem
Ed Hall
you). If it lands heads he adds ten to his number, if it is tails he
subtracts ten, then he gives you his answer. With what probability can you
guess his original number?
Ed
Bruce Bowler
are two possible answers. x+10 and x-10. given no restrictions on x,
it's just a guess which one of x+10 and x-10 is the correct answer.
David A Karr
Ed Hall <e...@copse.free-online.co.uk> wrote:
>You get a friend to think of a number and then flip a coin (without showing
>you). If it lands heads he adds ten to his number, if it is tails he
>subtracts ten, then he gives you his answer. With what probability can you
>guess his original number?
Assuming your friend picks a number according to a probability
distribution (whether he is aware of the distribution or not), and
assuming you can guess a number that lies between numbers he could
pick, your chance is somewhat better than 1/2.
The technique is similar to the technique for picking the better
envelope in the Two Envelopes problem (in the case where you are
permitted to count the money in one envelope). In fact this is
just the Two Enveloples problem converted to a logarithmic scale.
--
David A. Karr "Groups of guitars are on the way out, Mr. Epstein."
ka...@shore.net --Decca executive Dick Rowe, 1962
Eric Sosman
I think the probability of guessing the number is just a tiny
bit less than 50%, because there's a non-zero chance that either
I or my friend will botch the arithmetic ...
Bruce Bowler wrote:
>
> Seems to me that you've got a 50/50 chance. Given his result, x, there
> are two possible answers. x+10 and x-10. given no restrictions on x,
> it's just a guess which one of x+10 and x-10 is the correct answer.
>
> Ed Hall wrote:
> >
> > You get a friend to think of a number and then flip a coin (without showing
> > you). If it lands heads he adds ten to his number, if it is tails he
> > subtracts ten, then he gives you his answer. With what probability can you
> > guess his original number?
> >
> > Ed
David A Karr
>assuming you can guess a number that lies between numbers he could
>pick,
Oops, wrong assumption. To get a better than 1/2 chance, you need to
guess a number N such that your friend's guess had a positive prior
probability to be in the range [N - 10, N + 10) (or in the range
(N - 10, N + 10] when you use an obvious variation on the guessing
strategy).
I hope the above is a useful clue for anyone who hasn't seen the
solution already.
Ed Bauer
Ok, the number I guess is -498,184,204.
I flip a coin. It changes my number to either
-498,184,214 or -498,184,194.
What are the odds that you're going to guess
that exact number, given the range of
negative-infinity through positive-infinity?
Where does it say I have to tell you the
number I pick? Or is it the typical
rec.puzzles sloppily-worded question?
Ed. B.
David A Karr wrote in message ...
mark edward hardwidge
> Where does it say I have to tell you the number I pick?
"If it lands heads he adds ten to his number, if it is tails he
subtracts ten, then he gives you his answer."
Doesn't that "then he gives you his answer" part imply that
you'd have to tell the number?
--
Mark E. Hardwidge
hard...@uiuc.edu
Nick Wedd
<ka...@shore.net> writes
>Assuming your friend picks a number according to a probability
>distribution (whether he is aware of the distribution or not), and
>assuming you can guess a number that lies between numbers he could
>pick, your chance is somewhat better than 1/2.
I agree. If my friend says "-3", I think it far more likely that he
started with 7, than that he started with -13.
Nick
--
Nick Wedd ni...@maproom.co.uk
MR
> Whoa---
>
> Ok, the number I guess is -498,184,204.
>
> I flip a coin. It changes my number to either
> -498,184,214 or -498,184,194.
> What are the odds that you're going to guess
> that exact number, given the range of
> negative-infinity through positive-infinity?
>
> rec.puzzles sloppily-worded question?
If the number you guess is an integer, with no restriction, the probability of guessing a concrete number into a infinite group of
numbers is ZERO.
Nick Wedd
<mr.wanadoo.es@NOSPAM.?> writes
>If the number you guess is an integer, with no restriction, the probability of
>guessing a concrete number into a infinite group of
>numbers is ZERO.
Really? Please think of an integer and then tell us what it is.
David A Karr
>Ok, the number I guess is -498,184,204.
>
>I flip a coin. It changes my number to either
>-498,184,214 or -498,184,194.
>
>Where does it say I have to tell you the
>number I pick?
It doesn't. It _does_ say you have to tell me the number you got
in the second step (after the coin flip).
If the number you tell me is -498,184,214, for example, then I know
that the correct answer is either -498,184,224 or -498,184,204.
Looks like my odds of guessing correctly are pretty good.
> Or is it the typical
>rec.puzzles sloppily-worded question?
No, it's a typical case of an exactly-worded question that will trip
you up if you read it sloppily.
MR
> MR writes
>
> >If the number you guess is an integer, with no restriction, the probability of
> >guessing a concrete number into a infinite group of
> >numbers is ZERO.
>
> Really? Please think of an integer and then tell us what it is.
I've thinked a number.....
.
.
.
.
.
It's -654335645444
Have you guessed it?
Nick Wedd
<mr.wanadoo.es@NOSPAM.?> writes
No, I did not guess it. But I am surprised how small it is. You had
the whole infinite range of integers to choose from, and you chose one
with size less than 10^12. This makes me wonder if you really assigned
an equal probability to each integer.
dp
number, if it is tails he subtracts ten, then he gives you his
----
I'd say that its 100% if a tail is thrown. Providing he can
subtract 10 from his number correctly.
If he throws a head then the odds are pretty remote.
-----------------------------------------------------------
Got questions? Get answers over the phone at Keen.com.
Up to 100 minutes free!
http://www.keen.com
macavity
>This makes me wonder if you really assigned
>an equal probability to each integer.
>
>Nick
>--
You don't need to wonder, there is _no_ way that an equal
probability could have been assigned to each integer in Z, or Z+.
I don't think anywhere in the problem it was mentioned that the
probability is a uniform distribution - if mentioned, the
problem is wrong. In the absence of mentioning, if you assume a
particular distribution, you do so at your own peril. In this
case, your choice of uniform is infeasible.
MR
> MR
> >I've thinked a number.....
> >
> >
> >
> >Have you guessed it?
>
> No, I did not guess it. But I am surprised how small it is. You had
> the whole infinite range of integers to choose from, and you chose one
> an equal probability to each integer.
Hhhhhmmmmm.... do you mean that we have to keep in mind that the probability of each digit is different (someone wrote about it time
ago... 1 appear more times than 2 and so on...) ???
Gerry Quinn
>In article <sCEf5.2050$ph.41533@stones>,
>Ed Hall <e...@copse.free-online.co.uk> wrote:
>>you). If it lands heads he adds ten to his number, if it is tails he
>>subtracts ten, then he gives you his answer. With what probability can you
>>guess his original number?
>
>distribution (whether he is aware of the distribution or not), and
>assuming you can guess a number that lies between numbers he could
>pick, your chance is somewhat better than 1/2.
>
>envelope in the Two Envelopes problem (in the case where you are
>permitted to count the money in one envelope). In fact this is
>just the Two Enveloples problem converted to a logarithmic scale.
>
There is a difference, because there is a value (ten) incorporated in
the puzzle which he can use to scale his probability distribution. Let
us use real numbers (or probably computable numbers) rather than
integers. In the envelope paradox, the envelope can contain an
infinitesimal quantity with probability 50%. In this problem, it should
contain a number approacting infinity with 100% probability. The
existence of the value 'ten', and the additive nature of the change,
mean that zero is not an option.
- Gerry Quinn
ro...@my-deja.com
"Ed Hall" <e...@copse.free-online.co.uk> wrote:
>You get a friend to think of a number and then flip a coin
>(without showing you). If it lands heads he adds ten to his
>number, if it is tails he subtracts ten, then he gives you his
>answer. With what probability can you guess his original number?
Hi,
This reminds me a nice variant of the famous "surprise exam" problem
that can be formulated with almost the same words, namely :
You get a friend to think of a (strictly) positive number and then flip
a coin (without showing you). If it lands heads he adds one to his
number, if it is tails he doesn't (he keeps his number unchanged), then
he gives you his answer. Furthermore, your friend must choose his
number in such a way that there will be no logical proof for you to
guess his original number. Is it possible ?
Well, obviously your friend cannot choose "one", because if he does so
and gets tails he must tell you "one". If you hear "one" you know his
original number was "zero" or "one". Since the rules don't allow him to
choose zero, you could prove his original number must be "one".
Now, your friend cannot choose "two", because if he does so and gets
tails he must tell you "two". If you hear "two" you know his original
number was "one" or "two". Since you can prove he's not allowed to
choose "one" according to the paragraph above, you could prove his
original number must be "two".
Now, your friend cannot choose "three", because if he does so and gets
tails he must tell you "three". If you hear "three" you know his
original number was "two" or "three". Since you can prove he's not
allowed to choose "two" according to the previous paragraph, you could
prove his original number must be "three".
And so on, to infinity...
Do you agree ?
--
Dominique
Sent via Deja.com http://www.deja.com/
Before you buy.
David A Karr
>[David Karr wrote:]
>>In fact this is
>>just the Two Enveloples problem converted to a logarithmic scale.
>
>In the envelope paradox, the envelope can contain an
>infinitesimal quantity with probability 50%. In this problem, it should
>contain a number approacting infinity with 100% probability.
I wrote "logarithmic scale," hoping you'd figure out the conversion.
Now I'll be more explicit. Choose an initial number according to your
assumed distribution in the Two Envelopes problem. Take the logarithm
of that number, base 2^(1/10). You now have a number that (according
to your assumptions) has magnitude approaching infinity with
probability 1. (Note that nothing prevents you from picking negative
numbers in the number-guessing problem.)
On the other hand, you've just picked one of the nastiest possible
assumptions about the prior probability of the chosen number, to which
others have already objected (do you really think that with
probability 1 you will pick a number that you couldn't even specify to
me before we all die?). My assumption was that the number was
distributed according to a probability distribution (over integers
_or_ reals, or even complex numbers) as defined in any of the standard
texts--I just don't know in advance what that distribution is.
Whatever it is, though, it's the distribution of the logarithm of the
amount in the Two Envelopes, for some legal distribution of that amount.
>The
>existence of the value 'ten', and the additive nature of the change,
>mean that zero is not an option.
Not an option? Only in the sense that you have zero probability
of hearing zero (or any other predetermined number, for that matter).
Or perhaps I just didn't understand your sentence.
Wesley Clifford
>You get a friend to think of a (strictly) positive number and then flip
>a coin (without showing you). If it lands heads he adds one to his
>number, if it is tails he doesn't (he keeps his number unchanged), then
>he gives you his answer. Furthermore, your friend must choose his
>number in such a way that there will be no logical proof for you to
>guess his original number. Is it possible ?
>
>Well, obviously your friend cannot choose "one", because if he does so
>and gets tails he must tell you "one". If you hear "one" you know his
>original number was "zero" or "one". Since the rules don't allow him to
>choose zero, you could prove his original number must be "one".
>
>Now, your friend cannot choose "two", because if he does so and gets
>tails he must tell you "two". If you hear "two" you know his original
>number was "one" or "two". Since you can prove he's not allowed to
>choose "one" according to the paragraph above, you could prove his
>original number must be "two".
>
>Now, your friend cannot choose "three", because if he does so and gets
>tails he must tell you "three". If you hear "three" you know his
>original number was "two" or "three". Since you can prove he's not
>allowed to choose "two" according to the previous paragraph, you could
>prove his original number must be "three".
>
>And so on, to infinity...
>Do you agree ?
I flipped my coin.
17
What was my original number?
--Wesley
The only stupid question is the unasked question.
And that's a good thing, because I hate stupid questions!
macavity
>In article <sCEf5.2050$ph.41533@stones>,
> "Ed Hall" <e...@copse.free-online.co.uk> wrote:
>
>>You get a friend to think of a number and then flip a coin
>>(without showing you). If it lands heads he adds ten to his
>>number, if it is tails he subtracts ten, then he gives you his
>>answer. With what probability can you guess his original
number?
>
>Hi,
>This reminds me a nice variant of the famous "surprise exam"
problem
>that can be formulated with almost the same words, namely :
>
I don't think the example you talk about is related to the
referred problem. Your example is part of a class
of "incremental counting" things ... some described below.
This is sort of like the argument that there is no large
positive number. Take 0, definitely not large. Next, 1 cannot
be large, as it is just a small number plus 1. ... If n is not
large, then n+1 can't be large. So we have "proved" that there
is no large number.
There is a similar example to do with a prisoner condemned to
death if he cannot predict which day of the next week he is to
be hung. In this case, you start with the last day and work
backwords. I think the standard quip for this is that the
prisoner cannot arrive at any logical answer, and hence could be
executed on any day.
The general rule is - this sort of thing just doesn't work. In
the current example, as long as the guesser thinks of
some "large enough" number and then flips the coin, I am sure
that you have _no_ logical way of pinpointing _exactly_ what was
the original number. (Please note that if your logic doesn't
work, I do not have to explain to you why it doesn't. I just
have to ensure that your logic doesn't work, and by giving you
the number, I destroy your logic which builds on the "fact" that
no number can be given.;)
Jens Brix Christiansen
Nice version. It is fairly obvious (to me at least) that this paradox
is equivalent to the paradox of the Byzantine Generals. And I usually
claim that that in turn is a version of the paradox of the heap, but I
usually fail to convince astute rec.puzzlers of that.
macavity
>Nice version. It is fairly obvious (to me at least) that this
paradox
>is equivalent to the paradox of the Byzantine Generals. And I
usually
>claim that that in turn is a version of the paradox of the
heap, but I
>usually fail to convince astute rec.puzzlers of that.
>
>
I tend to agree that this is similar to the heap paradox - (I
assume equivalent to the "large number" problem described).
However I don't think they are identical - in "heap", "large
number" etc., the definitions of "heap" "large number" etc.
themselves are vague and this creates the paradox - for e.g., if
I had clearly said that "large numbers are ones above a
million", the paradox disappears.
In this example, (and the condemned prisoner), there is a sort
of consistent circular logic that does not allow you to decide;
but you can twist the fact that no choice is logical to give you
choices which cannot be logically distinguished - and presto,
you have a range of choices.
So while there is a sequential element to both, one is stemming
from a gray definition and incremental progression from white to
black, the other is a chain of consequential logic that is
perfect - which leads to its own pitfall.
MR
>
> You don't need to wonder, there is _no_ way that an equal
> probability could have been assigned to each integer in Z, or Z+.
>
> I don't think anywhere in the problem it was mentioned that the
> probability is a uniform distribution - if mentioned, the
> problem is wrong. In the absence of mentioning, if you assume a
> particular distribution, you do so at your own peril. In this
> case, your choice of uniform is infeasible.
I can think on a normal curve probability distribution of the integers...
People often thinks on small numbers, don't they?
Jens Brix Christiansen
I like this distinction. Let me simplify to G for "gray definition" and
L for "logic leads to contradiction".
The original Surprise Hanging (or exam or test of civil defense siren)
paradox is certainly of type L.
The Heap paradox is of type G. So is an equivalent given by the
definition "a large number is defined as a number so large that that
number of pebbles constiute a heap when piled up." And so, I think, is
the Byzantine Generals paradox: "We attack when a large number of
messages has been exchanged."
The paradox described in this thread (choose positive integer, flip coin
add one if tails, tell number, ask player what coin showed) is somewhere
in between. It has the logic characteristics of type L, but only by
induction. And the answer to "what should I choose?" is "choose some
large number". In that way, we are right back in G.
macavity
Well, what is a "small" number anyway?
Even with a "normal curve on integers", you won't get very far
if you want to cover "large (whatever that is)" numbers - you
will fix a mean for the normal (which is bound to be small for
someone's standards), and a sigma (ditto) and hence you can't
expect to get large numbers again,.
Tell you what, on positive integers, randomly choose an integer
using the probability distribution P(k) = 6/(k pi)^2. I'll
guarantee that you can expect a large number on the average ;).
macavity
In both your L and G variety, there are inductive steps in the
development of the paradox. This is the similarity.
In this example (for L), you need not actually choose something
large - mea culpa. Anything non-trivial is possible - if I say
2, as the final answer, _still_ you will not be able to say what
my original number was (it could have been 1 or 2, and there is
no logic which will help you prefer one over the other). So you
see, L really does not fall into G when you come to resolution
of the paradox. Same for the "condemned prisoner" business -
there are 7 days in the D-week, and he can be executed any day.
George Smith
Ed Bauer wrote:
> Whoa---
>
> Ok, the number I guess is -498,184,204.
>
> I flip a coin. It changes my number to either
> -498,184,214 or -498,184,194.
> What are the odds that you're going to guess
> that exact number, given the range of
> negative-infinity through positive-infinity?
>
> Where does it say I have to tell you the
> number I pick? Or is it the typical
> rec.puzzles sloppily-worded question?
>
> Ed. B.
>
> David A Karr wrote in message ...
> >David A Karr <ka...@shore.net> wrote:
> >>assuming you can guess a number that lies between numbers he could
> >>pick,
> >
> >Oops, wrong assumption. To get a better than 1/2 chance, you need to
> >guess a number N such that your friend's guess had a positive prior
> >probability to be in the range [N - 10, N + 10) (or in the range
> >(N - 10, N + 10] when you use an obvious variation on the guessing
> >strategy).
> >
> >I hope the above is a useful clue for anyone who hasn't seen the
> >solution already.
> >
Kevin Buhr
>
> There is a difference, because there is a value (ten) incorporated in
> the puzzle which he can use to scale his probability distribution. Let
> us use real numbers (or probably computable numbers) rather than
> mean that zero is not an option.
I don't understand.
Suppose, in the envelope paradox, after writing a positive, computable
number "x" on one envelope and "2x" on the other, I compute the
logarithm base 2^(1/20) of each number, jotting it down in red ink
below. Note that the red-ink numbers (each of which may be positive,
negative, or even zero) differ by 20.
How is this different from picking a computable number "y", putting
"y+10" and "y-10" in red on the envelopes, and computing 2^(1/20) to
the power of the red numbers to get an "x" and "2x"?
Any distribution for "x" with support on the positive real numbers can
be rewritten as a distribution for "y" with support on the real line.
Computability doesn't seem to be an issue, and I don't know what
you're trying to say about "infinitesimals" and "infinities"---you
seem to be saying something about the distributions, but I don't
know what.
--
Kevin <bu...@stat.wisc.edu>
Ed Hall
strategy I had been told to get a better than 50% chance was as follows:
While the other person is thinking of a number you also think of a number.
His number will be either 10 more or 10 less than the number he quotes. If
the number he quotes is less than the number you thought of, guess that his
actual number was 10 more. If the number is more, guess his actual number is
10 less. Assuming your numbers are completely different, you have a simple
50/50 chance, however if you (by chance?!?) pick a number within 10 of his,
you will pick the correct number every time. This means your chance is
slightly more than 50% (an amount that increases considering the points
people have raised about the chances of people picking low numbers.
If this works, why? Surely it is highly unintuitive.
Ed
MR
> "MR" <mr.(arroba)wanadoo.es@NOSPAM> wrote:
>
> Well, what is a "small" number anyway?
>
> Even with a "normal curve on integers", you won't get very far
> if you want to cover "large (whatever that is)" numbers - you
> will fix a mean for the normal (which is bound to be small for
> someone's standards), and a sigma (ditto) and hence you can't
> expect to get large numbers again,.
>
> Tell you what, on positive integers, randomly choose an integer
> using the probability distribution P(k) = 6/(k pi)^2. I'll
> guarantee that you can expect a large number on the average ;).
Bbbffffff......
I have to realize it, i'm overwhelmed with this problem.
Monwhea Jeng
> You get a friend to think of a number and then flip a coin (without showing
> you). If it lands heads he adds ten to his number, if it is tails he
> subtracts ten, then he gives you his answer. With what probability can you
> guess his original number?
>
> Ed
This is most clearly not a simple problem. Or rather, it is simply
impossible to state a single correct probability that you will be correct.
Your correct stratgy depends on your friend's probability distribution
for picking numbers, which will in turn depend on unstated assumptions,
like whether your friend knows that you are about to add or subtract 10
and then guess, and whether your friend is in competition with you.
For example, if you try this with a "typical" person, and they do not
know that you will add or subtract 10, and are not in competition with you,
I would guess that they have something like a 2/3rds chance of
picking 3, 5, or 7 (or at least some single digit number). If they pick
a double digit number, it will probably not begin with 1. There number
will not be negative. I would then guess that the probability that
I can guess their original number would be around 75%
On the other hand, if the person realizes what you are about to do, is
in competition with you, and is reasonably intelligent, they will pick
a number that is 20 or 30 digits long, and randomly make it positive
or negative. Then your chance of guessing correctly is arbitrarily
close to 50% (your friend makes it as close to 50% as he or she wishes,
by making the length of the number sufficiently long).
Momo
Monwhea Jeng
This works (sort of -- see below), but does not answer your initial
phrasing of the problem, which asked for the probability that you would
guess correctly. What you have shown is that you have a strategy which
does better than 50%, which is interesting, but not the original problem.
Incidentally, this is essentially the same as a "decision" puzzle in
the archive, which asks if you can have two cards with distinct numbers,
flip over one of them, and then state whether the other card is higher or
lower with better than 50% probability. There, the answer is also yes, for
identical reasons. However, also for identical reasons, your opponent can
make your probability 50%+epsilon, where epsilon is as small as they like.
The archive strategy, which is a little more robust than yours, is
to write an arbitrary continuous monotonically increasing function p(x),
which is 0 at x=-Infinity, and 1 at x=+Infinity, and upon hearing
your friend's final number x, decrease by 10 with probability p(x),
and increase by 10 with probability 1-p(x). This gives you a better than
50% chance of success regardless of what number your friend picks.
Momo
Moddy Te'eni
>
> You get a friend to think of a number and then flip a coin (without showing
> you). If it lands heads he adds ten to his number, if it is tails he
> subtracts ten, then he gives you his answer. With what probability can you
> guess his original number?
>
> Ed
I'd say a bit more than 50/50. You can suppose that most people ( then
ones who
don't read rec.puzzle ), will think of a rather small, positive number.
So my strategy will is this : Suppose my friend says N.
If -10 <= N <= 0 , I say N+10.
Otherwise, I flip a coin between N+10 and N-10.
-Moddy.
Andrew Bolt
> ro...@my-deja.com wrote:
> >Hi,
> >This reminds me a nice variant of the famous "surprise exam"
> >problem that can be formulated with almost the same words, namely :
> >
> >then flip a coin (without showing you). If it lands heads he adds
> >unchanged), then he gives you his answer. Furthermore, your friend
> >must choose his number in such a way that there will be no logical
> >proof for you to guess his original number. Is it possible ?
> >
> >Well, obviously your friend cannot choose "one", because if he
> >does so and gets tails he must tell you "one". If you hear "one"
> >you know his original number was "zero" or "one". Since the rules
> >don't allow him to choose zero, you could prove his original number
> >must be "one".
> >
> >Now, your friend cannot choose "two", because if he does so and
> >gets tails he must tell you "two". If you hear "two" you know his
> >allowed to choose "one" according to the paragraph above, you could
> >prove his original number must be "two".
> >
> >Now, your friend cannot choose "three", because if he does so
> >and gets tails he must tell you "three". If you hear "three" you
> >he's not allowed to choose "two" according to the previous paragraph,
> >you could prove his original number must be "three".
>
> the current example, as long as the guesser thinks of
> some "large enough" number and then flips the coin, I am sure
> that you have _no_ logical way of pinpointing _exactly_ what was
> the original number. (Please note that if your logic doesn't
> work, I do not have to explain to you why it doesn't. I just
> have to ensure that your logic doesn't work, and by giving you
> the number, I destroy your logic which builds on the "fact" that
> no number can be given.;)
Let's take a sample run, between you and your friend. But first,
I'd like to read the restriction in the rules as 'your friend must
choose his number in such a way that there will be no logical
proof for you to guess his original number -- for any possible
combination of coin tosses' - this makes it clear that he isn't
allowed to start with '1' and gamble on getting a series of heads
to start with.
F: (after flipping the coin in private) My final number is two.
Y: There are two possible ways of getting that: you either started
with '1' and flipped heads then tails. Or you started at '2',
and flipped tails first time.
If it wasn't for that 'logical proof' restriction, I would have
nothing further to go on. But assuming you obeyed that restriction,
you wouldn't be allowed to start with a '1'. That implies that
you would have had to have chosen '2'. But by proving this, we've
invalidated the original assumption.
Therefore you couldn't have legitimately chosen either number by
the rules. So you have cheated.
F: I didn't cheat: you still can't tell me which of the two numbers I
originally chose, so I have followed the rules.
Y: I bet you chose '2'.
F: I did - but that was only a bet - you haven't proven it. In fact,
you just prove that I couldn't have chosen '2', and yet I did. So
your proof is invalid.
Y: That's exactly what I'm saying: my proof is water-tight, within
the rules of the game. The only way you could have chosen '2'
would be by breaking the rules. If you break the rules, how am I
meant to stand a chance of proving anything about what you did?
F: So you're saying that you can't guess my original number, but
that doesn't matter: you're claiming that I've broken the rules
so I can't expect you to guess my number.
Hmm - I'm not sure about that. Let's play again.
(clips the coin in private a few times)
F: Ok, my final number is 3029838305.
What happened next?
--
Andrew Bolt, Andre...@arm.com
110 Fulbourn Road, Cambridge, CB1 9NJ, ENGLAND, +44 1223 400650
Gerry Quinn
>
>Suppose, in the envelope paradox, after writing a positive, computable
>number "x" on one envelope and "2x" on the other, I compute the
>logarithm base 2^(1/20) of each number, jotting it down in red ink
>below. Note that the red-ink numbers (each of which may be positive,
>negative, or even zero) differ by 20.
>
>How is this different from picking a computable number "y", putting
>"y+10" and "y-10" in red on the envelopes, and computing 2^(1/20) to
>the power of the red numbers to get an "x" and "2x"?
>
It's not. But how is the logarithmic distribution the same as the
envelope paradox, in which your winnings are proportional to the
contents of the envelopes, rather than their logarithms?
- Gerry Quinn
David A Karr
>>How is this different from picking a computable number "y", putting
>>"y+10" and "y-10" in red on the envelopes, and computing 2^(1/20) to
>>the power of the red numbers to get an "x" and "2x"?
>
>It's not. But how is the logarithmic distribution the same as the
>envelope paradox, in which your winnings are proportional to the
>contents of the envelopes, rather than their logarithms?
The main difference between the number-guessing game and Two Envelopes
is that in the latter, one argument says you should guess whether you
have the smaller or larger envelope in order to maximize the expected
value of your final choice; in the latter, you're told explicitly to
guess y, which is just equivalent to guessing whether you have the
smaller or the larger of y - 10 and y + 10.
Certainly the expected value of a random variable X is of course not
analogous to the expected value of (log X), so the two problems are
_not_ the same with respect to expected-value calculations. But if
you accept that what you really want to do in Two Envelopes is to end
up with the larger envelope, then you don't need to compute expected
values. In particular, a Two Envelopes tactic that relies exclusively
on guessing which envelope you have, without consideration of expected
values, maps just fine onto the number-guessing game.
In other words, it's just like the engineer, the mathematician, and
the tea kettle.
Kevin Buhr
>
> It's not. But how is the logarithmic distribution the same as the
> envelope paradox, in which your winnings are proportional to the
> contents of the envelopes, rather than their logarithms?
Oh. Yes, the payoff structure is different, but the "solution" is
still basically the same.
In the two-envelope problem, the "neat" thing is that using the
following strategy:
- pick some positive number "t";
- if the envelope amount is bigger, keep;
- if the envelope amount is smaller, switch.
always does as well as, and sometimes does better, than always
keeping, always switching, or randomly keeping or switching without
regard to the amount found in the envelope. The reason is that the
probability of picking the larger amount using this strategy is bigger
than 1/2 as long as the distribution of envelope amounts leads to the
smaller and lower values bracketting "t" with positive probability.
The same thing applies to the present problem. If you use the
following strategy:
- pick some real number "t";
- listen to your friend's announced number "z";
- if z>t, then guess z-10;
- if z<=t, then guess z+10.
then you always guess the number with probability at least 1/2 and
sometimes with probability exceeding 1/2. As long as the random
distribution of your friend's initial number allows it to come within
10 units of your "t" with positive probability, you will do better
than 1/2. If it doesn't, you still won't do worse than 1/2.
To see why, take t=0.
If your friend picks a number "x" greater than 10, then he'll use a
coin to determine if he should tell you "z=x+10" or "z=x-10". In
either case, the numbers are positive, so z>t and you guess z-10.
Because the coin is random, half the time you're wrong and half the
time you're right.
If your friend picks a number "x" less than or equal to -10, the same
analysis applies, except you always guess z+10. Again, you're right
and wrong with equal probability.
Obviously, then, if your friend never picks numbers in the half-open
interval (-10,10], you'll always have probability of 1/2 of guessing
right---you're just competing with the coin flip.
On the other hand, if your friend picks a number in (-10,10] you'll
always guess right! For example, if he picks 9, then if the coin is
heads, he'll tell you z=19, and you'll guess x=19-10=9. If the coin
is tails, he'll tell you z=-1, and you'll still guess x=-1+10=9.
Therefore, the probability you guess right is given by the formula:
1/2 + 1/2 p
where p>=0 is the probability of your friend picking an initial number
in the interval (t-10,t+10].
Kevin <bu...@stat.wisc.edu>
Gerry Quinn
>Gerry Quinn <ger...@indigo.ie> wrote:
>>>How is this different from picking a computable number "y", putting
>>>"y+10" and "y-10" in red on the envelopes, and computing 2^(1/20) to
>>>the power of the red numbers to get an "x" and "2x"?
>>
>>envelope paradox, in which your winnings are proportional to the
>>contents of the envelopes, rather than their logarithms?
>
>is that in the latter, one argument says you should guess whether you
>have the smaller or larger envelope in order to maximize the expected
>value of your final choice; in the latter, you're told explicitly to
>guess y, which is just equivalent to guessing whether you have the
>smaller or the larger of y - 10 and y + 10.
>
>Certainly the expected value of a random variable X is of course not
>analogous to the expected value of (log X), so the two problems are
>_not_ the same with respect to expected-value calculations. But if
>you accept that what you really want to do in Two Envelopes is to end
>up with the larger envelope, then you don't need to compute expected
>values. In particular, a Two Envelopes tactic that relies exclusively
>on guessing which envelope you have, without consideration of expected
>values, maps just fine onto the number-guessing game.
>
I see your point, though I think that it is the very non-logarithmic
expectation values that make the envelope paradox interesting.
In any case, it seems to me that the only consistent distribution of a
truly unknown real (or computable) is of a form proportional to 1/x.
This means that getting an amount between zero and infinity is
impossible.
In other words, the problem of what to do when you have opened the
envelope and found a finite amount is a red herring. The problem thus
stated contains an impossibility.
- Gerry Quinn
macavity
>In any case, it seems to me that the only consistent
distribution of a
>truly unknown real (or computable) is of a form proportional to
1/x.
>This means that getting an amount between zero and infinity is
>impossible.
>
Just curious, but what did you mean (if anything) by the above?
I get the feeling I maybe missing something profound.
Gerry Quinn
>ger...@indigo.ie (Gerry Quinn) wrote:
>>In any case, it seems to me that the only consistent
>distribution of a
>>truly unknown real (or computable) is of a form proportional to
>1/x.
>>This means that getting an amount between zero and infinity is
>>impossible.
>>
>
>Just curious, but what did you mean (if anything) by the above?
>I get the feeling I maybe missing something profound.
The envelope paradox has as a starting condition "you know nothing about
what amount to expect". Therefore, what is the chance that the amount
will be greater than $0.01? 50% obviously. What is the chance that the
amount will be greater than $1000000000000? Still 50%. Anything other
than 50% violates the stated condition.
Consequently, the probability is zero that the amount will be finite and
non-zero. It has an equal chance of being infinitesimal or infinite. A
function of the form k/x, although mathematically somewhat ill-formed,
is symmetrical and consistent with this.
And therefore, also, the variation in which you open the envelope and
find a finite amount contains a contradiction, or at least an unstatedly
infinitely improbable event. (Interestingly, though, the function k/x
solves that too, as if the cheque-writer picked a value according to
this distribution, and doubled it for the other envelope, there is a
2/3 chance that the envelope you open is the larger of the two. For any
finite amount, there is double the chance that it came about as the
larger cheque, although the chance is infinitesimal anyway.)
- Gerry Quinn
macavity
>In article <0b5e0384...@usw-ex0102-015.remarq.com>,
macavity <mxyptlk...@hotmail.com.invalid> wrote:
>>ger...@indigo.ie (Gerry Quinn) wrote:
>>>In any case, it seems to me that the only consistent
>>distribution of a
>>>truly unknown real (or computable) is of a form proportional
to
>>1/x.
>>>This means that getting an amount between zero and infinity is
>>>impossible.
>>>
>>
>>Just curious, but what did you mean (if anything) by the above?
>>I get the feeling I maybe missing something profound.
>
>The envelope paradox has as a starting condition "you know
nothing about
>what amount to expect". Therefore, what is the chance that the
amount
>will be greater than $0.01? 50% obviously. What is the chance
that the
>amount will be greater than $1000000000000? Still 50%.
Anything other
>than 50% violates the stated condition.
>
The way you have interpreted this is interesting. If P(x > a) =
0.5, and P(x > b) = 0.5 for any given a and b, then P(a<x<b)=0,
which means that the probability distribution you are taking
about is zero in any finite interval. Which in turn means that
you are talking about P(x)=0 everywhere, which is a
contradiction to P(x>a) = 0.5. This means, P(x>a) = P(x>b) for
all a, b is NOT possible.
If that is your interpretation of the envelope question, know
for sure that it is a non-starter. No one can say that there
exists such a distribution from which he will generate a number.
>Consequently, the probability is zero that the amount will be
finite and
>non-zero. It has an equal chance of being infinitesimal or
infinite. A
>function of the form k/x, although mathematically somewhat ill-
formed,
>is symmetrical and consistent with this.
>
If P(x)= k/x,
(1) it cannot be valid for an infinite interval of reals - P(x
over all possibilities) = 1 is a rather basic fact to define
probability distributions.
(2) Put x = 0.0001 and x = 10000 (or better still, test limits
as x --> 0 and oo) and see if it gives you an equal chance for
getting infinitesimal or infinite.
>And therefore, also, the variation in which you open the
envelope and
>find a finite amount contains a contradiction, or at least an
unstatedly
>infinitely improbable event.
With _any_ continuous distribution function, the probability of
picking any particular number is infinitesimal. So what?
> (Interestingly, though, the function k/x
>solves that too, as if the cheque-writer picked a value
according to
>this distribution, and doubled it for the other envelope,
This isn't a distribution, as pointed out earlier, for any
infinite interval. Forget picking anything using it. If you
start with one wrong statement, you can prove practically
anything, as learned chaps have already said.
there is a
>2/3 chance that the envelope you open is the larger of the
two. For any
>finite amount, there is double the chance that it came about as
the
>larger cheque, although the chance is infinitesimal anyway.)
>
>- Gerry Quinn
>
>
I guess either you didn't mean much, or you have a lot of
concepts that need strengthening.
Gerry Quinn
>ger...@indigo.ie (Gerry Quinn) wrote:
>>The envelope paradox has as a starting condition "you know
>nothing about
>>what amount to expect". Therefore, what is the chance that the
>amount
>>will be greater than $0.01? 50% obviously. What is the chance
>that the
>>amount will be greater than $1000000000000? Still 50%.
>Anything other
>>than 50% violates the stated condition.
>>
>The way you have interpreted this is interesting. If P(x > a) =
>0.5, and P(x > b) = 0.5 for any given a and b, then P(a<x<b)=0,
>which means that the probability distribution you are taking
>about is zero in any finite interval. Which in turn means that
>you are talking about P(x)=0 everywhere, which is a
>contradiction to P(x>a) = 0.5. This means, P(x>a) = P(x>b) for
>all a, b is NOT possible.
On the contrary: the envelope will contain an infinitesimal or an
infinite amount, each with 50% probability.
There is no other interpretation of the statement "You know nothing
about the expected distribution of the envelope contents". If you
assess that there is other than a 50% chance of the envelope containing
more than $X, where X is any named non-zero finite amount, you
contradict this statement. If you don't like my distribution, fine, but
then the envelope paradox still contains a contradiction in its stated
terms - you just don't have an interesting way to resolve it.
>
>If that is your interpretation of the envelope question, know
>for sure that it is a non-starter. No one can say that there
>exists such a distribution from which he will generate a number.
>
Nobody has to. You can think of it as a scaling factor. The currency
that the writer uses is likely incomensurable with yours, being either
infinitely large or infinitely small. The k/x distribution represents
the ratio of two arbitrary reals.
>>Consequently, the probability is zero that the amount will be
>finite and
>>non-zero. It has an equal chance of being infinitesimal or
>infinite. A
>>function of the form k/x, although mathematically somewhat ill-
>formed,
>>is symmetrical and consistent with this.
>>
>
>If P(x)= k/x,
>(1) it cannot be valid for an infinite interval of reals - P(x
>over all possibilities) = 1 is a rather basic fact to define
>probability distributions.
>(2) Put x = 0.0001 and x = 10000 (or better still, test limits
>as x --> 0 and oo) and see if it gives you an equal chance for
>getting infinitesimal or infinite.
>
It's not _standard_, because k must be infinitesimal. The distribution
curve has the shape of f(x)=1/x which clearly has 100% of its density
outside the range [x1..x2], where x1 and x2 are non-zero and finite.
Furthermore, 50% of the density is below x1 and 50% above x2.
>>And therefore, also, the variation in which you open the
>envelope and
>>find a finite amount contains a contradiction, or at least an
>unstatedly
>>infinitely improbable event.
>
>With _any_ continuous distribution function, the probability of
>picking any particular number is infinitesimal. So what?
>
Okay, but the probability of the number being finite at all, which is
the essence of the problem (and not it being a particular amount) is
infinitesimal.
>> (Interestingly, though, the function k/x
>>solves that too, as if the cheque-writer picked a value
>according to
>>this distribution, and doubled it for the other envelope,
>
>This isn't a distribution, as pointed out earlier, for any
>infinite interval. Forget picking anything using it. If you
>start with one wrong statement, you can prove practically
>anything, as learned chaps have already said.
>
It is a non-standard distribution, and it is also the only distribution
that makes any sense in the context of the problem, whether you like it
or not.
>there is a
>>2/3 chance that the envelope you open is the larger of the
>two. For any
>>finite amount, there is double the chance that it came about as
>the
>>larger cheque, although the chance is infinitesimal anyway.)
>>
>>- Gerry Quinn
>
>I guess either you didn't mean much, or you have a lot of
>concepts that need strengthening.
>
Mathematicians have always whined when physicists and others used
formulae in an area where they thought they were unsound. Generally the
'unsound' usages were validated, and later incorporated in mathematical
theory. The k/x function works too well for this problem to be
meaningless. I believe it has also occurred in some contexts related to
fractals and noise distributions, probably because it is the scaling
factor of last resort.
- Gerry Quinn
macavity
>>ger...@indigo.ie (Gerry Quinn) wrote:
>
>>>The envelope paradox has as a starting condition "you know
>>nothing about
>>>what amount to expect". Therefore, what is the chance that
the
>>amount
>>>will be greater than $0.01? 50% obviously. What is the
chance
>>that the
>>>amount will be greater than $1000000000000? Still 50%.
>>Anything other
>>>than 50% violates the stated condition.
>>>
>>The way you have interpreted this is interesting. If P(x > a)
=
>>0.5, and P(x > b) = 0.5 for any given a and b, then P(a<x<b)=0,
>>which means that the probability distribution you are taking
>>about is zero in any finite interval. Which in turn means that
>>you are talking about P(x)=0 everywhere, which is a
>>contradiction to P(x>a) = 0.5. This means, P(x>a) = P(x>b) for
>>all a, b is NOT possible.
>
>On the contrary: the envelope will contain an infinitesimal or
an
>infinite amount, each with 50% probability.
Repeatedly asserting something without proof might work in
oratory, (and sometimes even in history if you are the winner).
A question away from the topic, ever been in politics?
>
>There is no other interpretation of the statement "You know
nothing
>about the expected distribution of the envelope contents". If
you
>assess that there is other than a 50% chance of the envelope
containing
>more than $X, where X is any named non-zero finite amount, you
>contradict this statement. If you don't like my distribution,
fine, but
>then the envelope paradox still contains a contradiction in its
stated
>terms - you just don't have an interesting way to resolve it.
>
>>
>>If that is your interpretation of the envelope question, know
>>for sure that it is a non-starter. No one can say that there
>>exists such a distribution from which he will generate a
number.
>>
>
>Nobody has to. You can think of it as a scaling factor. The
currency
>that the writer uses is likely incomensurable with yours, being
either
>infinitely large or infinitely small. The k/x distribution
represents
>the ratio of two arbitrary reals.
This is like saying "My 2 is bigger than your 3". It is just
that we are from different worlds. Now k/x has become the ratio
of two reals, as opposed to a pdf?
>
>>>Consequently, the probability is zero that the amount will be
>>finite and
>>>non-zero. It has an equal chance of being infinitesimal or
>>infinite. A
>>>function of the form k/x, although mathematically somewhat
ill-
>>formed,
>>>is symmetrical and consistent with this.
>>>
>>
>>If P(x)= k/x,
>>(1) it cannot be valid for an infinite interval of reals - P(x
>>over all possibilities) = 1 is a rather basic fact to define
>>probability distributions.
>>(2) Put x = 0.0001 and x = 10000 (or better still, test limits
>>as x --> 0 and oo) and see if it gives you an equal chance for
>>getting infinitesimal or infinite.
>>
>
>It's not _standard_, because k must be infinitesimal. The
distribution
>curve has the shape of f(x)=1/x which clearly has 100% of its
density
>outside the range [x1..x2], where x1 and x2 are non-zero and
finite.
>Furthermore, 50% of the density is below x1 and 50% above x2.
>
So you agree k must be infinitesimal. Good. I just proved
earlier it has to be zero. We aren't that far off, maybe.
I don't know what context of the problem you are working with.
If someone gave you the problem saying "A is going to pick two
random (+ve?) reals from a distribution s.t. prob(x>a)=0.5 for
every a", then the statement of the problem is at best unclear.
Maybe you should specify _exactly_ what you heard as the
statement of the problem - i.e. about the distribution. Pls
note that just saying the distribution is not known is not the
same as the Very Strong (and absurd) statement that p(x>a)is 0.5
for every a in the set.
If you are serious about this, and not just arguing to save
face, check http://www.seanet.com/~ksbrown/kmath225.htm which
contains a reasonably good statement of the problems wrong
contexts can create in such problems.
>>there is a
>>>2/3 chance that the envelope you open is the larger of the
>>two. For any
>>>finite amount, there is double the chance that it came about
as
>>the
>>>larger cheque, although the chance is infinitesimal anyway.)
>>>
>>>- Gerry Quinn
>>
>>I guess either you didn't mean much, or you have a lot of
>>concepts that need strengthening.
>>
>
>Mathematicians have always whined when physicists and others
used
>formulae in an area where they thought they were unsound.
Generally the
>'unsound' usages were validated, and later incorporated in
mathematical
>theory. The k/x function works too well for this problem to be
>meaningless. I believe it has also occurred in some contexts
related to
>fractals and noise distributions, probably because it is the
scaling
>factor of last resort.
>
I know, mathematicians are always whining that things should be
perfect, that pi is not 22/7, etc. Terrible chaps, for sure.
In defense, though, try taking a sample of mathematical truths
or theorems that have been overturned (some - at least
conjectures have been overturned), vs. in other disciplices - as
you brought it, maybe physics. You maybe amazed at the result.
>- Gerry Quinn
Gerry Quinn
Your offensive tone just makes you look a fool. The analysis I make
depends on the concept of an infinitesimal, non-zero quantity that
permits a distribution of the form k/x with an area of 1. What I stated
is a consequence of that.
>>
>>It's not _standard_, because k must be infinitesimal. The
>distribution
>>curve has the shape of f(x)=1/x which clearly has 100% of its
>density
>>outside the range [x1..x2], where x1 and x2 are non-zero and
>finite.
>>Furthermore, 50% of the density is below x1 and 50% above x2.
>>
>
>So you agree k must be infinitesimal. Good. I just proved
>earlier it has to be zero. We aren't that far off, maybe.
>
I am really _sick_ of people who answer long posts by instantly spewing
out their uncomprehending garbage based on the bit they have read so
far, reading a bit more, and pontificating on their new comprehension
without deleting the nonsense they have previously typed. If you
knew that I was proposing an infinitesimal k, what do you mean by the
drivel you posted above? If you just mean to say "No probability
distributions involving infinitesimals can be meaningful" then why
did you not come straight out with it? I freely admit I am going beyond
the usual definitions. Perhaps you are uncertain whether this is
"allowed" and prefer to indulge in sarcasm rather than pretend you can
hear the question and risk the possibility of not knowing it all.
[--]
>>
>>It is a non-standard distribution, and it is also the only
>distribution
>>that makes any sense in the context of the problem, whether you
>like it
>>or not.
>>
>
>I don't know what context of the problem you are working with.
>If someone gave you the problem saying "A is going to pick two
>random (+ve?) reals from a distribution s.t. prob(x>a)=0.5 for
>every a", then the statement of the problem is at best unclear.
>Maybe you should specify _exactly_ what you heard as the
>statement of the problem - i.e. about the distribution. Pls
>note that just saying the distribution is not known is not the
>same as the Very Strong (and absurd) statement that p(x>a)is 0.5
>for every a in the set.
>
The envelope paradox, unfortunately, does not have a precise
mathematical statement. It is understood, however, that, the recipient
knows *nothing* about the possible distribution of the (non-negative
real) amounts in the envelopes. It is plainly the case that for any
positive real value a, p(x>a) must be 0.5. One not too unreasonable
approach is to say that this is on the face of things absurd and
therefore the problem as stated contains a contradiction. That is not
the option I take.
>If you are serious about this, and not just arguing to save
>face, check http://www.seanet.com/~ksbrown/kmath225.htm which
>contains a reasonably good statement of the problems wrong
>contexts can create in such problems.
>
Do yourself a favour - get off your high horse and realise that you are
the one who doesn't understand the argument here.
>>
>>Mathematicians have always whined when physicists and others
>used
>>formulae in an area where they thought they were unsound.
>Generally the
>>'unsound' usages were validated, and later incorporated in
>mathematical
>>theory. The k/x function works too well for this problem to be
>>meaningless. I believe it has also occurred in some contexts
>related to
>>fractals and noise distributions, probably because it is the
>scaling
>>factor of last resort.
>>
>I know, mathematicians are always whining that things should be
>perfect, that pi is not 22/7, etc. Terrible chaps, for sure.
>In defense, though, try taking a sample of mathematical truths
>or theorems that have been overturned (some - at least
>conjectures have been overturned), vs. in other disciplices - as
>you brought it, maybe physics. You maybe amazed at the result.
>
Consider cases like the impulse function, spurned by mathematicians
until physicists had long utilised it with success. f(x)=k/x is such a
function. I don't care about your prissy objections about the form a
probability distribution, in your equally ill-formed and ill-mannered
opinion, should take.
Do you want to stand up and say density distributions of the shape 1/x
have no uses? I'm pretty sure I have come across them in contexts such
as fractal noise! Do you admit that IF valid density distributions of
this form exist in some theories, your arguments are false and your
insults become self-inflicted wounds?
- Gerry Quinn
Monwhea Jeng
> The envelope paradox has as a starting condition "you know nothing about
> what amount to expect". Therefore, what is the chance that the amount
> will be greater than $0.01? 50% obviously. What is the chance that the
> amount will be greater than $1000000000000? Still 50%. Anything other
> than 50% violates the stated condition.
This certainly does not follow. Saying "you know nothing about what
to expect" does not imply that any question you might ask has an
answer of 50%. You will need to pick some "reasonable" and consistent
Bayesian probabilities, not just 50% for everything.
The classic example of why you can't just answer 50% for everything, is
to ask if you know anything about extraterrestial life. I do not (I can't
speak for you, of course). But I can't consistently say that there's
a 50% chance of extraterrestrial life in the Galaxy, and also a 50%
chance of life on Tau Ceti, and also a 50% chance of life on Alpha
Centauri. The last two statements imply at least a 75% chance of
aliens somewhere in the Galaxy. If I say "50% chance" each time you
ask me about a new star, and you ask me about 100 different stars, I've
essential said that there must be life somewhere in the Galaxy (with
probability 1-(1/2)^100).
Momo
Nick Wedd
du>, Monwhea Jeng <mo...@hal.physics.ucsb.edu> writes
>On Sat, 29 Jul 2000, Gerry Quinn wrote:
>
>> The envelope paradox has as a starting condition "you know nothing about
>> what amount to expect". Therefore, what is the chance that the amount
>> will be greater than $0.01? 50% obviously. What is the chance that the
>> amount will be greater than $1000000000000? Still 50%. Anything other
>> than 50% violates the stated condition.
>
>This certainly does not follow. Saying "you know nothing about what
>to expect" does not imply that any question you might ask has an
>answer of 50%. You will need to pick some "reasonable" and consistent
>Bayesian probabilities, not just 50% for everything.
I am with Gerry here. I believe that no finite set of Bayesian
probabilities can be consistent with the statement "you know nothing
about what amount to expect".
Here is an analogy.
"You are told that all the taxis in the city of Yoost are numbered
consecutively, from 1 upwards. You know nothing about the size of
Yoost, to which you have just flown. As you leave the air terminal, you
see a taxi. What is you probability distribution function for its
number?"
I believe that the only sensible answer is "0, for every number". I
believe that the statement "You know nothing about the size of Yoost" is
counterfactual: I do in fact know that all cities to which I can fly
have populations less than 10,000,000,000. It is not surprising that my
accepting this counterfactual statement leads me to unusual results.
Nick
--
Nick Wedd ni...@maproom.co.uk
macavity
Well, sick of yourself?
What I wrote would be false if:
1. You could show me a case where k/x has been used over an
infinite interval of reals as a probability function.
AND
2. If you could prove that "B knowing nothing about a
probability function" => probability function "A" is using is p
(x>y)=0.5 for every y, => prob density is k/x.
If I don't see an answer - read proof, not assertions or
opinions - for 1 and 2 (or at least one of them), this is my
last word to you on this subj.
macavity
>I am with Gerry here. I believe that no finite set of Bayesian
nothing
>about what amount to expect".
>
> "You are told that all the taxis in the city of Yoost are
numbered
>consecutively, from 1 upwards. You know nothing about the size
of
>Yoost, to which you have just flown. As you leave the air
terminal, you
>see a taxi. What is you probability distribution function for
its
>number?"
> I believe that the only sensible answer is "0, for every
number". I
>believe that the statement "You know nothing about the size of
Yoost" is
>counterfactual: I do in fact know that all cities to which I
can fly
>have populations less than 10,000,000,000. It is not
surprising that my
>accepting this counterfactual statement leads me to unusual
results.
>
>Nick
>--
>Nick Wedd ni...@maproom.co.uk
>
>
Hi Nick,
There is an issue here on whose distribution you are talking
about. If A tells B, "I am going to pick a positive real number
using a probability distribution that YOU know nothing about",
there is nothing wrong with the statement. It does not impose
non-Bayesian charecteristics on A's distribution - for example,
suppose A has decided to use e^(-t) as a pdf.
However, as you rightly pointed out, there is no distribution
that B can propose that will satisfy A's statement and still be
Bayesian. But IIRC, the envelope paradox does not require B to
assume what A's distribution is.
An exact statement of the paradox would make the discussion more
objective, perhaps.
David A Karr
>On the contrary: the envelope will contain an infinitesimal or an
>infinite amount, each with 50% probability.
>
I don't know about "no other," but surely the above is not itself
an interpretation of that statement--unless you mean that from
"you know nothing about the distribution" you can deduce that you
know that the distribution is bimodal with the other properties
you listed.
A better interpretation for the distribution would be "undefined."
On the other hand, a better statement of the paradox would avoid
foolish wording such as "you know nothing."
David A Karr
>I see your point, though I think that it is the very non-logarithmic
>expectation values that make the envelope paradox interesting.
Yes, that's what gives you the argument that envelope A is better than
envelope B which in turn is better than envelope A.
The similarity I was thinking of occurs when you get past this
obviously bogus conclusion, and start to think about what would be a
winning strategy in any remotely practical version of the problem
(meaning that you have _some_ notion of the prior distribution--e.g.
the amount was highly unlikely to be over a quadrillion dollars--but
only the very vaguest idea). A winning (but not necessarily optimal)
strategy has been described in other recent posts.
Gerry Quinn
>On Sat, 29 Jul 2000, Gerry Quinn wrote:
>
>> The envelope paradox has as a starting condition "you know nothing about
>> will be greater than $0.01? 50% obviously. What is the chance that the
>> amount will be greater than $1000000000000? Still 50%. Anything other
>> than 50% violates the stated condition.
>
>to expect" does not imply that any question you might ask has an
>answer of 50%. You will need to pick some "reasonable" and consistent
>Bayesian probabilities, not just 50% for everything.
>
>to ask if you know anything about extraterrestial life. I do not (I can't
>speak for you, of course). But I can't consistently say that there's
>a 50% chance of extraterrestrial life in the Galaxy, and also a 50%
>chance of life on Tau Ceti, and also a 50% chance of life on Alpha
>Centauri. The last two statements imply at least a 75% chance of
>aliens somewhere in the Galaxy. If I say "50% chance" each time you
>ask me about a new star, and you ask me about 100 different stars, I've
>essential said that there must be life somewhere in the Galaxy (with
>probability 1-(1/2)^100).
>
Different problem. You do know something about Tau Ceti and Alpha
Centauri - they are stars, that may have planets. The question is a
hard one, but you do have data to help you form an opinion - it's just
very difficult to work out an answer given what you know of chemistry,
stellar and galactic history, the kinds of things that may be alive,
etc.
A better example would be the question "Is there life in SQUOGGLE" when
you don't know whether SQUOGGLE is a galaxy, the universe, or an atom.
Even then it's not as perfect as a question about the real numbers!
- Gerry Quinn
Gerry Quinn
I would think a density function of any sort would be adequate to prove
the point, insofar as a probability function is a class of density
function. It wouldn't necessarily prove me right, but it would prove
that your dismissal of the concept was incorrect. Candidly I don't have
an example to hand, but I am certain they have been used. I'll file it
when I next come across the concept.
>AND
>
>2. If you could prove that "B knowing nothing about a
>probability function" => probability function "A" is using is p
>(x>y)=0.5 for every y, => prob density is k/x.
>
That's all pretty obvious, I should think, though it should be put a
little more formally:
If a value r represents an arbitrary positive real, then p(x>r)=0.5 for
any positive real x.
The above is really a matter of definition. Formal proofs are not my
game, but I defy anyone to produce an alternative function that is zero
on any closed interval [x1..x2] where x1 and x2 are named positive
reals, and that has an area of 0.5 above and below any such interval.
>If I don't see an answer - read proof, not assertions or
>opinions - for 1 and 2 (or at least one of them), this is my
>last word to you on this subj.
>
I am not interested in formal proofs. I like solving paradoxes, though.
The non-standard function I propose works remarkably well to solve this
one. Like Newton's infinitesimals, if it works it's good.
- Gerry Quinn
macavity
>In article <01a97cff...@usw-ex0102-015.remarq.com>,
to prove
>the point, insofar as a probability function is a class of
density
>function. It wouldn't necessarily prove me right, but it would
prove
>that your dismissal of the concept was incorrect. Candidly I
don't have
>an example to hand, but I am certain they have been used. I'll
file it
>when I next come across the concept.
>
I would be really surprised if you do this. BTW, not ANY
probability density function, I am talking specifically about
k/x, (for k "infinitesimal" or not) on any infinite interval
(say the + reals).
>>AND
>>
>>2. If you could prove that "B knowing nothing about a
>>probability function" => probability function "A" is using is p
>>(x>y)=0.5 for every y, => prob density is k/x.
>>
>
>That's all pretty obvious, I should think, though it should be
put a
>little more formally:
>
>If a value r represents an arbitrary positive real, then p(x>r)
=0.5 for
>any positive real x.
This does _not_ follow. If you know nothing about the
distribution someone else is using, how do you know that p(x>1)
=0.5, for e.g.? You know _nothing_ about it.
Look at the following statement: X thinks picks a positive real
number - say using a density e^(-t), and tells Y, I have picked
a number from the positive reals using a distribution I am not
telling you.
There is nothing wrong with the above statement. Y merely does
not know what distribution X is using. X could be using any
distribution. This does _not_ imply p(x>r)=0.5.
If you interpret the statement (Y does not know what
distribution X is using) in the form you have done, it is a
logical paradox in the statement itself. It is like saying, "I
have thought of a name that you know nothing about." What name
satisfies this condition? The definitition itself precludes
_any_ name you know.
>
>The above is really a matter of definition. Formal proofs are
not my
>game, but I defy anyone to produce an alternative function that
is zero
>on any closed interval [x1..x2] where x1 and x2 are named
positive
>reals, and that has an area of 0.5 above and below any such
interval.
>
Non one can, that is precisely the point. Other than repeatedly
asserting this, can you prove/ show/ indicate how k/x does this
amazing feat?
Gerry Quinn
>>That's all pretty obvious, I should think, though it should be
>put a
>>little more formally:
>>
>>If a value r represents an arbitrary positive real, then p(x>r)
>=0.5 for
>>any positive real x.
>
>This does _not_ follow. If you know nothing about the
>distribution someone else is using, how do you know that p(x>1)
>=0.5, for e.g.? You know _nothing_ about it.
>
The figure represents my assessment of the probability that x>r based on
what I know. By definition, I have all the information required to make
that assessment.
From my perspective, the probability that x>r is equal to that of x<r,
while that of the third possibility (x=r) is zero. Conclusion,
p(x>r) = p(x<r) = 0.5.
>Look at the following statement: X thinks picks a positive real
>number - say using a density e^(-t), and tells Y, I have picked
>a number from the positive reals using a distribution I am not
>telling you.
>
>There is nothing wrong with the above statement. Y merely does
>not know what distribution X is using. X could be using any
>distribution. This does _not_ imply p(x>r)=0.5.
>
p(x>r) is Y's assessment of the chance that a particular real selected
by Y is greater than the secret value r selected by X. For convenience,
let us assume that Y has selected x=10. Y asks himself these two
questions:
1. What is the chance (from my perspective) that 10>r?
2. What is the chance that 10<r?
Why would you (as Y) say that the two answers are not the same, or would
be different if Y had selected, say, 100?
>If you interpret the statement (Y does not know what
>distribution X is using) in the form you have done, it is a
>logical paradox in the statement itself. It is like saying, "I
>have thought of a name that you know nothing about." What name
>satisfies this condition? The definitition itself precludes
>_any_ name you know.
>
I can work with my interpretation and get interesting results, and I
have found no logical paradox yet.
>>
>>The above is really a matter of definition. Formal proofs are
>not my
>>game, but I defy anyone to produce an alternative function that
>is zero
>>on any closed interval [x1..x2] where x1 and x2 are named
>positive
>>reals, and that has an area of 0.5 above and below any such
>interval.
>>
>
>Non one can, that is precisely the point. Other than repeatedly
>asserting this, can you prove/ show/ indicate how k/x does this
>amazing feat?
>
Now that's perfectly easy!
First note that the area under k/x for any closed interval [x1..x2], x1
and x2 being positive reals, is k times ( log x2 - log x1), which is
infinitesimal (since k is).
Now note the shape of the distribution which is symmetrical about the
line y=x. Cut out the region [x1..x2], and cut out a further finite
region such that two identical infinite regions are left, one hugging
the x-axis and one the y-axis. (If x1=2 and x2=3, for example, we cut
out the region remaining below y=3 and left of x=3: the exact choice
depends on which number, or its reciprocal, is larger.)
So we have two identical infinite regions, compared to which the finite
area between x1 and x2, and the other part we removed, are
insignificant. One area is contained in f(x),x<x1 and the other in
f(x),x>x2.
Another way to see this is to notice that y=x^-1 represents a curve in
the series y=x^n, in which all curves with n<1 have their infinite area
concentrated at infinity while all those with n>1 have their infinite
area concentrated at zero. y=1/x hovers on the cusp of the two groups.
- Gerry Quinn
macavity
>
>I can work with my interpretation and get interesting results,
and I
>have found no logical paradox yet.
>
I am sure _you_ can. I prefer logic of a more general nature.
As you seem to be awake anyway, there is no point in trying
further to wake you up, and I leave you to your dreams.
David A Karr
>In article <Pine.LNX.4.10.100073...@hal.physics.ucsb.e
>du>, Monwhea Jeng <mo...@hal.physics.ucsb.edu> writes
>>to expect" does not imply that any question you might ask has an
>>answer of 50%. You will need to pick some "reasonable" and consistent
>>Bayesian probabilities, not just 50% for everything.
>
>about what amount to expect".
It's worse than that. _No_ probability distribution, finite or
infinite, is consistent with that assumption.
>Here is an analogy.
> "You are told that all the taxis in the city of Yoost are numbered
>consecutively, from 1 upwards. You know nothing about the size of
>Yoost, to which you have just flown. As you leave the air terminal, you
>see a taxi. What is you probability distribution function for its
>number?"
My answer is there is no such probability distribution.
Here's another analogy: Suppose I asked you for the greatest real root
of x^2 + x + 1?
In the version of the taxi problem I recall, you have already seen the
number of your taxi, and the problem is to estimate how many taxis
there are in the city. Since there was not a prior distribution of
the number of taxis, this devolves into various competing non-Bayesian
estimates, none of which (IIRC) emerges a clear winner.
Gerry Quinn
>Nick Wedd <Ni...@maproom.co.uk> wrote:
>>In article <Pine.LNX.4.10.100073...@hal.physics.ucsb.e
>>du>, Monwhea Jeng <mo...@hal.physics.ucsb.edu> writes
>>>This certainly does not follow. Saying "you know nothing about what
>>>to expect" does not imply that any question you might ask has an
>>>answer of 50%. You will need to pick some "reasonable" and consistent
>>>Bayesian probabilities, not just 50% for everything.
>>
>>I am with Gerry here. I believe that no finite set of Bayesian
>>probabilities can be consistent with the statement "you know nothing
>>about what amount to expect".
>
>It's worse than that. _No_ probability distribution, finite or
>infinite, is consistent with that assumption.
But surely you have sufficient information to answer a question such as:
"An arbitrary positive real has been selected. What odds should you
give that it is greater than 10?"
The only alternative is that it is meaningless to speak of an arbitrary
positive real. But as I pointed out to "macavity", we can work in a
non-standard fashion, and at least make some progress without obvious
contradictions. That, rather than a sterile attachment to textbook
math, is what I favour.
- Gerry Quinn
Gerry Quinn
>ger...@indigo.ie (Gerry Quinn) wrote:
>>
>and I
>>have found no logical paradox yet.
>>
>
>I am sure _you_ can. I prefer logic of a more general nature.
>As you seem to be awake anyway, there is no point in trying
>further to wake you up, and I leave you to your dreams.
>
You do not appear to have found a paradox either, just whinges and
insults. I leave you to your indignation at the sight of someone going
beyond textbook interpretations of probability distributions.
- Gerry Quinn
David A Karr
>But surely you have sufficient information to answer a question such as:
>"An arbitrary positive real has been selected. What odds should you
>give that it is greater than 10?"
>
>The only alternative is that it is meaningless to speak of an arbitrary
>positive real.
Not at all. The alternative is that it's meaningless to ask _that_
_question_ about and "arbitrary" positive real. Plenty of mathematics
is done with arbitrary positive reals that have no probabilities
attached to them.
> But as I pointed out to "macavity", we can work in a
>non-standard fashion, and at least make some progress without obvious
>contradictions. That, rather than a sterile attachment to textbook
>math, is what I favour.
We certainly can work in a non-standard fashion. What I object to is
your _deduction_ of your non-standard distribution from the statement
"you know nothing about the distribution." You've just deduced P
from (not P).
Monwhea Jeng
> Nick Wedd <Ni...@maproom.co.uk> wrote:
> >In article <Pine.LNX.4.10.100073...@hal.physics.ucsb.e
> >du>, Monwhea Jeng <mo...@hal.physics.ucsb.edu> writes
> >>This certainly does not follow. Saying "you know nothing about what
> >>to expect" does not imply that any question you might ask has an
> >>answer of 50%. You will need to pick some "reasonable" and consistent
> >>Bayesian probabilities, not just 50% for everything.
> >
> >I am with Gerry here. I believe that no finite set of Bayesian
> >probabilities can be consistent with the statement "you know nothing
> >about what amount to expect".
>
> It's worse than that. _No_ probability distribution, finite or
> infinite, is consistent with that assumption.
I'm not sure that we have any substantive disagreement here. It seems
like the difference is basically semantic. I would take "you know nothing
about what to expect" to mean that you don't have any real data directly
relevant to the subject at hand. But I would say that you still
can (and should) have some "reasonable and consistent" Bayesian
probabilities. That's because you will always have some vague, wild-ass
estimates for things.
Admittedly, I can see that if I have this Bayesian distribution, you will
claim that I do know something "about what to expect." But since
(1) I don't see any circumstances in which I will understand a question,
yet not have some (very poor) Bayesian probabilities before seeing data.
(2) I both hear and say "know nothing about what to expect."
(3) I would prefer to pick a meaning for that phrase such that the statement
is actually meaningful.
Then I would use the definition above when I hear that phrase.
> >Here is an analogy.
> > "You are told that all the taxis in the city of Yoost are numbered
> >consecutively, from 1 upwards. You know nothing about the size of
> >Yoost, to which you have just flown. As you leave the air terminal, you
> >see a taxi. What is you probability distribution function for its
> >number?"
>
> My answer is there is no such probability distribution.
This is a good example. It is reasonable, in everyday language, for me to
say that I know nothing about the size of Yoost, because I've never even
heard of the city. However, I do really do have some minimally decent
guesses about the size of Yoost, because I know something about Earth, and
cities. I know it's less than 6 billion, because it doesn't have more
people than th Earth. -- actually, I'm sure it's less than 10 million,
or else I would have heard of it. I'm pretty sure it's over 10000, or
else it wouldn't even have an air terminal or taxis. So even knowing
nothing about Yoost, I can still form some sort of Bayesian estimates (for
it's population and thus for it's number of taxis). Those estimates will
improve as I see specific taxis and gather other information on
Yoost, in the normal Bayesian fashion. A lot of my original distribution
will be a complete guess, but that's true for most Bayesian estimates.
Perhaps you will object that, strictly speaking, I have violated the
condition that "You know nothing about the size of Yoost." But I don't
see a situation in which I will know absolutely nothing. In the case
of the envelopes, I know that there is more money in the world than
in the two envelopes. If Yoost is a newly-discovered city circling a
distant star, I know something about how many humans can fit on a planet.
Momo
Gerry Quinn
>Gerry Quinn <ger...@indigo.ie> wrote:
>>But surely you have sufficient information to answer a question such as:
>>"An arbitrary positive real has been selected. What odds should you
>>give that it is greater than 10?"
>>
>>The only alternative is that it is meaningless to speak of an arbitrary
>>positive real.
>
>Not at all. The alternative is that it's meaningless to ask _that_
>_question_ about and "arbitrary" positive real. Plenty of mathematics
>is done with arbitrary positive reals that have no probabilities
>attached to them.
Okay, that's what I meant. I should have said a random positive real.
>
>> But as I pointed out to "macavity", we can work in a
>>non-standard fashion, and at least make some progress without obvious
>>contradictions. That, rather than a sterile attachment to textbook
>>math, is what I favour.
>
>We certainly can work in a non-standard fashion. What I object to is
>your _deduction_ of your non-standard distribution from the statement
>"you know nothing about the distribution." You've just deduced P
>from (not P).
>
I think this is a semantic issue. If you prefer, let us say "What is
your estimate of p(x>r) for a given x and a random positive real r?" or,
again, "What is your estimate of p(x>r) for a given x and a positive
real r selected in a completely unknown manner by a creature whose
nature and abilities you do not know?"
- Gerry Quinn
Tom Houser
macavity wrote in message <0b5e0384...@usw-ex0102-015.remarq.com>...
>ger...@indigo.ie (Gerry Quinn) wrote:
>>In any case, it seems to me that the only consistent
>distribution of a
>>truly unknown real (or computable) is of a form proportional to
>1/x.
>>This means that getting an amount between zero and infinity is
>>impossible.
>>
>
>Just curious, but what did you mean (if anything) by the above?
>I get the feeling I maybe missing something profound.
>
Maybe not profound, but the paradox (in any of its many forms) must assume
that the range of possible amounts in the envelopes is infinite. The
common-sense de-paradoxer then is to consider that the money got there
somehow... the choice of how much to put in the envelope has been made. If
we can assume that choice to be a finite process (and to me this is
axiomatic, but probably unsatisfying to many) then we must admit that the
range from which it chose was also finite. Any finite distribution is of
course non-paradoxical; the distribution is axiomatically finite... no
paradox.
Personally, this way of thinking about it gives me a warm fuzzy closure
feeling, but I got into an argument about it around a year ago on this NG
and it didn't seem that too many here shared my feelings. :)
Tom Houser
>In article <sCEf5.2050$ph.41533@stones>,
> "Ed Hall" <e...@copse.free-online.co.uk> wrote:
>
>>You get a friend to think of a number and then flip a coin
>>(without showing you). If it lands heads he adds ten to his
>>number, if it is tails he subtracts ten, then he gives you his
>>answer. With what probability can you guess his original number?
>
>Hi,
>This reminds me a nice variant of the famous "surprise exam" problem
>that can be formulated with almost the same words, namely :
>
>You get a friend to think of a (strictly) positive number and then flip
>a coin (without showing you). If it lands heads he adds one to his
>number, if it is tails he doesn't (he keeps his number unchanged), then
>he gives you his answer. Furthermore, your friend must choose his
>number in such a way that there will be no logical proof for you to
>guess his original number. Is it possible ?
>
>Well, obviously your friend cannot choose "one", because if he does so
>and gets tails he must tell you "one". If you hear "one" you know his
>original number was "zero" or "one". Since the rules don't allow him to
>choose zero, you could prove his original number must be "one".
>
>Now, your friend cannot choose "two", because if he does so and gets
>tails he must tell you "two". If you hear "two" you know his original
>number was "one" or "two". Since you can prove he's not allowed to
>choose "one" according to the paragraph above, you could prove his
>original number must be "two".
>
>Now, your friend cannot choose "three", because if he does so and gets
>tails he must tell you "three". If you hear "three" you know his
>original number was "two" or "three". Since you can prove he's not
>allowed to choose "two" according to the previous paragraph, you could
>prove his original number must be "three".
>
>And so on, to infinity...
>Do you agree ?
I don't know how satisfying this'll be, but I'll put it out there.
It seems like the faulty (hidden) premise to the problem is that your friend
eventually makes the choice of a first number. If he was infallibly logical
and followed the rules, he could not ever choose, due to the rules you
outlined above. You may as well ask him to prove "1=2" before choosing,
it's just not possible. So if your friend DOES choose, you've established
that he may well have chosen the number one (since he demonstrably cheats or
at least is logically fallible), which frees up the logic of the problem and
de-paradoxes it.
The surprise exam question, it seems to me, is a little more satisfyingly
de-paradoxed. If your teacher tells you that the test might happen any day
next week, from monday to friday, and that on the day of the test you won't
be sure that it will happen, then the logical inconsistency is in the
explicit premises themselves -- the test can NOT happen on friday, clearly
contradicting the first premise. The problem is impossible to pose
"paradoxically" if the premises are logically sound.
>--
>Dominique
>
>
>Sent via Deja.com http://www.deja.com/
>Before you buy.
David A Karr
>I think this is a semantic issue.
Looking over some of the other posts, I think maybe you're right.
The question is what does it mean that "you know nothing" about
the probability distribution.
I've been taking the position that "nothing" means really, really
nothing. In that case, well, you can't get something from nothing.
But an alternative is that it really means "nothing else
except the obvious real-world constraints given what the random value
represents," for instance only a limited number of taxis can possibly
exist in any city on Earth, and there are only a limited number of
reals that could possibly be represented in writing inside a business
envelope. These constraints might be enough to come up with some
kind of prior probability estimate.
>If you prefer, let us say "What is
>your estimate of p(x>r) for a given x and a random positive real r?" or,
>again, "What is your estimate of p(x>r) for a given x and a positive
>real r selected in a completely unknown manner by a creature whose
>nature and abilities you do not know?"
This is open-ended enough that I don't feel I have enough real-world
constraints to make such an estimate. I do think it's a reasonable
(but not necessary) guess that the probability decreases from 1 to 0
monotonically as x goes from 0 to infinity. I wouldn't conclude that
p(r = 0) = p(r > all reals) = 1/2.
Rich Grise
13.14159 or -6.85841, then it's pretty much a sure thing.
;-}
Cheers!
Rich
Bruce Bowler wrote:
>
> Seems to me that you've got a 50/50 chance. Given his result, x, there
> are two possible answers. x+10 and x-10. given no restrictions on x,
> it's just a guess which one of x+10 and x-10 is the correct answer.
>
> Ed Hall wrote:
> >
> > You get a friend to think of a number and then flip a coin (without showing
> > you). If it lands heads he adds ten to his number, if it is tails he
> > subtracts ten, then he gives you his answer. With what probability can you
> > guess his original number?
> >
> > Ed
Monwhea Jeng
> In article <Pine.LNX.4.10.100073...@hal.physics.ucsb.edu>, Monwhea Jeng <mo...@hal.physics.ucsb.edu> wrote:
> >On Sat, 29 Jul 2000, Gerry Quinn wrote:
> >
> >> The envelope paradox has as a starting condition "you know nothing about
> >> what amount to expect". Therefore, what is the chance that the amount
> >> will be greater than $0.01? 50% obviously. What is the chance that the
> >> amount will be greater than $1000000000000? Still 50%. Anything other
> >> than 50% violates the stated condition.
> >
> >This certainly does not follow. Saying "you know nothing about what
> >to expect" does not imply that any question you might ask has an
> >answer of 50%. You will need to pick some "reasonable" and consistent
> >Bayesian probabilities, not just 50% for everything.
> >
> >The classic example of why you can't just answer 50% for everything, is
> >to ask if you know anything about extraterrestial life. I do not (I can't
> >speak for you, of course). But I can't consistently say that there's
> >a 50% chance of extraterrestrial life in the Galaxy, and also a 50%
> >chance of life on Tau Ceti, and also a 50% chance of life on Alpha
> >Centauri. The last two statements imply at least a 75% chance of
> >aliens somewhere in the Galaxy. If I say "50% chance" each time you
> >ask me about a new star, and you ask me about 100 different stars, I've
> >essential said that there must be life somewhere in the Galaxy (with
> >probability 1-(1/2)^100).
>
> Different problem. You do know something about Tau Ceti and Alpha
> Centauri - they are stars, that may have planets. The question is a
> hard one, but you do have data to help you form an opinion - it's just
> very difficult to work out an answer given what you know of chemistry,
> stellar and galactic history, the kinds of things that may be alive,
> etc.
>
> A better example would be the question "Is there life in SQUOGGLE" when
> you don't know whether SQUOGGLE is a galaxy, the universe, or an atom.
> Even then it's not as perfect as a question about the real numbers!
>
> - Gerry Quinn
I'm not sure what your argument is. You've deleted your own example
of "taxis in Yoost." My point was that for any situation in which
we can make sense of the question, we have some, perhaps very poor, basis
for making a Bayesian estimate. If we take the phrase "you know nothing
about X" to be meaningful, it must mean that we have no direct information
on X, not that we know absolutely nothing about any subject that relates
to X. We do know something about envelopes and money, so similar relevant
examples are life in the universe or, to use your own example, taxis in
Yoost. How is life in Alpha Centauri more of a "different problem" than
taxis in Yoost?
On the other hand, "life in SQUOGGLE" IS a very different question than
the envelopes with money, because I don't even understand the SQUOGGLE
question.
Momo
Nick Wedd
du>, Monwhea Jeng <mo...@hugh.physics.ucsb.edu> writes
>I'm not sure what your argument is. You've deleted your own example
>of "taxis in Yoost." My point was that for any situation in which
>we can make sense of the question, we have some, perhaps very poor, basis
>for making a Bayesian estimate. If we take the phrase "you know nothing
>about X" to be meaningful, it must mean that we have no direct information
>on X, not that we know absolutely nothing about any subject that relates
>to X. We do know something about envelopes and money, so similar relevant
>examples are life in the universe or, to use your own example, taxis in
>Yoost. How is life in Alpha Centauri more of a "different problem" than
>taxis in Yoost?
The example of "taxis in Yoost" was mine - so I feel entitled to respond
to this, even though it was probably addressed to Gerry.
Sarah says "I am going to think of an integer".
I think "She'll probably think of quite a small one, such that she can
fully specify it in less than 1000 pages".
She continues "You know nothing about what integer to expect".
I think "If Sarah were an ordinary member of the public, I would take
that statement to be meaningless. But as she is a mathematician, she
must intend to convey something by it. She must mean that any ideas I
have about its distribution, such as my conjecture a couple of
paragraphs ago, are mistaken. I deduce that there can be no integer N
and positive probability P such that the size of Sarah's integer is less
than N with probability at least P."
macavity
>Sarah says "I am going to think of an integer".
>
>I think "She'll probably think of quite a small one, such that
she can
>fully specify it in less than 1000 pages".
This line is a seperate paradox in its own right IIRC. Which
is "the first number that takes more than 1000 pages to
specify?" Of course, the quotes themselves form a
specification, and hence the paradox.
>
>She continues "You know nothing about what integer to expect".
>
>I think "If Sarah were an ordinary member of the public, I
would take
>that statement to be meaningless. But as she is a
mathematician, she
>must intend to convey something by it. She must mean that any
ideas I
>have about its distribution, such as my conjecture a couple of
>paragraphs ago, are mistaken. I deduce that there can be no
integer N
>and positive probability P such that the size of Sarah's
integer is less
>than N with probability at least P."
>
There is a are two paradoxes built into the last para I think.
One is the heap paradox, and the second is this - suppose you
make a conclusion S (say like the last statement). Sarah knows
you are going to conclude S, and hence she knows you will
know "something" about the distribution, and hence S cannot be
correct as she sayd you know "nothing" about it. The last is at
best a paradox, at worst a simple contradiction.
>Nick
>--
>Nick Wedd ni...@maproom.co.uk
>
>
Gerry Quinn
>On Mon, 31 Jul 2000, Gerry Quinn wrote:
>> A better example would be the question "Is there life in SQUOGGLE" when
>> you don't know whether SQUOGGLE is a galaxy, the universe, or an atom.
>> Even then it's not as perfect as a question about the real numbers!
>>
>> - Gerry Quinn
>
>I'm not sure what your argument is. You've deleted your own example
>of "taxis in Yoost." My point was that for any situation in which
>we can make sense of the question, we have some, perhaps very poor, basis
>for making a Bayesian estimate. If we take the phrase "you know nothing
>about X" to be meaningful, it must mean that we have no direct information
>on X, not that we know absolutely nothing about any subject that relates
>to X. We do know something about envelopes and money, so similar relevant
>examples are life in the universe or, to use your own example, taxis in
>Yoost. How is life in Alpha Centauri more of a "different problem" than
>taxis in Yoost?
>
>On the other hand, "life in SQUOGGLE" IS a very different question than
>the envelopes with money, because I don't even understand the SQUOGGLE
>question.
>
As Nick pointed out, taxis in Yoost was not my analogy.
What do you not understand about SQUOGGLE? It is the name of a
location. It might be the Moon, it might be the Milky Way, it might be
the Universe, it might be a particular atom of Boron. It's an unknown
place, just as previously we spoke of an unknown number.
Consider the following two statements:
1. There is life in SQUOGGLE.
2. There is no life in SQUOGGLE.
One is true. One is false. The probabilities add up to one.
Given your current knowledge, is one more likely to be true, or have
they both the same likelihood? If the latter, you are assessing the
chances of life in SQUOGGLE at 50%.
But really, it's better to stick to the numbers. What is the
probability that an arbitrary real is greater than 10? What is the
difference between that and the probability that an arbitrary real is
greater than 100? There's nothing *special* about either number. By
the same token, the answer must be the same for 0.0001, 10, 100 or
googolplex. It has to be 50%.
- Gerry Quinn
David A Karr
>What do you not understand about SQUOGGLE? It is the name of a
>location. It might be the Moon, it might be the Milky Way, it might be
>the Universe, it might be a particular atom of Boron. It's an unknown
>place, just as previously we spoke of an unknown number.
>
>Consider the following two statements:
>1. There is life in SQUOGGLE.
>2. There is no life in SQUOGGLE.
>
>One is true. One is false.
Assuming appropriate definitions of "life" and "in", yes.
I'm with you far.
> The probabilities add up to one.
Only if there _are_ probabilities. Why should there be?
How high is "up"?
Gerry Quinn
>Gerry Quinn <ger...@indigo.ie> wrote:
>>What do you not understand about SQUOGGLE? It is the name of a
>>location. It might be the Moon, it might be the Milky Way, it might be
>>the Universe, it might be a particular atom of Boron. It's an unknown
>>place, just as previously we spoke of an unknown number.
>>
>>Consider the following two statements:
>>1. There is life in SQUOGGLE.
>>2. There is no life in SQUOGGLE.
>>
>>One is true. One is false.
>
>Assuming appropriate definitions of "life" and "in", yes.
>I'm with you far.
>
>> The probabilities add up to one.
>
>Only if there _are_ probabilities. Why should there be?
You are offered attractive odds of 2:1 if you guess correctly whether
there is life in SQUOGGLE. Surely you will make a guess (if your
utility function demands it you can always take side bets as insurance).
Which do you choose, 1 or 2? If you choose 1 you are implicitly saying
that from your current perspective, the chance of there being life in
SQUOGGLE exceeds 33%.
The rest is left as an exercise!
Remember, the probability in question is just your assessment. It
requires no more comprehension of SQUOGGLE (or, in the original case, of
the real numbers) than you yourself can bring to bear. By definition
you have enough information to make an assessment based on your current
knowledge of the problem. In the limit of zero knowledge that
assessment goes to 50%.
>
>How high is "up"?
That's not a question of fact like the example.
- Gerry Quinn
Andreas Gunnarsson
> But really, it's better to stick to the numbers. What is the
> probability that an arbitrary real is greater than 10? What is the
> difference between that and the probability that an arbitrary real is
> greater than 100? There's nothing *special* about either number. By
> the same token, the answer must be the same for 0.0001, 10, 100 or
> googolplex. It has to be 50%.
It depends on how you define "how to arbritrarily pick a real number". You
define it as being picked from your probability distribution (any number -
including infinitesimals and infinitely big - is equally likely), and from
that you deduce that that very distribution must have been used. It's like
assuming that a = 1 and from there prove that a = 1.
As for the probability that there is life on Squoggle, I don't know if
anyone really argued that it is 50% since you don't know anything about
it. If so, that is obviously wrong. If you don't know anything you can't
say anything about the probability:
I let you know that I have one thing on my desk. According to the "don't
know means 50%" there is a 50% probability that it is a yellow pen and 50%
probability of a blue pen. But there is also a 50% probability of a green
pen, 50% probability of a rock, 50% probability of a mailbox...
Andreas
--
Andreas Gunnarsson - zzl...@dd.chalmers.se
Using a metaphor as proof is like selling water and charging for single malt
QED
David A Karr
>ka...@shore.net (David A Karr) wrote:
>>> The probabilities add up to one.
>>
>>Only if there _are_ probabilities. Why should there be?
>
>You are offered attractive odds of 2:1 if you guess correctly whether
>there is life in SQUOGGLE.
I'm not sure how you mean this to be interpreted. Does it mean I can
risk, say, $10, make my guess, and if I'm wrong I lose my $10 but if
I'm right I gain $20?
> Surely you will make a guess
Under my interpretation above, yes. I'll flip a coin to choose
between "life" and "no life", thereby giving myself 1/2 chance to win
the $20 whether or not it is rational to assign probability to the
guess "life". Note that I am using a probability assigned to a
well-understood event (the coin toss) and _no_ _other_ probability
in my calculations.
>Which do you choose, 1 or 2? If you choose 1 you are implicitly saying
>that from your current perspective, the chance of there being life in
>SQUOGGLE exceeds 33%.
I'm not sure how to read this but it seems to be based on the
assumption that I have a choice between a 2:1 bet on a guess that
there _is_ life, or a 1:2 bet on a guess that there is _not_ life.
Is that what you meant in the first place?
In that case, I no longer agree with "Surely." If someone is holding
a gun to my head and threatens to shoot if I don't bet, then maybe,
but I would probably still flip the coin in order to bring the
expected final gain for me to 0, because the person posing the choice
presumably knows stuff about SQUOGGLE that I don't, and I don't like
sucker bets or other mind games of that sort. Given a choice to bet
or not to bet, I would not bet at all.
>By definition
>you have enough information to make an assessment based on your current
>knowledge of the problem.
Says who?
>>How high is "up"?
>
>That's not a question of fact like the example.
The "probability of life in SQUOGGLE" is not a question of fact
either. "Life in SQUOGGLE" may be, if anyone has any idea what
SQUOGGLE is, but the probability is another question with a different
answer, just as "is there a brass dollar in my pocket" is a question
of simple fact and "what probability _should_ you think there is that
I have a brass dollar in my pocket" is another quite different
question.
tonyp
Andreas Gunnarsson <zzl...@dd.chalmers.se> wrote
> As for the probability that there is life on Squoggle, I don't know if
> anyone really argued that it is 50% since you don't know anything about
> it. If so, that is obviously wrong. If you don't know anything you can't
> say anything about the probability:
> I let you know that I have one thing on my desk. According to the "don't
> know means 50%" there is a 50% probability that it is a yellow pen and
50%
> probability of a blue pen. But there is also a 50% probability of a
green
> pen, 50% probability of a rock, 50% probability of a mailbox...
Surely if "life" and "Squoggle" map onto "blue pen" and "desk", then
yellow pens and mailboxes, etc. are irrelevant. "Is there a blue pen on
Andreas's desk?" is a question that, like life on Squoggle, or "Does the
coin on Tony's desk show heads or tails?" has a population of exactly two
answers. Your "probability" of _choosing_ the correct answer "at random"
is 0.5, but _the_ "probability" of each answer is of course meaningless.
Note that the "answer" lies in the past; your _choice_ lies in the
future. Perhaps we should say that the past is the region of space-time
to which probability does not apply, but that would be too high-fallutin'.
-- Tony Prentakis
Gerry Quinn
>On Fri, 4 Aug 2000, Gerry Quinn wrote:
>
>> But really, it's better to stick to the numbers. What is the
>> probability that an arbitrary real is greater than 10? What is the
>> difference between that and the probability that an arbitrary real is
>> greater than 100? There's nothing *special* about either number. By
>> the same token, the answer must be the same for 0.0001, 10, 100 or
>> googolplex. It has to be 50%.
>
>It depends on how you define "how to arbritrarily pick a real number". You
>define it as being picked from your probability distribution (any number -
>including infinitesimals and infinitely big - is equally likely), and from
>that you deduce that that very distribution must have been used. It's like
>assuming that a = 1 and from there prove that a = 1.
>
No! I don't say what probability distribution it is picked from. I am
asking what the probability is from MY perspective given that I don't
know anything about the distribution. My k/x distribution represents
all possible distributions.
>As for the probability that there is life on Squoggle, I don't know if
>anyone really argued that it is 50% since you don't know anything about
>it. If so, that is obviously wrong. If you don't know anything you can't
>say anything about the probability:
I use a straighforward definition of subjective probability which is
defined in terms of odds that should be accepted by a rational betting
person with a flat utility function, in cases where tricks or special
predictive powers (e.g. Newcomb's paradox) of the person offering the
bet are not an issue.
Not everyone likes this definition but it is powerful, consistent, and
decisive. For example, it instantly makes me a 'thirder' in the
Sleeping Beauty paradox.
And the rational bet with regard to Squoggle is to take any odds better
than 1:1. Therefore my assessment of the probability of life in
Squoggle is exactly 50%.
>I let you know that I have one thing on my desk. According to the "don't
>know means 50%" there is a 50% probability that it is a yellow pen and 50%
>probability of a blue pen. But there is also a 50% probability of a green
>pen, 50% probability of a rock, 50% probability of a mailbox...
>
> Andreas
Incorrect inference. When you give me more information about the
problem, the probabilities change. I say there is 50% probability of
life in "Squoggle". I would say 90% for "Galaxy X" (I think life is
probably common and most galaxies are old enough to develop it). When
you talk about concrete things like pens etc. I can assess the
probability, and also in this case I will think about the possible
procedures by which you have selected an object to place on your desk.
In the envelope paradox, there are really three possibilities: number
less than A, number equal to A, and number greater than A. I assess the
second as zero, the others at 50% each.
For the desk problem, put yourself in my position. Should I accept 1:1
bets on a yellow pen, a blue pen and a mailbox? Clearly these would be
likely to lose me money.
- Gerry Quinn
Gerry Quinn
>Gerry Quinn <ger...@indigo.ie> wrote:
>>ka...@shore.net (David A Karr) wrote:
>>>> The probabilities add up to one.
>>>
>>>Only if there _are_ probabilities. Why should there be?
>>
>>You are offered attractive odds of 2:1 if you guess correctly whether
>>there is life in SQUOGGLE.
>
>I'm not sure how you mean this to be interpreted. Does it mean I can
>risk, say, $10, make my guess, and if I'm wrong I lose my $10 but if
>I'm right I gain $20?
>
Yes.
>> Surely you will make a guess
>
>Under my interpretation above, yes. I'll flip a coin to choose
>between "life" and "no life", thereby giving myself 1/2 chance to win
>the $20 whether or not it is rational to assign probability to the
>guess "life". Note that I am using a probability assigned to a
>well-understood event (the coin toss) and _no_ _other_ probability
>in my calculations.
>
And I only talk about the probability _from_my_perspective_ of "life in
Squoggle, also a well understood concept. For example, if I wake with
amnesia, I can assign a probability to "Today is Monday".
>>Which do you choose, 1 or 2? If you choose 1 you are implicitly saying
>>that from your current perspective, the chance of there being life in
>>SQUOGGLE exceeds 33%.
>
>I'm not sure how to read this but it seems to be based on the
>assumption that I have a choice between a 2:1 bet on a guess that
>there _is_ life, or a 1:2 bet on a guess that there is _not_ life.
>Is that what you meant in the first place?
>
Not really, what I'm saying is that if you have a flat utility function
and you trust that no trickery is involved, you will accept any odds
better than 1:1 on either of the two possibilities. By my definition of
probability, then, it is a factual matter whether there is or is not
life in Squoggle, but your assessment of the probability is 50%.
>In that case, I no longer agree with "Surely." If someone is holding
>a gun to my head and threatens to shoot if I don't bet, then maybe,
>but I would probably still flip the coin in order to bring the
>expected final gain for me to 0, because the person posing the choice
>presumably knows stuff about SQUOGGLE that I don't, and I don't like
>sucker bets or other mind games of that sort. Given a choice to bet
>or not to bet, I would not bet at all.
>
This is not an issue in my argument.
>>By definition
>>you have enough information to make an assessment based on your current
>>knowledge of the problem.
>
>Says who?
>
How can you have less knowledge than your current knowledge? To say
"you don't have enough information to make an assessment based on your
current knowledge" is like saying "you are unable to do something that
is in your power".
>>>How high is "up"?
>>
>>That's not a question of fact like the example.
>
>The "probability of life in SQUOGGLE" is not a question of fact
>either. "Life in SQUOGGLE" may be, if anyone has any idea what
>SQUOGGLE is, but the probability is another question with a different
>answer, just as "is there a brass dollar in my pocket" is a question
>of simple fact and "what probability _should_ you think there is that
>I have a brass dollar in my pocket" is another quite different
>question.
>
Yes, but in the envelope paradox we are talking about the second kind.
The first kind is rarely of interest. When I say "there is a 50% chance
of a sea battle" or "it's bright and the streets are quiet, it's
probably Sunday" I am speaking about a subjective probability. Likewise
in the envelope paradox, the contents of the envelopes are matters of
fact (at least if we make the questionable assumption that an
arbitrary real or computable number can be written down - counterfactual
in this universe because an arbitrarily long computable number would
require a physical embodiment that would undergo instant gravitational
collapse). The question of whether x exceeds r (with x chosen by me,
and r the arbitrary real) is one with two answers to each of which I can
assign a subjective probability. I argue that this must be 50%, from
which my k/x distribution follows.
So k/x, k infinitesimal is the distribution of an arbitrary real, as
seen by me. In shorthand, it's the distribution of an arbitrary real.
- Gerry Quinn
lta...@my-deja.com
ger...@indigo.ie (Gerry Quinn) wrote:
> In article
<Pine.GSO.4.21.000804...@uldor.dd.chalmers.se>,
Andreas Gunnarsson <zzl...@dd.chalmers.se> wrote:
> > As for the probability that there is life on Squoggle, I don't know
> > if anyone really argued that it is 50% since you don't know
> > anything about it. If so, that is obviously wrong. If you don't
> > know anything you can't
> > say anything about the probability:
>
> I use a straighforward definition of subjective probability which is
> defined in terms of odds that should be accepted by a rational
> person with a flat utility function, in cases where tricks or special
> predictive powers (e.g. Newcomb's paradox) of the person offering the
> bet are not an issue.
...
> And the rational bet with regard to Squoggle is to take any odds
> better than 1:1. Therefore my assessment of the probability of life
> in Squoggle is exactly 50%.
>
> >...But there is also a 50% probability of a green
> > pen, 50% probability of a rock, 50% probability of a mailbox...
>
> Incorrect inference. When you give me more information about the
> problem, the probabilities change. I say there is 50% probability of
> life in "Squoggle". I would say 90% for "Galaxy X" (I think life is
> probably common and most galaxies are old enough to develop it).
> When you talk about concrete things like pens etc. I can assess the
> probability, and also in this case I will think about the possible
> procedures by which you have selected an object to place on your
> desk.
>
> In the envelope paradox, there are really three possibilities: number
> less than A, number equal to A, and number greater than A. I assess
> the second as zero, the others at 50% each.
>
> For the desk problem, put yourself in my position. Should I accept
> 1:1 bets on a yellow pen, a blue pen and a mailbox? Clearly these
> would be likely to lose me money.
>
> - Gerry Quinn
Surely Gerry Quinn is right about the yellow pen bet, but I don't see
how Life on Squoggle qualifies as a subjective probability assignment
based on total ignorance. Even if all we know about Squoggle is that
it's a physical location, it still behooves us (as rational bettors) to
do what we can with this little scrap of information. For example the
basis for the subjective probability assignment could be:
p = number of inhabited locations / number of named locations
Or alternatively:
p = number of inhabited locations I've never heard of / number of named
locations I've never heard of
(the second is more subtle but admittedly hard to estimate!)
In this sense Andreas Gunnarsson has a point about ignorance and
real-world probability assessments. If we understand an event
description at all, we know enough about it to speculate (at least a
little bit) about what equivalence classes it might belong to.
This is just a minor observation about practical examples such as
Squoggle. I'm not claiming that it has much to do with more explicitly
mathematical problems, such as those that are stated in terms of
probability distributions over the reals.
Larry Tapper
Nis Jørgensen
wrote:
>I use a straighforward definition of subjective probability which is
>defined in terms of odds that should be accepted by a rational betting
>person with a flat utility function, in cases where tricks or special
>predictive powers (e.g. Newcomb's paradox) of the person offering the
>bet are not an issue.
>
>Not everyone likes this definition but it is powerful, consistent, and
>decisive. For example, it instantly makes me a 'thirder' in the
>Sleeping Beauty paradox.
Simple probability theory instantly made me a thirder in the
Sleeping Beauty problem (I refuse to call it a paradox, until I
get a consistent argument from a halfer)
>And the rational bet with regard to Squoggle is to take any odds better
>than 1:1. Therefore my assessment of the probability of life in
>Squoggle is exactly 50%.
OK. I have made a sealed envelope containing my definition of
Squoggle. Will you take a 2:1 bet that there is life in Squoggle.
Why not?
Courtenay Footman
Because you aren't offering him the alternative bet. He will take
one of the bets if you say "You can pick between a 2:1 bet that there
is life on Squoggle, and a 2:1 bet that there isn't, but you can't
pick both."
--
Courtenay Footman I have again gotten back on the net, and
c...@lightlink.com again I will never get anything done.
(All mail from non-valid addresses is automatically deleted by my system.)
Gerry Quinn
>Surely Gerry Quinn is right about the yellow pen bet, but I don't see
>how Life on Squoggle qualifies as a subjective probability assignment
>based on total ignorance. Even if all we know about Squoggle is that
>it's a physical location, it still behooves us (as rational bettors) to
>do what we can with this little scrap of information. For example the
>basis for the subjective probability assignment could be:
>
>p = number of inhabited locations / number of named locations
>
>Or alternatively:
>
>p = number of inhabited locations I've never heard of / number of named
>locations I've never heard of
>
>(the second is more subtle but admittedly hard to estimate!)
>
>In this sense Andreas Gunnarsson has a point about ignorance and
>real-world probability assessments. If we understand an event
>description at all, we know enough about it to speculate (at least a
>little bit) about what equivalence classes it might belong to.
>
>This is just a minor observation about practical examples such as
>Squoggle. I'm not claiming that it has much to do with more explicitly
>mathematical problems, such as those that are stated in terms of
>probability distributions over the reals.
I don't disagree - in fact I noted earlier that the Squoggle problem is
less 'pure' than the 'real number' problem.
- Gerry Quinn
Gerry Quinn
>In article <h2dooskdp79uj42gs...@4ax.com>, Nis Jørgensen wrote:
>>OK. I have made a sealed envelope containing my definition of
>>Squoggle. Will you take a 2:1 bet that there is life in Squoggle.
>>Why not?
>>
>Because you aren't offering him the alternative bet. He will take
>one of the bets if you say "You can pick between a 2:1 bet that there
>is life on Squoggle, and a 2:1 bet that there isn't, but you can't
>pick both."
>
Correct.
- Gerry Quinn
Nis Jørgensen
Footman) wrote:
>>OK. I have made a sealed envelope containing my definition of
>>Squoggle. Will you take a 2:1 bet that there is life in Squoggle.
>>Why not?
>>
>
>Because you aren't offering him the alternative bet. He will take
>one of the bets if you say "You can pick between a 2:1 bet that there
>is life on Squoggle, and a 2:1 bet that there isn't, but you can't
>pick both."
How will he choose between them? Flipping a coin I guess. Now try
the above for any statement - "The sky is blue", "Pigs have
wings", anything - and note how the same strategy gives you a
win. Does this mean that the chances of these being true are 50%.
I guess this "subjective probability" doesn't work with mixed
strategies ... at least not this way.
Gerry Quinn
>On 5 Aug 2000 18:55:26 -0500, c...@adore.lightlink.com (Courtenay
>Footman) wrote:
>
>>>OK. I have made a sealed envelope containing my definition of
>>>Squoggle. Will you take a 2:1 bet that there is life in Squoggle.
>>>Why not?
>>>
>>
>>Because you aren't offering him the alternative bet. He will take
>>one of the bets if you say "You can pick between a 2:1 bet that there
>>is life on Squoggle, and a 2:1 bet that there isn't, but you can't
>>pick both."
>
>How will he choose between them? Flipping a coin I guess. Now try
>the above for any statement - "The sky is blue", "Pigs have
>wings", anything - and note how the same strategy gives you a
>win. Does this mean that the chances of these being true are 50%.
>
No. And I can optimise my choices by taking into account that, for
example, pigs don't have wings (except maybe in some freaky genetic
laboratory).
The fact that the strategy still gives you a win doesn't prove that the
situation is the same. Indeed, the fact that I can improve on the
strategy proves that it isn't!
>I guess this "subjective probability" doesn't work with mixed
>strategies ... at least not this way.
>
If I make money, it's working!
- Gerry Quinn
Nis Jørgensen
wrote:
>>I guess this "subjective probability" doesn't work with mixed
>>strategies ... at least not this way.
>
>If I make money, it's working!
So how do you calculate the "subjective probability" when mixed
strategies are involved?
David A Karr
> Nis Jørgensen <n...@dkik.dk> wrote:
>>I guess this "subjective probability" doesn't work with mixed
>>strategies ... at least not this way.
>
>If I make money, it's working!
Well, I post to rec.puzzles, and I got a raise in my salary.
So posting to rec.puzzles helps me make money--it's working!
The fact is that with the mixed strategy (coin flipping),
your subjective probability of "life in SQUOGGLE" had zero,
zilch, nada, indeed absolutely _nothing_ to do with the success
of the strategy, as evidenced by the fact that you make money
_just_ _as_ _fast_ no matter what that subjective probability is.
Naturally, if you actually know something abuot the meaning of the
question--better than someone saying "_I_ have written down the
meaning right here and will show you as soon as you guess"--then you
_may_ have enough information to estimate a probability. But I
thought we were asking what you do when you _don't_ have that
information.
David A Karr
>c...@adore.lightlink.com (Courtenay Footman) wrote:
>> Nis Jørgensen wrote:
>>>OK. I have made a sealed envelope containing my definition of
>>>Squoggle. Will you take a 2:1 bet that there is life in Squoggle.
>>>Why not?
>>>
>>Because you aren't offering him the alternative bet. He will take
>>one of the bets if you say "You can pick between a 2:1 bet that there
>>is life on Squoggle, and a 2:1 bet that there isn't, but you can't
>>pick both."
>
Do you really mean to say what I think you're saying here: that you
would take the bet under Courtenay's terms (you get to choose which
side to bet on), but not under Nis's terms (you must bet that there is
life, or not bet at all)?
If you do not accept Nis's bet precisely as he proposed it to you,
then I don't believe that you even believe that 50% business.
Gerry Quinn
>Gerry Quinn <ger...@indigo.ie> wrote:
>>c...@adore.lightlink.com (Courtenay Footman) wrote:
>>> Nis Jørgensen wrote:
>>>>OK. I have made a sealed envelope containing my definition of
>>>>Squoggle. Will you take a 2:1 bet that there is life in Squoggle.
>>>>Why not?
>>>>
>>>Because you aren't offering him the alternative bet. He will take
>>>one of the bets if you say "You can pick between a 2:1 bet that there
>>>is life on Squoggle, and a 2:1 bet that there isn't, but you can't
>>>pick both."
>>
>>Correct.
>
>Do you really mean to say what I think you're saying here: that you
>would take the bet under Courtenay's terms (you get to choose which
>side to bet on), but not under Nis's terms (you must bet that there is
>life, or not bet at all)?
>
>If you do not accept Nis's bet precisely as he proposed it to you,
>then I don't believe that you even believe that 50% business.
>
The only reason I don't accept Nis's bet is that he has chosen the
referent for Squoggle, and he has a financial interest in choosing a
region he thinks probably does not have life. This causes the chances
to go below 50%, and indeed, the higher odds he offers the lower the
chances (from my perspective) that he has picked somewhere with life.
So it looks like a sucker bet.
If I thought he had no interest in whether I guess right or wrong, I
would take the bet.
- Gerry Quinn
Gerry Quinn
>On Sun, 06 Aug 2000 12:42:31 GMT, ger...@indigo.ie (Gerry Quinn)
>
>>>I guess this "subjective probability" doesn't work with mixed
>>>strategies ... at least not this way.
>>
>>If I make money, it's working!
>
>strategies are involved?
>
It's harder, but sometimes you have to do it. Games (and wars) would be
boring if you didn't have to make guesses about what your opponent will
do.
- Gerry Quinn
Gerry Quinn
>Gerry Quinn <ger...@indigo.ie> wrote:
>
>>>I guess this "subjective probability" doesn't work with mixed
>>>strategies ... at least not this way.
>>
>>If I make money, it's working!
>
>So posting to rec.puzzles helps me make money--it's working!
>
>The fact is that with the mixed strategy (coin flipping),
>your subjective probability of "life in SQUOGGLE" had zero,
>zilch, nada, indeed absolutely _nothing_ to do with the success
>of the strategy, as evidenced by the fact that you make money
>_just_ _as_ _fast_ no matter what that subjective probability is.
>
I disagree. Consider the situation where I am offered just one of the
two options, but a trusted referee has determined that the offerer has
no interest in whether I win or lose. Then the 50% probability is the
correct one on which to accept or reject the offered bet. No other
probability works as well.
>Naturally, if you actually know something abuot the meaning of the
>question--better than someone saying "_I_ have written down the
>meaning right here and will show you as soon as you guess"--then you
>_may_ have enough information to estimate a probability. But I
>thought we were asking what you do when you _don't_ have that
>information.
Yes. I don't see where we diverged from that. I have estimated a
probability of 50% for each case. Any extra information (for example,
that the person selecting the location of Squoggle has an interest in
seeing me lose) changes the probability.
- Gerry Quinn
David A Karr
> ka...@shore.net (David A Karr) wrote:
>>The fact is that with the mixed strategy (coin flipping),
>>your subjective probability of "life in SQUOGGLE" had zero,
>>zilch, nada, indeed absolutely _nothing_ to do with the success
>>of the strategy, as evidenced by the fact that you make money
>>_just_ _as_ _fast_ no matter what that subjective probability is.
>
>I disagree.
About what? Was the above any less than completely accurate about
the mixed strategy that was described earlier?
> Consider the situation where I am offered just one of the
>two options, but a trusted referee has determined that the offerer has
>no interest in whether I win or lose.
Ah, so now we have a trusted referee who has examined the motivations
of the offerer of the bet. That's pretty powerful information. It's
also a completely new situation on that account.
In that case, why not assume the bet is exactly fair? That is, if the
odds are offered at 2:1, then take that as evidence that your best
estimate of the relative probabilities is 1:2. After all, someone who
goes around making bets at odds of 2:1 that pay off half the time is
helping people win money, and someone who has no interest in whether
you win or lose surely wouldn't do that (or would they?).
Gerry Quinn
>Gerry Quinn <ger...@indigo.ie> wrote:
>> ka...@shore.net (David A Karr) wrote:
>>>The fact is that with the mixed strategy (coin flipping),
>>>your subjective probability of "life in SQUOGGLE" had zero,
>>>zilch, nada, indeed absolutely _nothing_ to do with the success
>>>of the strategy, as evidenced by the fact that you make money
>>>_just_ _as_ _fast_ no matter what that subjective probability is.
>>
>>I disagree.
>
>About what? Was the above any less than completely accurate about
>the mixed strategy that was described earlier?
>
Yes, but when you have more information the mixed strategy becomes
suboptimal.
>> Consider the situation where I am offered just one of the
>>two options, but a trusted referee has determined that the offerer has
>>no interest in whether I win or lose.
>
>Ah, so now we have a trusted referee who has examined the motivations
>of the offerer of the bet. That's pretty powerful information. It's
>also a completely new situation on that account.
>
It's closer to the canonical envelope problem. The point is that you
are focusing on irrelevant aspects of my imperfect analogies. Useful
debating tactic, but gets us no closer to the meat of the issue.
>In that case, why not assume the bet is exactly fair? That is, if the
>odds are offered at 2:1, then take that as evidence that your best
>estimate of the relative probabilities is 1:2. After all, someone who
>goes around making bets at odds of 2:1 that pay off half the time is
>helping people win money, and someone who has no interest in whether
>you win or lose surely wouldn't do that (or would they?).
>
It's a point. But I think you know you are going off on an irrelevant
tangent here. I came up with 2:1 simply to create a concrete example of
a "better than evens" bet.
- Gerry Quinn
David A Karr
>ka...@shore.net (David A Karr) wrote:
>>Was the above any less than completely accurate about
>>the mixed strategy that was described earlier?
>
>Yes, but when you have more information the mixed strategy becomes
>suboptimal.
But the latter is a different question, isn't it?
Of _course_ when you have more information it changes the answer. In
fact, when you have more information, it may become possible
rationally to assign a probability (one worth betting on) to an event
to which you previously could not rationally assign a probability.
>>Ah, so now we have a trusted referee who has examined the motivations
>>of the offerer of the bet. That's pretty powerful information. It's
>>also a completely new situation on that account.
>
>It's closer to the canonical envelope problem.
That's making a claim about the envelope problem, one whose truth is not
obvious to me. We haven't had much discussion of the motivation of the
other player in Two Envelopes.
>The point is that you
>are focusing on irrelevant aspects of my imperfect analogies. Useful
>debating tactic, but gets us no closer to the meat of the issue.
No, the point is that every time you want to establish a probability,
you do so by adding new information to the situation. I've never
claimed that there is no information you can add that would allow a
probability to be assigned (that would be a silly claim, wouldn't it?)
so the fact that you _can_ add such information doesn't help. If it
seems we're going around in circles, _that_ is why.
>>In that case, why not assume the bet is exactly fair? That is, if the
>>odds are offered at 2:1, then take that as evidence that your best
>>estimate of the relative probabilities is 1:2. [...]
>
>It's a point. But I think you know you are going off on an irrelevant
>tangent here. I came up with 2:1 simply to create a concrete example of
>a "better than evens" bet.
Fine. Generalize to K:1 odds, K any real number. Put K in the place of
2 in my argument. Same thing.
But this discussion has been chock full of irrelevant tangents, as I've
already complained. Here's the fundamental statement of what I feel
is the point of argument:
For any situation in the class of all situations in which you could
possibly find yourself, for any statement Q about that situation,
there is a unique subjective probability p that you can rationally
assign to the truth of Q.
You appear to be using this statement as an axiom in your analysis of
the "random real" situation. I think the statement is absurd, because
there are many questions to which I could not assign such a unique
probability. (In fact, in real life, those questions seem to occur
more often than those in which I _can_ assign a unique probability.)
You've been telling me (in other posts) how I would behave in certain
situations (e.g., that I would take such-and-such a bet as long as the
odds were better than 1:1). I think I'm a better judge of how I would
behave than you are.
You seem to be claiming that if someone has chosen a random positive real
number under certain circumstances--but we don't know what the
procedure was for choosing the number--then the chance that the number
is greater than 10^10^10^100 is precisely the same as the chance that
it is greater than 1/10^10^10^100. I don't believe that. To be more
precise, you're free to believe such a subjective probability yourself,
but I think that would be an irrational choice, and the associated math
promises to be so ugly I'm not going to touch that either.
Gerry Quinn
>But this discussion has been chock full of irrelevant tangents, as I've
>already complained. Here's the fundamental statement of what I feel
>is the point of argument:
>
> For any situation in the class of all situations in which you could
> possibly find yourself, for any statement Q about that situation,
> there is a unique subjective probability p that you can rationally
> assign to the truth of Q.
Fair enough. And I assert that as my information drops to zero and the
situation becomes ununderstandable, p takes on a limiting value of 0.5.
>
>You appear to be using this statement as an axiom in your analysis of
>the "random real" situation. I think the statement is absurd, because
>there are many questions to which I could not assign such a unique
>probability. (In fact, in real life, those questions seem to occur
>more often than those in which I _can_ assign a unique probability.)
>
That's a bit like saying you don't know the answer to Goldbach's
Conjecture. Of course you don't, but if you were a perfectly rational
being you might.
>You've been telling me (in other posts) how I would behave in certain
>situations (e.g., that I would take such-and-such a bet as long as the
>odds were better than 1:1). I think I'm a better judge of how I would
>behave than you are.
>
I'm not talking about you, but about the perfectly rational individual
who features so frequently as the subject of posts on this newsgroup,
even if he never seems to post himself ;-)
>You seem to be claiming that if someone has chosen a random positive real
>number under certain circumstances--but we don't know what the
>procedure was for choosing the number--then the chance that the number
>is greater than 10^10^10^100 is precisely the same as the chance that
>it is greater than 1/10^10^10^100. I don't believe that. To be more
>precise, you're free to believe such a subjective probability yourself,
>but I think that would be an irrational choice, and the associated math
>promises to be so ugly I'm not going to touch that either.
>
But the math is very nice and simple when you accept my claim! On the
other hand, if I accept your claim it becomes quite horrific. Let's
take the value g to represent the probability that the number will be
between 1/googolplexplex and googolplexplex.
What is the chance that it will be between googolplexplex and
googolplexplexplex (a much larger range)? Greater or lesser than g?
How soon before your totals add up to more than 1?
Needless to say we are talking about numbers selected by rational beings
- if selected by humans, a positive g would be reasonable.
I would also suggest that the cheque in the envelope should be written
in a currency whose value is an arbitrary real multiple of the dollar.
Again, we move away from the concept of "selecting a number" towards
the concept of an "arbitrary positive real". You could say I'm sitting
on the fence, but I don't see any errors or paradox arising from my
arguments here. Your argument seems to be basically "give up, this
extension of our concepts is meaningless". But mathematics doesn't work
that way. All the extension has to do is be consistent, and it's
legitimate. At a simple level, it's consistent, and the record suggests
that when that happens the foundational stuff can be made to fall in
line, despite much grumbling among conservatives...
- Gerry Quinn
lta...@my-deja.com
ger...@indigo.ie (Gerry Quinn) wrote:
> In article <Rvzj5.74$%L1.2...@news.shore.net>, ka...@shore.net (David
A Karr) wrote:
>
> >Here's the fundamental statement of what I feel
> >is the point of argument:
> >
> > For any situation in the class of all situations in which you
> > could possibly find yourself, for any statement Q about that
> > situation,
> > there is a unique subjective probability p that you can rationally
> > assign to the truth of Q.
>
> Fair enough. And I assert that as my information drops to zero and
> the situation becomes ununderstandable, p takes on a limiting value
> of 0.5.
I see what you mean, but I'm not convinced that this is an accurate
description of rational behavior when you find yourself in the
neighborhood of complete ignorance. There are two cases to consider:
(a) When you know only a tiny bit about a proposition (as in the
Squoggle example) it does not follow that your estimated subjective
probability should be somewhere close to 0.5. Your probability should
be based on whatever partition of events follows from the tiny bit you
know. If the proposition is "Squoggles can fly", it would seem
perfectly reasonable to assign a low probability on the basis of the
fact that few things can fly.
(b) When you have reached the limit and cannot make any sense at all
out of the proposition, it seems just as reasonable to say all bets are
off as it is to say you should go with the 0.5 default. I'm not even
sure I know what it means to be discussing this case. The exchange
between you and David Karr on the subject has been complicated by the
fact that someone always enters the picture offering to make a bet. In
that case I would say that you aren't in a situation of complete
ignorance, after all, and we're back to case (a).
It may be lack of imagination on my part, but I can't think of a really
good example that's much purer than Squoggle. For example, how about
making a bet on the truth-value of a declarative statement in some
language you don't understand? (For example, betting on "Schnee ist
weiss" if you don't know German.) Even this doesn't seem to work
because you could easily have some empirically based theory about the
distribution of truth values of randomly encountered sentences.
Complete ignorance seems awfully hard to come by.
ma...@mimosa.csv.warwick.ac.uk
ger...@indigo.ie (Gerry Quinn) writes:
>In article <Rvzj5.74$%L1.2...@news.shore.net>, ka...@shore.net (David A Karr) wrote:
>
>>
>>You appear to be using this statement as an axiom in your analysis of
>>the "random real" situation. I think the statement is absurd, because
>>there are many questions to which I could not assign such a unique
>>probability. (In fact, in real life, those questions seem to occur
>>more often than those in which I _can_ assign a unique probability.)
>>
>
>That's a bit like saying you don't know the answer to Goldbach's
>Conjecture. Of course you don't, but if you were a perfectly rational
>being you might.
I am inclined to think that 'perfectly rational being' is a meaningless
(or at least a completely unhelpful) concept. Could you say what you mean
by such a person?
Presumably it could compute infinitely fast (or at least arbitrarily fast).
Then it could settle Goldbach's conjecture by checking all cases - it could
do that by doubling its speed of operation every minute. It could also
solve provably unsolvable problems, etc.
>>You've been telling me (in other posts) how I would behave in certain
>>situations (e.g., that I would take such-and-such a bet as long as the
>>odds were better than 1:1). I think I'm a better judge of how I would
>>behave than you are.
>>
>
>I'm not talking about you, but about the perfectly rational individual
>who features so frequently as the subject of posts on this newsgroup,
>even if he never seems to post himself ;-)
That's because there is no such animal.
>>You seem to be claiming that if someone has chosen a random positive real
>>number under certain circumstances--but we don't know what the
>>procedure was for choosing the number--then the chance that the number
>>is greater than 10^10^10^100 is precisely the same as the chance that
>>it is greater than 1/10^10^10^100. I don't believe that. To be more
>>precise, you're free to believe such a subjective probability yourself,
>>but I think that would be an irrational choice, and the associated math
>>promises to be so ugly I'm not going to touch that either.
>>
>
>But the math is very nice and simple when you accept my claim! On the
>other hand, if I accept your claim it becomes quite horrific. Let's
>take the value g to represent the probability that the number will be
>between 1/googolplexplex and googolplexplex.
According to your claims, g = 0.
>What is the chance that it will be between googolplexplex and
>googolplexplexplex (a much larger range)? Greater or lesser than g?
>How soon before your totals add up to more than 1?
Still 0, according to you.
In fact, your number has a 1/2 chance of being infinitesimally small,
and a 1/2 chance of being infinitely large, so it cannot reasonably
be any real number at all. This is not a consistent mathematical
theory at all!
I think the basic problem here is that the expression 'arbitrary real
number' does not mean anything by iteslf and without the specification
of a distribution, any more than 'perfectly reasonable being' means anything,
without further qualification.
Derek Holt.
David A Karr
>Fair enough. And I assert that as my information drops to zero and the
>situation becomes ununderstandable, p takes on a limiting value of 0.5.
>
>>more often than those in which I _can_ assign a unique probability.)
>
>That's a bit like saying you don't know the answer to Goldbach's
>Conjecture. Of course you don't, but if you were a perfectly rational
>being you might.
"Might" because we don't actually know if the question is even
decidable. :-)
>I'm not talking about you, but about the perfectly rational individual
>who features so frequently as the subject of posts on this newsgroup,
>even if he never seems to post himself ;-)
That's the problem, isn't it? If you could just get this perfectly
rational person to post that *she* assigns the value 0.5 to the
no-information case, then I might have to believe it.
>But the math is very nice and simple when you accept my claim!
Is it? The only simple thing seems to be the claim that p = 1/2.
If all you needed was a value for p, that's simple. But as soon
as you ask what's the underlying distribution, it's non-standard.
> On the
>other hand, if I accept your claim it becomes quite horrific. Let's
>take the value g to represent the probability that the number will be
>between 1/googolplexplex and googolplexplex.
>
>What is the chance that it will be between googolplexplex and
>googolplexplexplex (a much larger range)? Greater or lesser than g?
I don't know. If I knew, that would be quite definite information
about the distribution.
>How soon before your totals add up to more than 1?
Never.
>Your argument seems to be basically "give up, this
>extension of our concepts is meaningless". But mathematics doesn't work
>that way. All the extension has to do is be consistent, and it's
>legitimate.
That's a bit of an oversimplification. To be accepted as _worthwhile_,
the extension has to produce some useful results. Of course "useful"
is a bit subjective.
So do something useful with the p = 1/2 assumption. Does it really
support the k/x distribution (however that is to be interpreted),
or was that another independent assumption?
Andreas Gunnarsson
> >> But really, it's better to stick to the numbers. What is the
> >> probability that an arbitrary real is greater than 10? What is the
> >> difference between that and the probability that an arbitrary real is
> >> greater than 100? There's nothing *special* about either number. By
> >> the same token, the answer must be the same for 0.0001, 10, 100 or
> >> googolplex. It has to be 50%.
The phrase "arbtrary real" is meaningless if you don't define how you
arbitrarily pick that real. It is not obvious how arbitrary reals are
distributed. I would say that a distribution of arbitrary reals in most
cases favors numbers that need little space when written down. If there is
a widely accepted definition that says something else I stand corrected.
With another distribution of arbitrary reals, any number could be equally
probable as any other number and the probability for a number between a
and a+n could be the same as the probability for a number between b and
b+n for any a, b and n. The probability distribution p(x)=k has those
properties. Unfortunately it is impossible to actually choose a number
using that distribution, that's one reason why I wouldn't define
"arbitrarily chosen real" as picked from that distribution.
[...]
> No! I don't say what probability distribution it is picked from. I am
> asking what the probability is from MY perspective given that I don't
> know anything about the distribution. My k/x distribution represents
> all possible distributions.
Pick x1 from probability distribution p1 and x2 from probability
distribution p2. If p1 and p2 are the same probability distributions, then
p(x1>x2)=p(x1<x2)=0.5-s and p(x1=x2)=s.
I haven't proved this, but it seems obvious to me. Actually, I would guess
that p(x1>x2)=p(x1<x2)=0.5-s if the average* of p1 is the same as the
average of p2. I'll assume that this is true without proving it. The proof
is left as an exercise for the interested.
I'll also assume that s=0 from now on, I don't think it makes a difference
for the argument.
Your statement that p(x>n)=0.5 for any positive n and an arbitrarily
chosen number x is true if x is chosen from the probability distribution
p(x) = 0 for x<=0 and p(x) = k/x for x>0. It is also true if p(x=0)=0.5,
p(x=infinity)=50% and p=0 for other x. Unfortunately it is impossible to
chose a number from those probability distributions.
Note the difference between the statements "p(x>n)=0.5 for any positive
real n" and "p(x>n)=0.5 for an arbitrary positive real n";
"p(x>n)=0.5 for any positive real n" is true if x is arbitrarily "chosen"
from the distributions above.
"p(x>n)=0.5 for an arbitrarily chosen positive real n" is true if x and n
are arbitratily chosen from probability distributions with the same
average.
If you don't know anything about the distribution used to arbitrarily pick
the number x and you pick the number y from a known distribution you don't
know anything about the probability p(x>y). You *do* have 50% chance to
guess whether x>y or x<y, for example by tossing a coin.
Let's assume that you'll guess that x>y if you get tails and x<y
otherwise. p(you guess that x>y)=p(you guess that x<y)=0.5.
Let's also say that p(x>y)=a and p(x<y)=1-a.
We have 4 cases:
x>y, you guess x>y - probability a*0.5
x>y, you guess x<y - probability a*0.5
x<y, you guess x>y - probability (1-a)*0.5
x<y, you guess x<y - probability (1-a)*0.5
In the first and last case you guess right, in case two and three you
guess wrong.
p(you guess right)=a*0.5 + (1-a)*0.5 = 0.5
p(you guess wrong)=a*0.5 + (1-a)*0.5 = 0.5
This does not depend on the value of a or the probability distributions
used.
If you assume that p(x<y) = p(x>y) = 0.5 you could just as well guess that
p(x>y) each time. However, if you do that you don't know how probable it
is that you guess right. The probability that you guess would be a and the
probability that you guess wrong is 1-a, not 0.5.
The probability that x>y only depends on what probability distributions
are used to pick x and y, not what you know about either or both of the
distributions. The probability that you can make a correct guess does
however depend on what you know about how they are chosen. Of course, when
the numbers actually have been chosen the probability is either 0 or 1,
not something in between but your chance of guessing right may not have
changed.
Am I correct when I believe that what you mean when you say "subjective
probability that x>y" is what I mean when I say "probability that I can
successfully guess whether x>y"? In that case we might mean the same
thing.
* I don't know if "average of a probability distribution" is the correct
term, but I use it for the integral from x=negative infinity to x=positive
infinity of x*p(x) (the integral of p(x) is of course 1).
Kevin Buhr
>
> Not everyone likes this definition but it is powerful, consistent, and
> decisive. For example, it instantly makes me a 'thirder' in the
> Sleeping Beauty paradox.
It doesn't seem consistent. That is, your assessment of the
probabilities appears to depend on your whims as you define your
alternatives.
For example, if the question is:
What is the probability of life on Squoggle?
your answer would seem to be 50%. But, if the question is:
What is the probability of intelligent life on Squoggle?
your answer would *still* seem to be 50%.
This would seem to imply that the probability of unintelligent life on
Squoggle is 0%, but that would seem to contradict the answer to the
third question:
What is the probability of unintelligent life on Squoggle?
You *don't* seem to have consistency, at least as I understand it.
The problem here doesn't seem to be that you have gained any
information. You've merely asked (or been asked) several yes-no
questions, and it's led you to a set of inconsistent probabilities.
To go back to your amnesia patient answering "What's the probability
that today is Monday?" from another post. Is the answer 50% or is it
1/7? If the latter, then I think the *real* argument should be:
(a) I've identified "n" alternatives;
(b) with my given information, the "n" alternatives are equally likely;
(c) therefore, the probability of any one alternative is 1/n.
You can argue that the probability of life on Squoggle is 50% if you
like, but you've assumed (b), namely that it's equally likely that
there is or isn't life on Squoggle. This assumption would seem to
preclude you from concluding that all other sets of alternatives are
equally likely, or else you quickly arrive at contradictions, as
above.
Kevin <bu...@stat.wisc.edu>
Gerry Quinn
>ger...@indigo.ie (Gerry Quinn) writes:
>>
>> Not everyone likes this definition but it is powerful, consistent, and
>> decisive. For example, it instantly makes me a 'thirder' in the
>> Sleeping Beauty paradox.
>
>It doesn't seem consistent. That is, your assessment of the
>probabilities appears to depend on your whims as you define your
>alternatives.
>
>For example, if the question is:
>
> What is the probability of life on Squoggle?
>
>your answer would seem to be 50%. But, if the question is:
>
> What is the probability of intelligent life on Squoggle?
>
>your answer would *still* seem to be 50%.
>
>This would seem to imply that the probability of unintelligent life on
>Squoggle is 0%, but that would seem to contradict the answer to the
>third question:
>
> What is the probability of unintelligent life on Squoggle?
>
>You *don't* seem to have consistency, at least as I understand it.
>
The problem, I think, is that the 'Squoggle' example is not a pure
mathematical one. While I agree it is not entirely defensible, let me
elaborate a little to show how the problems do arise from 'knowledge'.
By "the probability of unintelligent life" you mean the probability that
there is "life" but no living organisms are more intelligent than, say,
a human (or whatever criterion you pick).
What are the chances of this? Of all the possible places that have
life, be they caves, galaxies, portions of space containing spacecraft,
what is the probability that the life has intelligence within the range
"minimal response to environment" to "language and tool use"? Might it
be zero? After all, the majority of the regions that Squoggle might be
are vast beyond comprehension. And can we say that it is likely that a
galaxy would contain life of the given intelligence range, but no
higher?
The closer we bring our argument towards the "pure uncertainty" of "an
arbitrary real" the more sensible the proposition, that there is no
chance that life exists in such a way as to precisely satisfy this
intelligence range seems. And this is much stronger if we change the
question to "What is the probability that there is life in Squoggle with
an equivalent IQ of 10, but none with an equivalent IQ of 100?" Most
would put that close to zero, I think.
>The problem here doesn't seem to be that you have gained any
>information. You've merely asked (or been asked) several yes-no
>questions, and it's led you to a set of inconsistent probabilities.
>
>To go back to your amnesia patient answering "What's the probability
>that today is Monday?" from another post. Is the answer 50% or is it
>1/7? If the latter, then I think the *real* argument should be:
>
> (a) I've identified "n" alternatives;
> (b) with my given information, the "n" alternatives are equally likely;
> (c) therefore, the probability of any one alternative is 1/n.
>
>You can argue that the probability of life on Squoggle is 50% if you
>like, but you've assumed (b), namely that it's equally likely that
>there is or isn't life on Squoggle. This assumption would seem to
>preclude you from concluding that all other sets of alternatives are
>equally likely, or else you quickly arrive at contradictions, as
>above.
Well, I would argue that the problem is that a tiny amount of
information, attached to our notions about the meaning of terms such as
"life" and "intelligence" trickles through.
I still think that if it makes sense to assign - in the absense of other
information - a probability to the statement "An arbitrary selected real
value r is greater than A" where A is a value such as 10 or googol, the
probability assigned must be 0.5.
- Gerry Quinn