1/2<->1/3
Joe Keane
explained to you and you remember it at all times.
On Sunday you're put to sleep with drugs. That night, Geddy Lee flips a
fair coin and rolls a fair die.
On Monday and Tuesday, according to the conditions below, you may be
woken up by Nurse Chapel. After this, Will Smith pulls out his penlight
and pushes the history eraser button, and you forget being woken up.
On Monday, you're woken up if and only if the die shows 1.
On Tuesday, you're woken up if and only if: the coin shows heads and the
die shows 1; or, the coin shows tails and the die shows 2.
On Wednesday, Dr. Franklin shoots you up and sends you on your way.
...
You're woken up by Nurse Chapel.
What's the chance the coin is heads?
--
Joe Keane, amateur mathematician
Dennis Ugolini
2 possible coin outcomes X 6 possible die outcomes = 12 possible outcomes.
3 result in you being awakened _at least once_:
Heads and 1
Tails and 1
Tails and 2
Thus P(heads) = 1/3.
What a lovely setup..."halfer" logic gives you a result of 1/3, and
"thirder" logic gives you a result of 1/2 (two awakenings for heads and 1).
Dennis Ugolini
ugo...@leland.stanford.edu
Public User
>
> 2 possible coin outcomes X 6 possible die outcomes = 12 possible outcomes.
>
> 3 result in you being awakened _at least once_:
>
> Heads and 1
> Tails and 1
> Tails and 2
>
> Thus P(heads) = 1/3.
>
> What a lovely setup..."halfer" logic gives you a result of 1/3, and
> "thirder" logic gives you a result of 1/2 (two awakenings for heads and 1).
What is "halfer" and "thirder" logic?
Yours truly,
Will
Wei-Hwa Huang
As I think Dennis interprets it:
"Halfer" logic -- all outcomes of the fair randomizers (coins and dice) are
equally likely. Prune those outcomes that are known to be impossible.
"Thirder" logic -- all awakenings are equally likely. Every awakening
is considered an individual outcome.
Proability of an event is the number of all outcomes with that event
divided by the number of all outcomes.
--
Wei-Hwa Huang, whu...@ugcs.caltech.edu, http://www.ugcs.caltech.edu/~whuang/
---------------------------------------------------------------------------
If all my friends jumped off a cliff... what reason is there for me to live?
David A Karr
>> On Monday, you're woken up if and only if the die shows 1.
>>
>> On Tuesday, you're woken up if and only if: the coin shows heads and the
>> die shows 1; or, the coin shows tails and the die shows 2.
>
>
>Heads and 1
>Tails and 1
>Tails and 2
>
>Thus P(heads) = 1/3.
>
>What a lovely setup..."halfer" logic gives you a result of 1/3, and
>"thirder" logic gives you a result of 1/2 (two awakenings for heads and 1).
It is even lovelier than that.
Let's add some more details to the procedure. There is a clock in the
room where you are woken that always shows the correct time of day.
If you are woken on Monday or Tuesday, it will always be at exactly
1:00 pm. Then, at 1:05, Robbie the Robot will roll into the room and
tell you (truthfully) what day it is. You will be allowed to stay
awake continuously from 1:00 to 1:10. You know and remember this
schedule of events.
Now according to halfer logic, if Robbie has told you that it is
Monday, you can eliminate Tails and 2, leaving two equiprobable
states, so P(heads) = 1/2. Similarly, if Robbie has told you that it
is Tuesday, in that case too P(heads) = 1/2.
So during any awakening, at 1:04 pm you will believe that
P(heads) = 1/3, but also at 1:04 pm you know that in one minute Robbie
will enter and tell you either, "Today is Monday," or, "Today is
Tuesday." In either case, you know that by 1:06 pm (two minutes from
now) you will believe that P(heads) = 1/2.
At 1:04 pm, then, halfer logic therefore compels you to believe for
certain that in the future you will believe P(heads) = 1/2, yet this
logic forces you to believe P(heads) = 1/3 for the time being. That
is, you are prevented from updating your probability to 1/2 even
though you know you certainly _will_ update your probability to 1/2.
And just this sort of paradoxical result has been cited as evidence
against thirdism.
--
David A. Karr "Groups of guitars are on the way out, Mr. Epstein."
ka...@shore.net --Decca executive Dick Rowe, 1962
Tom Stevenson
Dennis Ugolini <ugo...@mail.lns.cornell.edu> wrote in message
news:7j0rlm$cgp$1...@lnsnews.lns.cornell.edu...
> In article <7j0qft$ad9$1...@rocky.jgk.org>, Joe Keane <j...@jgk.org> writes:
> > You visit Dr. Evil's lab for an experiment. The following protocol is
> > explained to you and you remember it at all times.
> >
> > On Sunday you're put to sleep with drugs. That night, Geddy Lee flips a
> > fair coin and rolls a fair die.
> >
> > On Monday and Tuesday, according to the conditions below, you may be
> > woken up by Nurse Chapel. After this, Will Smith pulls out his penlight
> > and pushes the history eraser button, and you forget being woken up.
> >
> >
> > On Tuesday, you're woken up if and only if: the coin shows heads and the
> > die shows 1; or, the coin shows tails and the die shows 2.
> >
> >
> > ...
> >
> > You're woken up by Nurse Chapel.
> >
> > What's the chance the coin is heads?
> >
> > --
> > Joe Keane, amateur mathematician
>
>
>
> Heads and 1
> Tails and 1
> Tails and 2
>
> Thus P(heads) = 1/3.
>
> What a lovely setup..."halfer" logic gives you a result of 1/3, and
> "thirder" logic gives you a result of 1/2 (two awakenings for heads and
1).
>
> ugo...@leland.stanford.edu
Sorry, but that's wrong. If the outcome is heads and 1 you would be awoken
twice, Monday, and Tuesday. Therefore, the probability remains 1/2
Jamie Dreier
> It is even lovelier than that.
>
> Let's add some more details to the procedure. There is a clock in the
> room where you are woken that always shows the correct time of day.
> If you are woken on Monday or Tuesday, it will always be at exactly
> 1:00 pm. Then, at 1:05, Robbie the Robot will roll into the room and
> tell you (truthfully) what day it is. You will be allowed to stay
> awake continuously from 1:00 to 1:10. You know and remember this
> schedule of events.
>
> Now according to halfer logic, if Robbie has told you that it is
> Monday, you can eliminate Tails and 2, leaving two equiprobable
> states, so P(heads) = 1/2. Similarly, if Robbie has told you that it
> is Tuesday, in that case too P(heads) = 1/2.
But that's not what I want to say. I think that Robbie has not really
given any new information, so that I should still think that P(heads) =
1/3, no matter what he says. That is to say, I don't think the two states
are equiprobable. Is there some reason I should think they are?
> So during any awakening, at 1:04 pm you will believe that
> P(heads) = 1/3, but also at 1:04 pm you know that in one minute Robbie
> will enter and tell you either, "Today is Monday," or, "Today is
> Tuesday." In either case, you know that by 1:06 pm (two minutes from
> now) you will believe that P(heads) = 1/2.
>
> At 1:04 pm, then, halfer logic therefore compels you to believe for
> certain that in the future you will believe P(heads) = 1/2, yet this
> logic forces you to believe P(heads) = 1/3 for the time being. That
> is, you are prevented from updating your probability to 1/2 even
> though you know you certainly _will_ update your probability to 1/2.
> And just this sort of paradoxical result has been cited as evidence
> against thirdism.
It seems to me that Thirders are stuck with this alarming consequence, for
this puzzle as well as the original. Halfers manage to avoid it
consistently (it may be the motivating intuition behind Halfism, or at
least one of them).
Let me fill in my story a little. Suppose that *before* you are put to
sleep, Sandra the, uh, Cyborg shows up and tells you that you will be
awakened, if in fact you will. (Otherwise she shows up and sings you a
song.) After she has told you that you will be awakened, she tells you
either that you will be awakened on Monday, or that you will be awakened
on Tuesday, with an equal chance of telling you each of those.
If/when Sandra tells you that you will be awakened, you judge pr(heads) =
1/3. When she then tells you (say) that you will be awakened on Monday, do
you change your view to pr(heads) = 1/2? I think not! You shouldn't.
Halfers (if I may presume to speak on their behalf [unintentional but not
unnoticed]) say that this situation, before you are put to sleep, is
closely analogous to the situation you (David) have described, in which
you have been awakened. We therefore give analogous answers.
-Jamie
--
SpamGard: For real return address replace "DOT" with "."
David A Karr
>ka...@shore.net (David A Karr) wrote:
>> Now according to halfer logic, if Robbie has told you that it is
>> Monday, you can eliminate Tails and 2, leaving two equiprobable
>> states, so P(heads) = 1/2. Similarly, if Robbie has told you that it
>> is Tuesday, in that case too P(heads) = 1/2.
>
>But that's not what I want to say. I think that Robbie has not really
>given any new information,
Yeesh. If he says "Monday" then you know (for example) that the die
came up 1, which you didn't know before. How can that *not* be new
information?
>That is to say, I don't think the two states
>are equiprobable. Is there some reason I should think they are?
I was just following the argument by Denis Ugolini, in which three
states (of which Robbie eliminated one) were judged equiprobable.
Did Denis misrepresent "halfer" logic?
>If/when Sandra tells you that you will be awakened, you judge pr(heads) =
>1/3. When she then tells you (say) that you will be awakened on Monday, do
>you change your view to pr(heads) = 1/2? I think not! You shouldn't.
I'm having a very hard time imagining any reason why I shouldn't
change my view based on this extra information.
Before Sandra's announcement, my view was that:
P(Heads & I'll wake at least once) = 1/12,
P((not Heads) & I'll wake at least once) = 1/6,
P(Heads & I'll wake on Monday) = 1/12, and
P((not Heads) & I'll wake on Monday) = 1/12.
It seems trivial to compute from this that
P(Heads | I'll wake at least once) = 1/3 and
P(Heads | I'll wake on Monday) = 1/2.
Why shouldn't I believe these facts? They don't have much bearing
on what I believe during an awakening (being told by Sandra that
I'll wake at least once is not the same as actually waking), but
they seem obvious enough.
>Halfers [...] say that this situation, before you are put to sleep, is
>closely analogous to the situation you (David) have described, in which
>you have been awakened.
Actually there's a big difference between the two situations, just
as there's a big difference between asking the woman with
two children if she has any daughters, and asking her if that girl
over there is her daughter. Guess what, that detail in the setup
also determines whether certain probabilities are 1/3 or 1/2.
Jamie Dreier
> >But that's not what I want to say. I think that Robbie has not really
> >given any new information,
>
> Yeesh. If he says "Monday" then you know (for example) that the die
> came up 1, which you didn't know before. How can that *not* be new
> information?
Yeah, sorry.
It is new information. It just doesn't happen to change the probability of
heads.
> >That is to say, I don't think the two states
> >are equiprobable. Is there some reason I should think they are?
>
> I was just following the argument by Denis Ugolini, in which three
> states (of which Robbie eliminated one) were judged equiprobable.
> Did Denis misrepresent "halfer" logic?
I suppose so. I didn't read that.
I personally do not see any clear sense in assigning probabilities to
these states.
(My Sandra the Cyborg addendum omitted, on the grounds that probably
nobody except David Karr is reading this.)
> >If/when Sandra tells you that you will be awakened, you judge pr(heads) =
> >1/3. When she then tells you (say) that you will be awakened on Monday, do
> >you change your view to pr(heads) = 1/2? I think not! You shouldn't.
>
> I'm having a very hard time imagining any reason why I shouldn't
> change my view based on this extra information.
Oooh.
Then, I think you are in big trouble, the worst kind.
Suppose I know no more than you do when Sandra enters.
When (if) she tells you that you will be awakened, I am going to sell you
a ticket worth $3.01 if the coin came up Tails. I'll sell it to you for
$2.00. Good deal, you think. The chance of Tails is 2/3.
So she does tell you that you will be awakened, and you buy my ticket.
Now Sandra is about to tell you either that you will be awakened Monday,
or that you will be awakened Tuesday. (We both know this.) As soon as she
does, I plan to offer you a ticket worth $3.01 if the coin came up Heads.
I will offer it to you $1.50. I inform you of this plan. You can see that
you are going to take the offer. You are going to think it's a good deal,
because you are going to think that the chance of Heads is 1/2. I suggest
that you might like to buy the ticket now, since if you do I will give you
a discount -- you can have the ticket for $1.49. Great, you think, a
discount, this way I save a penny.
You have now spent $3.49 on a pair of tickets guaranteed to return a total
of $3.01.
As far as I can see, there are only two possibilities. Either you have
acted irrationally, or you rationally concluded that as soon as Sandra
gave you some information you were going to act irrationally, so you acted
to minimize the damage you would do to yourself. In either case we must
conclude that the updating strategy you suggest is irrational.
> Before Sandra's announcement, my view was that:
>
> P(Heads & I'll wake at least once) = 1/12,
> P((not Heads) & I'll wake at least once) = 1/6,
> P(Heads & I'll wake on Monday) = 1/12, and
> P((not Heads) & I'll wake on Monday) = 1/12.
>
> It seems trivial to compute from this that
>
> P(Heads | I'll wake at least once) = 1/3 and
> P(Heads | I'll wake on Monday) = 1/2.
>
> Why shouldn't I believe these facts?
You should believe them.
Nevertheless, what you may not do is to believe that the chance of Heads
is 1/3 when Sandra tells you that you will be awakened once, and then
update to 1/2 as soon as she tells you that you will be awakened on
Monday. (At least, not if you are also planning to update to P(Heads) =
1/2 as soon as she tells you that you will be awakened on Tuesday.)
If you do, I can make dutch book against you. A pragmatist like you ;-)
should see that this means you suffer from a nasty incoherence.
The trick, of course, is that when Sandra tells you that you will wake on
Monday, the information you receive is not the information: that you will
wake on Monday. That's why you can't conditionalize and arrive at P(Heads)
= 1/2. This in itself seems somewhat paradoxical, but you provide the key
below....
> >Halfers [...] say that this situation, before you are put to sleep, is
> >closely analogous to the situation you (David) have described, in which
> >you have been awakened.
>
> Actually there's a big difference between the two situations, just
> as there's a big difference between asking the woman with
> two children if she has any daughters, and asking her if that girl
> over there is her daughter. Guess what, that detail in the setup
> also determines whether certain probabilities are 1/3 or 1/2.
I agree that the same detail is important. I think it is important in the
same way in both situations (the post-awakening one and the Sandra one).
So I don't agree that there is a big difference between the two
situations.
gbar...@my-deja.com
While the outcome probability of the coin toss remains 1/2, the
probability of being awakened by the nurse and that the coin is heads is
still 1/3. Independence of events is the key. Think about it as if you
had 12 sleeping subjects, one for each possible outcome of the combine
coin and die toss. What is the probability that for any random subject
awakened by the nurse that the coin would be heads?
Sent via Deja.com http://www.deja.com/
Share what you know. Learn what you don't.
David A Karr
flipped and a fair six-sided die is rolled, and the person subjected
to the experiment is woken on Monday if the flip and toss are
(heads, 1) or (tails, 1) and on Tuesday if (heads, 1) or (tails, 2).
Like Beauty, the person being awakened does not initially know what
day it is nor have any memory as to whether there was a previous
awakening.
Jamie Dreier <pl43...@brownvmDOTbrown.edu> wrote:
>ka...@shore.net (David A Karr) wrote:
>> I was just following the argument by Denis Ugolini, in which three
>> states (of which Robbie eliminated one) were judged equiprobable.
>> Did Denis misrepresent "halfer" logic?
>
>I suppose so. I didn't read that.
>I personally do not see any clear sense in assigning probabilities to
>these states.
The three states were:
Heads and 1
Tails and 1
Tails and 2
If you can't assign probabilities to these states, then it seems to
me that the ability to assign probability to heads is also cast into
doubt. I actually find this view somewhat attractive. (I'd stop short
of advising Beauty to kick or shoot the interviewer, but a simple
"Stop asking stupid questions like that" might well be in order.)
>> Actually there's a big difference between the two situations, just
>> as there's a big difference between asking the woman with
>> two children if she has any daughters, and asking her if that girl
>> over there is her daughter. Guess what, that detail in the setup
>> also determines whether certain probabilities are 1/3 or 1/2.
>
>I agree that the same detail is important. I think it is important in the
>same way in both situations (the post-awakening one and the Sandra one).
>So I don't agree that there is a big difference between the two
>situations.
Ha! Hoist on my own petard! I had the answer under my nose and
overlooked it. In your setup, Sandra was volunteering some information,
not answering questions. We can't simply conditionalize on her statements
until we've established relevant facts of the form
P(Sandra says P | P),
in particular when these facts are also conditioned on the interesting
outcome. This is analogous to meeting a woman in the street who says
out of the blue that she has a daughter; do all women with daughters
act like this? And are women with daughters and sons as likely to do
it as women with daughters only?
So suppose Sandra is programmed to answer yes, no, or the name of a
day of the week in truthful response to any question whenever she can.
I begin by asking her if I'll awake. Supposing she answers yes, I
should update my view to P(heads) = 1/3.
Next suppose I were to ask her if I will awake on Monday. If she
answered yes to this as well, I believe it is then correct for me to
update my view to P(heads) = 1/2 (though perhaps I should proceed more
cautiously in case you have another of those diachronic Dutch books up
your sleeve).
But in fact what happens in your scenario is not like asking Sandra if
I'll wake on Monday; it's like asking her, "What is one day in the set
{Monday, Tuesday} on which I'll be awakened?" Now if I'm to be
awakened on Monday because (Tails, 1) occurred, Sandra _must_ answer
"Monday," but if I'm to be awakened on Monday because (Heads, 1)
occurred, then she must choose between "Monday" and "Tuesday" since
both are truthful answers. Her probability of answering "Monday" when
Monday is a true answer is not independent of the coin toss, and
the conditionalizing on it is therefore more complex. Correct
conditionalizing would arrive at P(heads) = 1/3 still.
So I'm more than ever sure that P(heads | there is an awakening) = 1/3,
and that Denis Ugolini's main claims are correct: thirder methodology
gives P(heads) = 1/2 and halfer methodology gives P(heads) = 1/3 for
the setup I described at the start of this post.
The conditionalization prior to sleeping does not seem to map to the
situation during the awakening, however. If Sandra is compelled to
tell me what day it is during each awakening, then the chance that she
will tell me "Monday" during the experiment, if I am woken on Monday,
is 1 whether the coin came up heads or tails. Of course I believe it
can occur that during _this_ awakening she will tell me it's Tuesday
even though I was also woken up on Monday, and I will adjust
accordingly; but since you don't believe in doing probabilities on
"this awakening" I don't see what use this is to you.
In my view, if Sandra is scheduled to come in during each awakening
and tell me I will be woken, then a few minutes later to tell me what
day it is, I will maintain that P(heads) = 1/2 throughout, because I
expect an equal number of awakenings on heads or tails, and an equal
number of Monday awakenings on heads or tails. If I followed your
method to the best of my current understanding, I'd believe
P(heads) = 1/3 initially, and I don't know what to do after finding
out what day it is.
woodpus...@my-deja.com
wrote:
> >> On Monday, you're woken up if and only if the
die shows 1.
> >>
> >> On Tuesday, you're woken up if and only if:
the coin shows heads and the
> >> die shows 1; or, the coin shows tails and the
die shows 2.
> >
least once_:
> >
> >Heads and 1
> >Tails and 1
> >Tails and 2
> >
> >
> >What a lovely setup..."halfer" logic gives you
a result of 1/3, and
> >"thirder" logic gives you a result of 1/2 (two
awakenings for heads and 1).
>
>
> Let's add some more details to the procedure.
There is a clock in the
> room where you are woken that always shows the
correct time of day.
> If you are woken on Monday or Tuesday, it will
always be at exactly
> 1:00 pm. Then, at 1:05, Robbie the Robot will
roll into the room and
> tell you (truthfully) what day it is. You will
be allowed to stay
> awake continuously from 1:00 to 1:10. You know
and remember this
> schedule of events.
>
told you that it is
> Monday, you can eliminate Tails and 2, leaving
two equiprobable
> states, so P(heads) = 1/2. Similarly, if Robbie
has told you that it
> is Tuesday, in that case too P(heads) = 1/2.
>
believe that
> P(heads) = 1/3, but also at 1:04 pm you know
that in one minute Robbie
> will enter and tell you either, "Today is
Monday," or, "Today is
> Tuesday." In either case, you know that by 1:06
pm (two minutes from
> now) you will believe that P(heads) = 1/2.
>
> At 1:04 pm, then, halfer logic therefore compels
you to believe for
> certain that in the future you will believe
P(heads) = 1/2, yet this
the time being. That
> is, you are prevented from updating your
probability to 1/2 even
> though you know you certainly _will_ update your
probability to 1/2.
> And just this sort of paradoxical result has
been cited as evidence
> against thirdism.
Whether you get a contradiction depends on what
halfer logic you use. Some halfers say that
the probabilities are as follows.
P(Heads,Mon) = 1/2.
P(Tails,Mon) = 1/4.
P(Tails,Tue) = 1/4.
Using this logic on the above scenario we get
P(Tails,2,Tue) = 1/3.
P(Tails,1,Mon) = 1/3.
P(Heads,1,Mon) = 1/6.
P(Heads,1,Tue) = 1/6.
This implies P(Heads) = 1/3.
Now when Robbie the Robot tells you the day,
you can eliminate the appropriate states and
conditionalize using Bayes Theorem to get
P(Heads|Mon) = 1/3 and P(Heads|Tue) = 1/3.
There is no inconsistency here.
Furthermore, if you use this same logic on the
original Sleeping Beauty problem, you don't get
a problem due to the differing probabilities
on Monday and Tuesday. This logic implies that
P(Heads) = 1/2, P(Heads|Mon) = 2/3,
P(Heads|Tue) = 0. P(Heads) is between the
other two values just like they should be.
woodpus...@my-deja.com
woodpus...@my-deja.com
use. In the original Sleeping Beauty problem, some halfers say
that the following are the probabilities.
P(Heads,Mon) = 1/2
P(Tails,Mon) = 1/4
P(Tails,Tue) = 1/4
Using this logic on the above scenario, we get the following
probabilities.
P(Tails,2,Tue) = 1/3
P(Tails,1,Mon) = 1/3
P(Heads,1,Mon) = 1/6
P(Heads,1,Tue) = 1/6
These probabilities imply that P(Heads) = 1/3. Now when Robbie the
Robot tells you what day it is, you can eliminate the appropriate
states and conditionalize using Bayes theorem to get that
P(Heads|Mon) = 1/3 and P(Heads|Tue) = 1/3. There is no inconsistency
here.
Furthermore, if you apply this same logic to the original Sleeping
Beauty problem, there is no problem with the differing probabilities
on Monday and Tuesday. The probabilities are P(Heads) = 1/2,
P(Heads|Mon) = 2/3, and P(Heads|Tue) = 0. P(Heads) is between the