Was Sleeping Beauty wrongly convicted?
Nick Wedd
with rather small numbers, and one with very large numbers. Here I give
a form of the problem with moderate numbers, which may be easier to
understand intuitively.
You were present when a crime was committed, but you took no part in it.
You have been arrested, and are tried for this crime.
After you have given your evidence, the judge sentences you (this
country has an unusual judicial system). Her sentence is "You will now
be taken from this court and locked in a prison cell. Meanwhile the
trial will continue and we will hear the evidence of the witnesses. If
we find you not guilty, you will be released from prison at 10 p.m.
tomorrow - this short sentence will be for your failure to prevent or
report the crime. But if we find you guilty, you will be visited at 10
p.m. tomorrow by someone who will administer a soporific and 24-hour
memory-wipe. This treatment will repeated until you have passed 100
nights in prison, and then you will be released at 10 p.m.
Psychologists advise us that this will have a reformatory effect on
you."
You know that you are innocent of the crime. But there is a chance that
the witnesses will lie and the court will reach an unjust verdict - you
estimate this chance as 20%.
Some time later, you wake up in prison. How do you rate your chance of
being released the same evening?
As a thirder, I know that the answer is 5/104. But my gut feeling is
that it is .802 (ok, my guts aren't that precise - let's say somewhere
around 0.8). I am finding it hard to reconcile this at present.
Nick
--
Nick Wedd ni...@maproom.co.uk
Michael Will
>Some time later, you wake up in prison. How do you rate your chance of
>being released the same evening?
Hmm, I know that I would stay up past 10PM the first night. And even
if deprived of 24 hours of memory I'll still know this since my part
of the trial was more than 24 hours prior to the first possible drugging.
(The details of the sentencing make it obvious to stay up, as seen below.)
If I wake up and remember going to sleep the night before, I know it's
decision day and I place the odds at 80%. If I don't remember being
up past 10, I know I was convicted and had 24 hours of memory wiped,
so I then place the odds at 1%.
- Michael
Nick Wedd
<en...@cc.newcastle.edu.au> writes
>You are either found innocent (80%) and will be released that night,
>or guilty (20%). If guilty then you stay for 100 nights, so there is a
>1% chance you will be released that night. So the chance that you
>will be released is 80% plus 20% * 1% or 80.2% .
This is the view held by most halfers.
>> As a thirder, I know that the answer is 5/104. But my gut feeling is
>> that it is .802 (ok, my guts aren't that precise - let's say somewhere
>> around 0.8). I am finding it hard to reconcile this at present.
>
>Where did you get 5/104 from?
5/104 is derived as follows.
With p=0.8, you will wake up in prison once. With p=0.2, you will wake
up in prison 100 times. So the total expected number of prison-
awakenings is 0.8+100*0.2 = 20.8. Of these, 0.8+0.2 = 1 is a final
prison-awakening. So the probability that this is a final one is 1/20.8
= 5/104.
> As far as I can see both this and the
>previous post I answered in this thread are very simple problems
>in statistics. How on earth do they generate so much posting?
Because many people, "halfers", argue as you do. Others, "thirders",
argue as I do. There has been one conversion from the halfer to the
thirder viewpoint. The halfer viewpoint has more immediate intuitive
appeal.
James Smith
> You were present when a crime was committed, but you took no part in it.
> You have been arrested, and are tried for this crime.
>
> After you have given your evidence, the judge sentences you (this
> country has an unusual judicial system). Her sentence is "You will now
> be taken from this court and locked in a prison cell. Meanwhile the
> trial will continue and we will hear the evidence of the witnesses. If
> we find you not guilty, you will be released from prison at 10 p.m.
> tomorrow - this short sentence will be for your failure to prevent or
> report the crime. But if we find you guilty, you will be visited at 10
> p.m. tomorrow by someone who will administer a soporific and 24-hour
> memory-wipe. This treatment will repeated until you have passed 100
> nights in prison, and then you will be released at 10 p.m.
> Psychologists advise us that this will have a reformatory effect on
> you."
>
> You know that you are innocent of the crime. But there is a chance that
> the witnesses will lie and the court will reach an unjust verdict - you
> estimate this chance as 20%.
>
> Some time later, you wake up in prison. How do you rate your chance of
> being released the same evening?
You are either found innocent (80%) and will be released that night,
or guilty (20%). If guilty then you stay for 100 nights, so there is a
1% chance you will be released that night. So the chance that you
will be released is 80% plus 20% * 1% or 80.2% .
> As a thirder, I know that the answer is 5/104. But my gut feeling is
> that it is .802 (ok, my guts aren't that precise - let's say somewhere
> around 0.8). I am finding it hard to reconcile this at present.
Where did you get 5/104 from? As far as I can see both this and the
previous post I answered in this thread are very simple problems
in statistics. How on earth do they generate so much posting?
Jim
Nick Wedd
<mich...@tampabay.rr.com> writes
>Nick Wedd wrote:
>
>>Some time later, you wake up in prison. How do you rate your chance of
>>being released the same evening?
>
>
>if deprived of 24 hours of memory I'll still know this since my part
>of the trial was more than 24 hours prior to the first possible drugging.
>(The details of the sentencing make it obvious to stay up, as seen below.)
>
>If I wake up and remember going to sleep the night before, I know it's
>decision day and I place the odds at 80%. If I don't remember being
>up past 10, I know I was convicted and had 24 hours of memory wiped,
>so I then place the odds at 1%.
You are right. I specified the problem badly.
Let me try again. On the first evening, at 10 p.m., they administer a
soporific and a one-hour memory wipe. On subsequent evenings, they
administer a soporific and a 25-hour memory wipe. Does that work
better?
Michael Will
>You are right. I specified the problem badly.
Did you? I wasn't picking on you in particular, but this entire
theme continues to introduce real-world situations but expects
only pure math to find the answer. Perhaps you specified an
intended problem badly, but I liked the one I could answer.
I even had choices - I could have also announced that I would
never cut my hair/fingernails and would use those effects to
estimate my odds of release in the case of a guilty verdict.
Even in the legendary three door puzzle, concern was given in
this group to the motives, if any, of the host. Many if not most
puzzles submitted here allow or encourge lateral thinking. To
keep putting Beauty (I may start calling her Pauline now) into
various life-changing situations but ignoring exploitation of
real-world effects is getting old for my tastes.
(Although the clone bit was nice, but I got disallowed from that
one by trying to get a clone to send a message in a bottle or
build a raft to get word back to the prisoner.)
>Let me try again. On the first evening, at 10 p.m., they administer a
>soporific and a one-hour memory wipe. On subsequent evenings, they
>administer a soporific and a 25-hour memory wipe. Does that work
>better?
Well, that seems to put it back into a purely math-based form
again. If general puzzle-solving skills aren't to be used, perhaps
these questions can be better argued in a math group. Again,
nothing personal. Really. But how do the real-world details matter
if only one (expected) answer is correct?
Where the heck has Waldo gone? Is he sleeping too? There was a
man who got involved in the real world. What are the odds he
would respond positively if I woke him up at 4AM to ask him?
- Michael
Nick Wedd
<mich...@tampabay.rr.com> writes
>Nick Wedd wrote:
>
>>You are right. I specified the problem badly.
>
>Did you? I wasn't picking on you in particular, but this entire
>theme continues to introduce real-world situations but expects
>only pure math to find the answer. Perhaps you specified an
>intended problem badly, but I liked the one I could answer.
I am glad that you enjoyed the problem which I accidentally specified.
>I even had choices - I could have also announced that I would
>never cut my hair/fingernails and would use those effects to
>estimate my odds of release in the case of a guilty verdict.
>
>Even in the legendary three door puzzle, concern was given in
>this group to the motives, if any, of the host.
Only because many of the people who posted that problem specified it
inadequately.
>Many if not most
>puzzles submitted here allow or encourge lateral thinking. To
>keep putting Beauty (I may start calling her Pauline now) into
>various life-changing situations but ignoring exploitation of
>real-world effects is getting old for my tastes.
>
>(Although the clone bit was nice, but I got disallowed from that
> one by trying to get a clone to send a message in a bottle or
> build a raft to get word back to the prisoner.)
>
>>Let me try again. On the first evening, at 10 p.m., they administer a
>>soporific and a one-hour memory wipe. On subsequent evenings, they
>>administer a soporific and a 25-hour memory wipe. Does that work
>>better?
>
>Well, that seems to put it back into a purely math-based form
>again. If general puzzle-solving skills aren't to be used, perhaps
>these questions can be better argued in a math group. Again,
>nothing personal. Really. But how do the real-world details matter
>if only one (expected) answer is correct?
I am glad that I have got it right this time - at least, I hope I have.
I want to keep real-world details out of this problem, because it is
hard enough without them. You are probably right that it would be
better argued in a math group. But some of the leading arguers on both
sides are regular of this group, so it may be hard to shift it.
>Where the heck has Waldo gone? Is he sleeping too? There was a
>man who got involved in the real world. What are the odds he
>would respond positively if I woke him up at 4AM to ask him?
ObRealWorldPuzzle:
Here in England, car number plates carry year letters which indicate the
age of the car. I look out for these, and am interested in any car
which is around 30-40 years old. Surprisingly often, such cars are a
very pale blue color. Why is this?
Here are some hypotheses, none of them convincing.
1. In the 60s, most new cars were very pale blue.
2. All cars tend to go that color as they age.
3. People who drive very carefully like that colour.
4. Rust shows up particularly well against a pale blue background, so
cars of that color tend to receive better maintenance.
5. I have an unjustified belief that old cars tend to be very pale
blue, and only look out for such cars.
Michael Will
>I am glad that I have got it right this time - at least, I hope I have.
>I want to keep real-world details out of this problem, because it is
>hard enough without them. You are probably right that it would be
>better argued in a math group. But some of the leading arguers on both
>sides are regular of this group, so it may be hard to shift it.
Understood, even if the puzzle at times appears to be how many ways
people can express the same concept(s) without reaching consensus.
>ObRealWorldPuzzle:
>
>Here in England, car number plates carry year letters which indicate the
>age of the car. I look out for these, and am interested in any car
>which is around 30-40 years old. Surprisingly often, such cars are a
>very pale blue color. Why is this?
Touche. I am now doomed to membership in the hypothesis five club
until I can come up with a better explanation. I just suddenly realized
the old car that was parked next door until last year was, well, pale
blue. I fully expect to see the roads teeming with them now.
- Michael
Patrick Hamlyn
>ObRealWorldPuzzle:
>
>Here in England, car number plates carry year letters which indicate the
>age of the car. I look out for these, and am interested in any car
>which is around 30-40 years old. Surprisingly often, such cars are a
>very pale blue color. Why is this?
>
>Here are some hypotheses, none of them convincing.
>
>1. In the 60s, most new cars were very pale blue.
>2. All cars tend to go that color as they age.
>3. People who drive very carefully like that colour.
>4. Rust shows up particularly well against a pale blue background, so
>cars of that color tend to receive better maintenance.
>5. I have an unjustified belief that old cars tend to be very pale
>blue, and only look out for such cars.
How's this: All colours containing red fade much quicker than other colours. By
the time a car has got to 30-40 years old, if it had any red in it to start
with, the owner has become unhappy with the result and ditched it or re-sprayed
it. Only originally blue or similar colours still remain original - though
somewhat faded, they are still reasonable-looking colours, and the owners are
still happy with them.
James Smith
> >Where did you get 5/104 from?
>
> 5/104 is derived as follows.
>
> With p=0.8, you will wake up in prison once. With p=0.2, you will wake
> up in prison 100 times. So the total expected number of prison-
> awakenings is 0.8+100*0.2 = 20.8. Of these, 0.8+0.2 = 1 is a final
> prison-awakening. So the probability that this is a final one is 1/20.8
> = 5/104.
Your mathematics is wrong. 20.8 and 1 are the averages of probability
functions. You are saying that the average of the function obtained by
dividing those two functions is the same as the result of dividing the
two averages. It is only so if the two functions are constant, and here
they are clearly not. You could add or subtract the averages, but you
can't divide them.
Jim
Wei-Hwa Huang
>As a thirder, I know that the answer is 5/104. But my gut feeling is
>that it is .802 (ok, my guts aren't that precise - let's say somewhere
>around 0.8). I am finding it hard to reconcile this at present.
Perhaps because you gut realizes that there are a whole slew of potential
awakenings that you won't remember anyway, so it ignores them?
--
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?
Matthew T. Russotto
Wei-Hwa Huang <whu...@ugcs.caltech.edu> wrote:
}Nick Wedd <Ni...@maproom.co.uk> writes:
}>As a thirder, I know that the answer is 5/104. But my gut feeling is
}>that it is .802 (ok, my guts aren't that precise - let's say somewhere
}>around 0.8). I am finding it hard to reconcile this at present.
}
}Perhaps because you gut realizes that there are a whole slew of potential
}awakenings that you won't remember anyway, so it ignores them?
In that case, your gut should answer "100%"
--
Matthew T. Russotto russ...@pond.com
"Extremism in defense of liberty is no vice, and moderation in pursuit
of justice is no virtue."
John Rickard
: Nick Wedd <Ni...@maproom.co.uk> writes:
: >As a thirder, I know that the answer is 5/104. But my gut feeling is
: >that it is .802 (ok, my guts aren't that precise - let's say somewhere
: >around 0.8). I am finding it hard to reconcile this at present.
:
: Perhaps because you gut realizes that there are a whole slew of potential
: awakenings that you won't remember anyway, so it ignores them?
Indeed. The probability that you assign to being released this
evening is (as also in the lottery variant) unimportant, since there
is nothing you can use it for. (It's presumably no good even trying
to prepare your mind to accept staying in prison, since the effects of
that will be cancelled by the memory wipe.)
You might as well *assume* that you'll be released this evening and
use your time profitably planning what you'll do after the release, or
doing some other thinking that will be of value to you if your memory
is not wiped.
--
John Rickard <John.R...@virata.com>
Nick Wedd
<en...@cc.newcastle.edu.au> writes
>Nick Wedd wrote:
>> With p=0.8, you will wake up in prison once. With p=0.2, you will wake
>> up in prison 100 times. So the total expected number of prison-
>> awakenings is 0.8+100*0.2 = 20.8. Of these, 0.8+0.2 = 1 is a final
>> prison-awakening. So the probability that this is a final one is 1/20.8
>> = 5/104.
>
>Your mathematics is wrong. 20.8 and 1 are the averages of probability
>functions. You are saying that the average of the function obtained by
>dividing those two functions is the same as the result of dividing the
>two averages. It is only so if the two functions are constant, and here
>they are clearly not. You could add or subtract the averages, but you
>can't divide them.
You have a point. But the 0.8+0.2=1 isn't really an average. It is a
certainty. There will definitely be exactly one final prison-awakening.
It is acceptable to divide an average by a constant.
Here is a version of the Sleeping Beauty problem to which I expect most
people to give the thirder answer:
My name is on the mailing list of Spam Insurance. Every few months, it
comes to the top of the list, and the salesman tosses a coin. On heads,
he sends me a letter offering 40% off life insurance. On tails, he
sends two letters, posted a few days apart, each offering 20% off life
insurance.
I frequently receive letters selling life insurance, from many different
companies. Whenever I receive one, I put it straight in the rubbish bin
unopened, and immediately forget about it [this part is far more
plausible than the "memory-wiping drug" thing].
One day, my telephone line blows down in a storm. Without the use of
telephone or internet, I become extremely bored. Looking for something
to do, I extract the top item from my bin, and open it. It is a letter
from Spam Insurance. What is the probability that the offered discount
is 40%?
James Smith
> >> With p=0.8, you will wake up in prison once. With p=0.2, you will wake
> >> up in prison 100 times. So the total expected number of prison-
> >> awakenings is 0.8+100*0.2 = 20.8. Of these, 0.8+0.2 = 1 is a final
> >> prison-awakening. So the probability that this is a final one is 1/20.8
> >> = 5/104.
> >
> >Your mathematics is wrong. 20.8 and 1 are the averages of probability
> >functions. You are saying that the average of the function obtained by
> >dividing those two functions is the same as the result of dividing the
> >two averages. It is only so if the two functions are constant, and here
> >they are clearly not. You could add or subtract the averages, but you
> >can't divide them.
>
> You have a point. But the 0.8+0.2=1 isn't really an average. It is a
> certainty. There will definitely be exactly one final prison-awakening.
> It is acceptable to divide an average by a constant.
No, it is a function that has a value at two points and is zero
everywhere
else. The two points are at 1 and 100, the point at which the average
lies
just happens to fall at 1. (By the definition, of course, you expect
this.)
> Here is a version of the Sleeping Beauty problem to which I expect most
> people to give the thirder answer:
>
> My name is on the mailing list of Spam Insurance. Every few months, it
> comes to the top of the list, and the salesman tosses a coin. On heads,
> he sends me a letter offering 40% off life insurance. On tails, he
> sends two letters, posted a few days apart, each offering 20% off life
> insurance.
>
> I frequently receive letters selling life insurance, from many different
> companies. Whenever I receive one, I put it straight in the rubbish bin
> unopened, and immediately forget about it [this part is far more
> plausible than the "memory-wiping drug" thing].
>
> One day, my telephone line blows down in a storm. Without the use of
> telephone or internet, I become extremely bored. Looking for something
> to do, I extract the top item from my bin, and open it. It is a letter
> from Spam Insurance. What is the probability that the offered discount
> is 40%?
'A few days' isn't exact, so I can't give exact probabilities, but it
should
still be obvious how to get the answer, which is the way I used in the
previous question. Assuming a few months is a lot longer period than
a few days, the answer here would be about a 50% chance of either.
When you can do something two ways, and you get two different answers,
and there is an obvious flaw in one method, and that method also returns
what is obviously a wrong answer, surely it is not hard to see which
method must be wrong. :-)
Jim
Wei-Hwa Huang
>Here is a version of the Sleeping Beauty problem to which I expect most
>people to give the thirder answer:
>My name is on the mailing list of Spam Insurance. Every few months, it
>comes to the top of the list, and the salesman tosses a coin. On heads,
>he sends me a letter offering 40% off life insurance. On tails, he
>sends two letters, posted a few days apart, each offering 20% off life
>insurance.
>I frequently receive letters selling life insurance, from many different
>companies. Whenever I receive one, I put it straight in the rubbish bin
>unopened, and immediately forget about it [this part is far more
>plausible than the "memory-wiping drug" thing].
>One day, my telephone line blows down in a storm. Without the use of
>telephone or internet, I become extremely bored. Looking for something
>to do, I extract the top item from my bin, and open it. It is a letter
>from Spam Insurance. What is the probability that the offered discount
>is 40%?
It sure SEEMS like it should be 1/3, since 1/3 of the Spam insurance
envelopes you get are 40%.
But remember the question about the guy with two girlfriends in two
towns who decided which one to go to by hopping on whichever 10-minute
train came by first?
In this case, suppose that this guy sends you the first envelope on
the first of every month, then sends you the second envelope, if there is
one, on the fifteenth of every month. Then in all of the months, only
in half of them is the most recent Spam Insurance envelope 40%, so the
probability should be 1/2.
There is another factor. You mentioned you receive a lot of envelopes,
and you usually trash them quickly. And you picked the top item off
your bin -- which may not have been a Spam Insurance envelope. For
instance, if you get one piece of insurance junk mail every day, then
(using the previous scenarios) we know that it must be the first or
fifteenth of the month. If we assume that all days are equally probable,
then again the probability is 1/3.
But if you were going to dig down to the first Spam Insurance envelope
you saw (and it just happened to be the first) then the probability
becomes 1/2 again.
So, in my conclusion, the answer depends on whether we're saying "all
days have equal probability regardless of their aggregation" or "days
are only division points and the actual probability is based on the
continuous ranges between the days." Both interpretations are valid
but give different answers.
-----
Consider the girlfriend problem again. A person hops on the first
train he sees, normally -- it can be going to A (every ten minutes on
the 6s) or going to B (every ten minutes on the 7s).
The person goes to the train station, immediately sees a train,
and gets on it. While riding, he idly wonders: "What is the probability
that I am going to A?"
Answer 1: "For every ten-minute interval, I get on the A train only
in that one-minute interval, but I get on the B train in the nine-minute
interval. Obviously I am more likely to ride the B train on any given
day, and the probability that this is the A train is 1/10."
Answer 2: "But wait, today I didn't wait at the station -- there was a
train there immediately! So I can discount all those intervals where
I'm waiting 5 minutes for the B train, etc. And since there are an
equal number of A trains and B trains, the probability that this is
an A train must be 1/2."
Matt McLelland
> In this case, suppose that this guy sends you the first envelope on
> the first of every month, then sends you the second envelope, if there is
> one, on the fifteenth of every month. Then in all of the months, only
> in half of them is the most recent Spam Insurance envelope 40%, so the
> probability should be 1/2.
Your dispute with Nick's scenario is based on an assumption that he checks his
mail once a month? I think his problem was phrased well enough to give the
intended answer of 1/3. Most people check their mail on every mail day, and I
am willing to assume Nick is one of them. There could be a question of "a few
days" being short enough to straddle a holiday weekend, but lets just pretend
that there isn't. =)
> But if you were going to dig down to the first Spam Insurance envelope
> you saw (and it just happened to be the first) then the probability
> becomes 1/2 again.
Only if he checks his mail once a month..
> -----
>
> Consider the girlfriend problem again. A person hops on the first
> train he sees, normally -- it can be going to A (every ten minutes on
> the 6s) or going to B (every ten minutes on the 7s).
> The person goes to the train station, immediately sees a train,
> and gets on it. While riding, he idly wonders: "What is the probability
> that I am going to A?"
>
> Answer 1: "For every ten-minute interval, I get on the A train only
> in that one-minute interval, but I get on the B train in the nine-minute
> interval. Obviously I am more likely to ride the B train on any given
> day, and the probability that this is the A train is 1/10."
>
> Answer 2: "But wait, today I didn't wait at the station -- there was a
> train there immediately! So I can discount all those intervals where
> I'm waiting 5 minutes for the B train, etc. And since there are an
> equal number of A trains and B trains, the probability that this is
> an A train must be 1/2."
In this case, answer 2 is correct, but this isn't same problem. If you want
this situation to be analogous to the mail situation, then have two trains going
to point A and one to point B. Assume that their schedules have the same
period, and offset them however you like such that two trains never come within
the same minute. Now show up at the train station, find a train waiting for
you, and compute the probabilities.
Wei-Hwa Huang
>Wei-Hwa Huang <whu...@ugcs.caltech.edu> wrote:
>}Nick Wedd <Ni...@maproom.co.uk> writes:
>}>As a thirder, I know that the answer is 5/104. But my gut feeling is
>}>that it is .802 (ok, my guts aren't that precise - let's say somewhere
>}>around 0.8). I am finding it hard to reconcile this at present.
>}
>}Perhaps because you gut realizes that there are a whole slew of potential
>}awakenings that you won't remember anyway, so it ignores them?
>In that case, your gut should answer "100%"
No; there are two potential awakenings that you will remember, weighted
4:1 -- which is why your gut answers 80%.
Matthew T. Russotto
Wei-Hwa Huang <whu...@ugcs.caltech.edu> wrote:
}russ...@wanda.vf.pond.com (Matthew T. Russotto) writes:
}>Wei-Hwa Huang <whu...@ugcs.caltech.edu> wrote:
}>}Nick Wedd <Ni...@maproom.co.uk> writes:
}>}>As a thirder, I know that the answer is 5/104. But my gut feeling is
}>}>that it is .802 (ok, my guts aren't that precise - let's say somewhere
}>}>around 0.8). I am finding it hard to reconcile this at present.
}>}
}>}Perhaps because you gut realizes that there are a whole slew of potential
}>}awakenings that you won't remember anyway, so it ignores them?
}
}>In that case, your gut should answer "100%"
}
}No; there are two potential awakenings that you will remember, weighted
}4:1 -- which is why your gut answers 80%.
There are two potential awakenings you will remember, but in both, the
answer to the question "Will I be released from prison today" is
"yes".
Nick Wedd
<whu...@ugcs.caltech.edu> writes
>There is another factor. You mentioned you receive a lot of envelopes,
>and you usually trash them quickly.
Not "usually", "always". "Whenever I receive one, I put it straight in
the rubbish bin"
> And you picked the top item off
>your bin -- which may not have been a Spam Insurance envelope. For
>instance, if you get one piece of insurance junk mail every day, then
>(using the previous scenarios) we know that it must be the first or
>fifteenth of the month. If we assume that all days are equally probable,
>then again the probability is 1/3.
>
>But if you were going to dig down to the first Spam Insurance envelope
>you saw (and it just happened to be the first) then the probability
>becomes 1/2 again.
I had no intention of digging down. "Looking for something to do, I
extract the top item from my bin, and open it. It is a letter from Spam
Insurance."
>So, in my conclusion, the answer depends on whether we're saying "all
>days have equal probability regardless of their aggregation" or "days
>are only division points and the actual probability is based on the
>continuous ranges between the days." Both interpretations are valid
>but give different answers.
I thought that I had stated the problem in a way that only allowed one
interpretation.
>Consider the girlfriend problem again. A person hops on the first
>train he sees, normally -- it can be going to A (every ten minutes on
>the 6s) or going to B (every ten minutes on the 7s).
>The person goes to the train station, immediately sees a train,
>and gets on it. While riding, he idly wonders: "What is the probability
>that I am going to A?"
>
>Answer 1: "For every ten-minute interval, I get on the A train only
> in that one-minute interval, but I get on the B train in the nine-minute
> interval. Obviously I am more likely to ride the B train on any given
> day, and the probability that this is the A train is 1/10."
>
>Answer 2: "But wait, today I didn't wait at the station -- there was a
> train there immediately! So I can discount all those intervals where
> I'm waiting 5 minutes for the B train, etc. And since there are an
> equal number of A trains and B trains, the probability that this is
> an A train must be 1/2."
2 is the better answer - it makes use of the available information that
the wait on the platform was less than one minute.
Nick Wedd
>> Here is a version of the Sleeping Beauty problem to which I expect most
>> people to give the thirder answer:
>>
>> My name is on the mailing list of Spam Insurance. Every few months, it
>> comes to the top of the list, and the salesman tosses a coin. On heads,
>> he sends me a letter offering 40% off life insurance. On tails, he
>> sends two letters, posted a few days apart, each offering 20% off life
>> insurance.
>>
>> I frequently receive letters selling life insurance, from many different
>> companies. Whenever I receive one, I put it straight in the rubbish bin
>> unopened, and immediately forget about it [this part is far more
>> plausible than the "memory-wiping drug" thing].
>>
>> One day, my telephone line blows down in a storm. Without the use of
>> telephone or internet, I become extremely bored. Looking for something
>> to do, I extract the top item from my bin, and open it. It is a letter
>> from Spam Insurance. What is the probability that the offered discount
>> is 40%?
>
>'A few days' isn't exact, so I can't give exact probabilities, but it
>should
>still be obvious how to get the answer, which is the way I used in the
>previous question. Assuming a few months is a lot longer period than
>a few days, the answer here would be about a 50% chance of either.
OK, I'll be more precise. "A few days" is four working days. The
salesman works 240 days a year. My name comes to the top of his file
every 120 working days.
None of this makes any difference to my answer, which is 1/3.
>When you can do something two ways, and you get two different answers,
>and there is an obvious flaw in one method, and that method also returns
>what is obviously a wrong answer, surely it is not hard to see which
>method must be wrong. :-)
My view precisely. However I would not describe the flaw in your method
as obvious :-)
David A Karr
>>But if you were going to dig down to the first Spam Insurance envelope
>>you saw (and it just happened to be the first) then the probability
>>becomes 1/2 again.
>
>extract the top item from my bin, and open it. It is a letter from Spam
>Insurance."
>I thought that I had stated the problem in a way that only allowed one
>interpretation.
In my opinion, you did. The difficulty lies in comparing your problem
with Sleeping Beauty.
You mentioned that the memory lapse in your problem is more realistic
than in Sleeping Beauty, but in fact there's no real need for a memory
lapse at all. You need never have identified any mail from Spam
Insurance as being from Spam Insurance; you need merely have
identified it as junk mail. Perhaps you regard _all_ mail as junk
mail (this saves a lot of trouble when it comes time to pay the bills).
So instead of relying on a bizarre memory defect to unbalance your
probabilities, you rely on a setup that isn't guaranteed to produce
the described scenario even once in your lifetime. (Perhaps every time
you get bored, the top item in your wastebasket is from a company other
than Spam Insurance.) In contrast, prior even to beginning the
procedure, Beauty is _guaranteed_ to wake up in an interview.
So, unfortunately (and I do mean unfortunately, because a resolution
to Sleeping Beauty would be nice to reach), I don't think you've
addressed the Sleeping Beauty problem here.
--
David A. Karr "Groups of guitars are on the way out, Mr. Epstein."
ka...@shore.net --Decca executive Dick Rowe, 1962
Wei-Hwa Huang
I think you misread the problem, Matt. Nick throws away all mail immediately
into his rubbish bin. On this particular bored day, he is looking at the
top piece of mail in his rubbish bin, not mail he has gotten today.
Matt McLelland <mat...@flash.net> writes:
>Wei-Hwa Huang wrote:
>> In this case, suppose that this guy sends you the first envelope on
>> the first of every month, then sends you the second envelope, if there is
>> one, on the fifteenth of every month. Then in all of the months, only
>> in half of them is the most recent Spam Insurance envelope 40%, so the
>> probability should be 1/2.
>Your dispute with Nick's scenario is based on an assumption that he checks his
>mail once a month? I think his problem was phrased well enough to give the
>intended answer of 1/3. Most people check their mail on every mail day, and I
>am willing to assume Nick is one of them. There could be a question of "a few
>days" being short enough to straddle a holiday weekend, but lets just pretend
>that there isn't. =)
>> But if you were going to dig down to the first Spam Insurance envelope
>> you saw (and it just happened to be the first) then the probability
>> becomes 1/2 again.
>Only if he checks his mail once a month..
>> -----
>>
>> Consider the girlfriend problem again. A person hops on the first
>> train he sees, normally -- it can be going to A (every ten minutes on
>> the 6s) or going to B (every ten minutes on the 7s).
>> The person goes to the train station, immediately sees a train,
>> and gets on it. While riding, he idly wonders: "What is the probability
>> that I am going to A?"
>>
>> Answer 1: "For every ten-minute interval, I get on the A train only
>> in that one-minute interval, but I get on the B train in the nine-minute
>> interval. Obviously I am more likely to ride the B train on any given
>> day, and the probability that this is the A train is 1/10."
>>
>> Answer 2: "But wait, today I didn't wait at the station -- there was a
>> train there immediately! So I can discount all those intervals where
>> I'm waiting 5 minutes for the B train, etc. And since there are an
>> equal number of A trains and B trains, the probability that this is
>> an A train must be 1/2."
>In this case, answer 2 is correct, but this isn't same problem. If you want
>this situation to be analogous to the mail situation, then have two trains going
>to point A and one to point B. Assume that their schedules have the same
>period, and offset them however you like such that two trains never come within
>the same minute. Now show up at the train station, find a train waiting for
>you, and compute the probabilities.
--
Matt McLelland
> I think you misread the problem, Matt. Nick throws away all mail immediately
> into his rubbish bin. On this particular bored day, he is looking at the
> top piece of mail in his rubbish bin, not mail he has gotten today.
Whoops. When he said "into my bin", I interpreted it as "mail bin". I don't have a
"rubbish bin", I have a trash can. =)
Nevertheless, I still disagree with your claim that
> >> But if you were going to dig down to the first Spam Insurance envelope
> >> you saw (and it just happened to be the first) then the probability
> >> becomes 1/2 again.
Lets suppose that he has a bottomless rubbish bin which never gets emptied. He
throws away all kinds of junk mail every day. On a random day he gets bored and
decides to dig until he finds something from Spam Insurance. Let D be the depth at
which he finds such a letter. What is P[ 40% | D=1]? It is 1/3.
Nick Wedd
<j...@255.255.255.192> writes
>Indeed. The probability that you assign to being released this
>evening is (as also in the lottery variant) unimportant, since there
>is nothing you can use it for. (It's presumably no good even trying
>to prepare your mind to accept staying in prison, since the effects of
>that will be cancelled by the memory wipe.)
>
>You might as well *assume* that you'll be released this evening and
>use your time profitably planning what you'll do after the release, or
>doing some other thinking that will be of value to you if your memory
>is not wiped.
You are saying that when I wake up in prison, my actions and beliefs on
this day are irrelevant if I am to be memory-wiped.
I think I can get round that.
You wake up in prison, as described above. After breakfast, you are
visited by the Mad Vicar. This well-known character was once ordained
as a priest, and so is allowed to visit prisons. But his interest is in
a form of charitable gambling. He likes to offer deals to prisoners,
and visits prisons for this purpose. He visits them in an unpredictable
order, and never visits the same prison twice within one year. He is
totally honest and trustworthy. There is no-one else like him. All
these facts are known to you.
You ask the Mad Vicar whether you were found guilty. He says that he
does not know. He then asks you to guess whether you were found guilty:
if you guess correctly, he will pay $100 into your bank account. Of
course, you won't know whether you have won until you are released. In
fact you may not even remember the deal; but the $100 will be most
welcome.
How should you guess?
Matt McLelland
This is like my original Betty and Beauty scenario. Let's suppose that you
don't know how you should guess (you followed this thread and are utterly
confused). So, you *ask him* how you should bet. He, having found you in
the prison on a random day, will conditionalize and feel that you are
probably guilty. But, lets suppose that he is a halfer. As a halfer, he
knows that he would think himself probably innocent, were he in your shoes.
So how should he answer? Maybe he says "Well, I think that you are probably
guilty, but if I were in your position, I would think that I was probably
innocent." Now lets suppose for a moment that you don't find his bizarre
answer particularly helpful and so insist "Look. I have told you everything
I know, now give me an answer. WHAT SHOULD I CHOOSE?"
Halfers, please answer for the good priest, how should he advise this
prisoner?
Yuen, Tsz (EXCHANGE:WWP:1M42)
For the first time the salesman sent you letter(s):
P(2*20%) = 0.5
P(1*40%) = 0.5
So, for the first time, P(40%) = 0.5.
However, assume the salesman flip coin and sent you letter
again.
Then,
P(4*20%) = 0.25
P(2*20% + 1*40%) = 0.5
P(2*40%) = 0.25
So, after the saleman sent you letters twice, P(40%) =
P(40%|(2*20% + 1*40%)) + P(40%|2*40%)
=
0.5*0.33333 + 0.25 = 0.41666667
the third time the saleman sent you letters. it gets
complicated but bare with me here.
P(6*20%) = 0.125
P(4*20% + 1*40%) = .375
P(2*20% + 2*40%) = .375
P(3*40%) = 0.125
So after the third time the saleman sent you letters, P(40%)
=
P(40%|(4*20% + 1*40%)) + P(40%|(2*20% + 2*40%)) +
P(40%|(3*40%)) =
.375*1/5 + .375*2/4 + 0.125 = 0.3875
All answers are different depends on how many times the
saleman sent you letters. So, I think this puzzle missing
one important varible here is how many time the saleman has
sent you letters.
Please feel free to laugh at my gammar and/or line wrap, but
corrections are welcome!
> >Here is a version of the Sleeping Beauty problem to which I expect most
> >people to give the thirder answer:
>
> >My name is on the mailing list of Spam Insurance. Every few months, it
> >comes to the top of the list, and the salesman tosses a coin. On heads,
> >he sends me a letter offering 40% off life insurance. On tails, he
> >sends two letters, posted a few days apart, each offering 20% off life
> >insurance.
>
> >I frequently receive letters selling life insurance, from many different
> >companies. Whenever I receive one, I put it straight in the rubbish bin
> >unopened, and immediately forget about it [this part is far more
> >plausible than the "memory-wiping drug" thing].
>
> >One day, my telephone line blows down in a storm. Without the use of
> >telephone or internet, I become extremely bored. Looking for something
> >to do, I extract the top item from my bin, and open it. It is a letter
Nick Wedd
(EXCHANGE:WWP:1M42) <tsz...@americasm01.nt.com> writes
>Here is my two cents of this puzzles:
>
>For the first time the salesman sent you letter(s):
>P(2*20%) = 0.5
>P(1*40%) = 0.5
>
>So, for the first time, P(40%) = 0.5.
>
>However, assume the salesman flip coin and sent you letter
>again.
>Then,
>P(4*20%) = 0.25
>P(2*20% + 1*40%) = 0.5
>P(2*40%) = 0.25
>
>So, after the saleman sent you letters twice, P(40%) =
>P(40%|(2*20% + 1*40%)) + P(40%|2*40%)
> =
>0.5*0.33333 + 0.25 = 0.41666667
>All answers are different depends on how many times the
>saleman sent you letters. So, I think this puzzle missing
>one important varible here is how many time the saleman has
>sent you letters.
>
>Please feel free to laugh at my gammar and/or line wrap, but
>corrections are welcome!
Your English grammar is infinitely better than my Chinese grammar.
Your line wrap is fine too.
I understand your calculations if you assume that all the letters in
my rubbish bin get mixed up in a random order, and no-one ever empties
it.
I should have specified, EITHER that it acts as a stack, and the
newest letter is always put on top of it, OR that it is emptied every
day.
Matt McLelland
> In article <3756ED9C...@americasm01.nt.com>, Yuen, Tsz
> (EXCHANGE:WWP:1M42) <tsz...@americasm01.nt.com> writes
> >All answers are different depends on how many times the
> >saleman sent you letters. So, I think this puzzle missing
> >one important varible here is how many time the saleman has
> >sent you letters.
...
> Your English grammar is infinitely better than my Chinese grammar.
> Your line wrap is fine too.
Are my long lines still bothering anyone now that I have the latest
version of Netscape?
> I understand your calculations if you assume that all the letters in
> my rubbish bin get mixed up in a random order, and no-one ever empties
> it.
I don't.
> I should have specified, EITHER that it acts as a stack, and the
> newest letter is always put on top of it, OR that it is emptied every
> day.
Why does that matter? Let's suppose that the guy has flipped N coins, and
sent you one 40% or two 20% letters each time. You put all mail in a pile
and never throw any away. When you get bored, you pull one single piece
of mail out of the pile at random and notice that it just happens to be
from Spam Insurance. What are the odds that it is a 40% letter?
The odds are 1/3, for any N. For a simple proof imagine that he stamps
the outside of each envelope with a number indicating how many times he
had tossed the coin when he sent the letter. You get a letter marked 18.
Well, there are three equally likely possibilities: you found the single
40% letter #18, you found the first 20% letter #18, or you found the
second 20% letter #18. The odds of 40% are 1 in 3. Since this is true
regardless of the number stamped on the front, that number needn't even be
there.
Matt McLelland
> Here is my two cents of this puzzles:
>
> For the first time the salesman sent you letter(s):
> P(2*20%) = 0.5
> P(1*40%) = 0.5
>
> So, for the first time, P(40%) = 0.5.
I see now what you have done. You have assumed that the only mail in your bin
came from Spam Insurance! I don't think that this is consistent with the
problem statement, which implies that there is a large amount of mail in the
mail bin.
Jamie Dreier
> Halfers, please answer for the good priest, how should he advise this
> prisoner?
I may be the only Halfer left.
I wasn't following the thread. If you want an answer, you'll have to
summarize the set-up for me.
-Jamie
--
SpamGard: For real return address replace "DOT" with "."
Matt McLelland
> > Halfers, please answer for the good priest, how should he advise this
> > prisoner?
>
> I may be the only Halfer left.
>
> I wasn't following the thread. If you want an answer, you'll have to
> summarize the set-up for me.
Ok, no problem. Lets consider the variant in which Beauty is woken on a
random day (either Monday or Tuesday) when the coin flips heads, and on
both Monday and Tuesday on tails.
If you go to visit Beauty on a random day (M or T) and find her awake, then
you will think P[heads| she is awake]=1/3. Now suppose that she asks you
what she should think. Since you are a halfer, it looks to me like you are
torn. You think that P[heads]=1/3, but you think she should think that
P[heads]=1/2. What do you tell her the probability is if she asks?
James Smith
> >When you can do something two ways, and you get two different answers,
> >and there is an obvious flaw in one method, and that method also returns
> >what is obviously a wrong answer, surely it is not hard to see which
> >method must be wrong. :-)
>
> My view precisely. However I would not describe the flaw in your method
> as obvious :-)
You are trying to argue that if you toss a coin there is a 1/3 chance
that
it will come down heads. Doesn't that alert you to something?
To reiterate something I just posted:
Someone has suggested that the 'thirder' argument lies in the belief
that after SB removes one of the states the other three are
equi-probable.
But they are not. The probability of the removed state doesn't go away,
it accretes to the probability of one of the other states, and so the
three remaining states are not eqi-probable. That is the flaw in the
argument.
Jim
Nick Wedd
<en...@cc.newcastle.edu.au> writes
>You are trying to argue that if you toss a coin there is a 1/3 chance
>that
>it will come down heads.
I have never tried to argue that. I have argued that, from SB's
standpoint, when the coin was tossed there was a 1/3 chance that it came
down heads.
> Doesn't that alert you to something?
>To reiterate something I just posted:
>
>Someone has suggested that the 'thirder' argument lies in the belief
>that after SB removes one of the states the other three are
>equi-probable.
>But they are not. The probability of the removed state doesn't go away,
>it accretes to the probability of one of the other states, and so the
>three remaining states are not eqi-probable. That is the flaw in the
>argument.
You put this well. "Accrete" is a good word for it.
I think I understand the halfer position. Of the four cases (H/Mon,
H/Tue, T/Mon, T/Tue), we know that H/Tue is not possible. In my view it
"goes away". In your view, it "accretes" to H/Mon.
How do you decide that it accretes to H/Mon and not to T/Tue? I can
imagine a "quarterer" position. A quarterer SB would reason "I don't
know if it is Monday or Tuesday, both are equally likely. But if it's
Tuesday, it must be Tails. So
p( H, Mon ) = 0.25
p( T, Mon ) = 0.25
p( T, Tue ) = 0.5 "
This does not seem right to me, but nor does the halfer position. All
three positions, halfer, thirder and quarterer are internally
consistent. I can see no reason to prefer the halfer to the quarterer
position.
(I am about to go on holiday for a week. I shan't see any replies until
June 14th.)
Jamie Dreier
Oh!
That's an interesting question.
I would not be entirely sure what she means when she asks, but if I had to
answer the question as stated I would say that the probability is 1/3.
There are much more mundane cases like this.
Matt McLelland
> > Ok, no problem. Lets consider the variant in which Beauty is woken on a
> > random day (either Monday or Tuesday) when the coin flips heads, and on
> > both Monday and Tuesday on tails.
> >
> > If you go to visit Beauty on a random day (M or T) and find her awake, then
> > you will think P[heads| she is awake]=1/3. Now suppose that she asks you
> > what she should think. Since you are a halfer, it looks to me like you are
> > torn. You think that P[heads]=1/3, but you think she should think that
> > P[heads]=1/2. What do you tell her the probability is if she asks?
> I would not be entirely sure what she means when she asks, but if I had to
> answer the question as stated I would say that the probability is 1/3.
But you do agree with the conflict as I have described it? That is, you
acknowledge that in this situation, there will be two people who should not be
able to reach a consensus on the probability of heads? Do you further agree
that there is no piece of information which you could give Beauty [without
knowledge of the coin toss] which should convince her that P[heads]=1/3?
> There are much more mundane cases like this.
Hmmm. I'm not sure what you mean by this... could you elaborate, maybe give an
example?
Yuen, Tsz (EXCHANGE:WWP:1M42)
1) Since you picked the letter already and it's from Spam,
your condition that you have a large amount of mails in your
bin does not matter.
2) you are picking up the top(latest) mail only.
3) Assume no delay for the salesman to send you letter, you
receive mail right after the salesman send you the letter.
4) Further assume that the salesman only flip coin and send
letter at the first of the month and every month has 30
days(I know this is stupid but just assuming).
5) let the 'few days apart' = 15days. i.e. if salesman is
sending you those 20% discount letter, you will receive it
only on 1st and 16th. Otherwise you just receive a 40% at
the first of the month.
6) let today is the 1st of the month and the salesman is
flipping coin.
The chance you will have a 20% discount letter on the 'top'
of your bin by the first of next month = 0.5
The chance you will have a 40% discount letter on the 'top'
of your bin by the first of next month = 0.5
7) Since I assume the salesman only send letter on the 1st
or 16th.
The chance you have a 20% on 'top' from 1st to 15th of the
month = 0.5
The chance you have a 40% on 'top' from 1st to 15th of the
month = 0.5
8) If the salesman send you a 20%, he will send you another
one on 15th. No more letter for the month otherwise.
=> From 15th to 30th, this event depends on whether you
receive a 20% or 40% at the first of the month.
a.) If you receive a 20% at the 1st, 20% at the 'top' = 100%
from 15th to 30th.
b.) If you receive a 40% at the 1st, 40% at the 'top' = 100%
from 15th to 30th.
which are sure events.
9) Now forget the assumption that the salesman only send you
letter on the 1st and 15th. That the salesman would only
send letter 'a few' days apart at any day of the month(let
it be the x-th day of the month).
Right after the salesman flipped coin and send you letter.
Let me copy and paste here:
The chance you have a 20% on 'top' from 'x-th' to the 'x-th
+ a few' of the month = 0.5
The chance you have a 40% on 'top' from 'x-th' to the 'x-th
+ a few' of the month = 0.5
(meaning the 2nd letter has not come in yet, if there is a
2nd coming.)
and from the 'x-th + a few' to the salesman decide to flip
another coin(meaning just before another letter cycle):
Given (20% on top from the 'x-th' to 'x-th + a few'), 20% on
top = 1(because you receive another 20% letter), 20% on top
= 0 otherwise.
given (40% on top from the 'x-th' to 'x-th + a few'), 40% on
top = 1, = 0 otherwise.
So, I think P(40% off) = 0.5.
Again, feel free to laugh at my grammar(I even spell it
right this time!).
Again, any corrections are welcome.
Yuen, Tsz (EXCHANGE:WWP:1M42)
the puzzle says the letter picked is from Spam already, so
the fact that there are many mails in the bin is irrelevant
after you know that the letter is from Spam.
John Rickard
1 day if found not guilty, 100 days if found guilty; memory wipes
between each day; after release, will remember last day.]
Nick Wedd <Ni...@maproom.co.uk> wrote:
: >Indeed. The probability that you assign to being released this
: >evening is (as also in the lottery variant) unimportant, since there
: >is nothing you can use it for.
[...]
: You wake up in prison, as described above. After breakfast, you are
: visited by the Mad Vicar. [...] never visits the same prison twice
: within one year. He is totally honest and trustworthy. There is
: no-one else like him. All these facts are known to you.
[...]
: You ask the Mad Vicar whether you were found guilty. He says that he
: does not know. He then asks you to guess whether you were found guilty:
: if you guess correctly, he will pay $100 into your bank account. Of
: course, you won't know whether you have won until you are released. In
: fact you may not even remember the deal; but the $100 will be most
: welcome.
(Assume that the utility to me of $100 is the same whether I was found
guilty or not.)
Good example!
: How should you guess?
Being a thirder, I obviously guess "yes", since there is a 25/26
chance that I was found guilty. The question that interests me is
whether your gut feeling is at all changed now that the probability
does have some real relevance to your actions; if not, my previous
article was irrelevant. I don't know what *my* gut feeling would be
in this situation.
If I had to decide my strategy beforehand, deciding to guess "yes"
(should the Mad Vicar ever make me such an offer) would be optimal. I
think that halfers would therefore want guessing "yes" to be correct
even if I haven't thought about it beforehand; but if so, how do they
justify this if they think I should judge the probability that I was
found guilty as 1/5? If the Mad Vicar makes the offer every day, then
the halfers' justification (if I can generalize from other scenarios)
is that although there is only a 1/5 chance that I was found guilty, I
stand to gain $10000 if I was found guilty and guess "yes" as opposed
to only $100 if I was found not guilty and guess "no". But with the
Mad Vicar visiting only once, this seems less tenable. Halfers?
Going back to the scenario without the Mad Vicar, I can't say that my
gut feeling differs from yours. But does your gut feeling change if
you can communicate with other people in the same situation? Forget
about guilt or innocence and possible lying witnesses: just suppose
that the police arrest five people at random every day, and keep one
of them in prison for 100 days and the remaining four for one day.
You wake up in prison in a room with 103 other people; you talk among
yourselves and find that you are all in similar situations and have
similar memories of being arrested. (But prison rules strictly forbid
revealing any information about the date of your arrest!) You know
that five of you will be released that evening. Is your gut feeling
still that you are probably going to be one of those five? I think my
gut gets confused in this scenario.
--
John Rickard <John.R...@virata.com>
Matt McLelland
> let me try this again:
>
> 1) Since you picked the letter already and it's from Spam,
> your condition that you have a large amount of mails in your
> bin does not matter.
This is wrong. =)
> [...] 9) Now forget the assumption that the salesman only send you
> letter on the 1st and 15th. That the salesman would only
> send letter 'a few' days apart at any day of the month(let
> it be the x-th day of the month).
> Right after the salesman flipped coin and send you letter.
> Let me copy and paste here:
> The chance you have a 20% on 'top' from 'x-th' to the 'x-th
> + a few' of the month = 0.5
No. This is why it matters that you get other mail. There isn't a range of
days in which the Spam Insurance letter is on top - there is only a single
day per letter. After that, other junk mail gets put on the pile.
Suppose that you receive his first letter on the 'x-th' day (instead of
assuming that this is when he sent the letter). Also, assume that you would
receive the second letter on the 'xth + afew' day. When you pick a letter
off of the top, it could only be from Spam Insurance if the day is the 'x-th'
or 'xth + afew'. If it is any other day, then you would get junk mail from
someone else. The three possibilities of (40%, 'xth day'), (20%, 'xth day'),
and (20%, 'xth + afew day') are equally likely. Thus P[40%] = 1/3.
Patrick A. O'Donnell
> > Halfers, please answer for the good priest, how should he advise this
> > prisoner?
>
> I may be the only Halfer left.
Well, if a Waffler can count as a part-time Halfer, then you're not
alone.
I have definitely decided to be a Waffler. I am convinced that there
is no single correct answer to SB, but that the probability to be
assigned depends on the question asked. This was, of course,
discussed at the beginning of the whole thread. I'm not sure that
dedicated Thirders or Halfers at this point still recognize that
dependence. It seems that alternate scenarios that Halfers and
Thirders come up with to explicate their point of view always seem to
ask an appropriate question supporting the desired outcome.
Basically, any question which weights the number of awakenings or the
span of time the awakenings occur over supports the Thirder camp. If
the number of awakenings is not important to the question, the Halfer
camp prevails.
From a frequentist point of view, if we perform the experiment many
times, we can observe that over all the experiments, the fraction of
heads flipped is approximately 1/2. In all the experiments the number
of awakenings due to heads is approximately 1/3. Different questions;
different answers; no surprises.
If I visit SB just once during her original experiment and she's
awake, then we both should decide that p(H | I visit) = 1/3. That I
am twice as likely to visit her if the coin came up tails does count.
Of course, if the coin is terribly unfair and p(H) = 99/100, then I
find P(H | I visit) jumps to over 98%. No surprise -- it's much more
likely that I happened to visit during the one heads awakening.
If SB is simply asked what probability she should assign to heads
without any further information beyond the fact that she's awake, she
should answer 1/2, _as long as there is no weighting based on the
number of answers_. Once it "matters" how many right answers are
given, then 1/3 looks more attractive.
Or, put another way, let's say we're scoring SB in the experiment. We
ask her, "heads or tails?" She gets one point every time she answers
correctly, and zero points if she answers incorrectly. Her strategy
will depend on how we accumulate her points. If we sum them, then
certainly she should answer, "heads," with probability 1/3. If we
take the max (i.e. she wins if she answers correctly at least once),
then she should answer, "heads," with probability 1/2. Different
questions; different answers; no surprises.
If we merely awaken her and ask her, "what is your credence that the
coin landed 'heads'" (as in the original problem statement), then her
best response (as a Waffler) at that point is to reply, "why do you
want to know?"
- Patrick O'Donnell
p...@alum.mit.edu
Patrick A. O'Donnell
> Matt McLelland wrote:
> >
> > I see now what you have done. You have assumed that the only mail in your bin
> > came from Spam Insurance! I don't think that this is consistent with the
> > problem statement, which implies that there is a large amount of mail in the
> > mail bin.
>
> the puzzle says the letter picked is from Spam already, so
> the fact that there are many mails in the bin is irrelevant
> after you know that the letter is from Spam.
Not really. What it tells you is that you are in one of three
intervals of time:
between the receipt of a 40% letter and the next junk letter
between the receipt of a 20% letter and the next junk letter
between the receipt of a 20% letter and the next junk letter
With the given probabilities, it is twice as likely to be a 20% letter
as it is a 40% letter.
If Nick kept looking in his rubbish bin until he found a letter from
Spam, then 20% and 40% letters would be equally likely. This,
however, was not the question asked.
- Patrick O'Donnell, Waffler
p...@alum.mit.edu
Matt McLelland
> Basically, any question which weights the number of awakenings or the
> span of time the awakenings occur over supports the Thirder camp. If
> the number of awakenings is not important to the question, the Halfer
> camp prevails.
For the record, I have not encountered a single question in which the halfer
camp prevails. Some cases have been troubling to varying degrees, but I
think that the thirder position is consistently the correct one.
> If I visit SB just once during her original experiment and she's
> awake, then we both should decide that p(H | I visit) = 1/3. That I
> am twice as likely to visit her if the coin came up tails does count.
> Of course, if the coin is terribly unfair and p(H) = 99/100, then I
> find P(H | I visit) jumps to over 98%. No surprise -- it's much more
> likely that I happened to visit during the one heads awakening.
You can rightly conditionalize on the event "She was awake" to reach
P[heads]=1/3, but she cannot rightly conditionalize on the event "Patrick
visited me" to change her probability assessment. Suppose that I visit her
on one day, and you visit her on the other. Either of us who finds her awake
will conditionalize to P[heads]=1/3. Look at it from her point of view.
When she wakes up, she knows that one of us is going to arrive. If she can
conclude that P[heads]=1/3 no matter which it is, then she must be able to
conclude that P[heads]=1/3 before either of us arrive, or for that matter,
even if we never do arrive.
Yuen, Tsz (EXCHANGE:WWP:1M42)
>
> > Matt McLelland wrote:
> > >
> > > I see now what you have done. You have assumed that the only mail in your bin
> > > came from Spam Insurance! I don't think that this is consistent with the
> > > problem statement, which implies that there is a large amount of mail in the
> > > mail bin.
> >
> > the puzzle says the letter picked is from Spam already, so
> > the fact that there are many mails in the bin is irrelevant
> > after you know that the letter is from Spam.
>
> Not really. What it tells you is that you are in one of three
> intervals of time:
>
> between the receipt of a 40% letter and the next junk letter
> between the receipt of a 20% letter and the next junk letter
> between the receipt of a 20% letter and the next junk letter
if you receive a 40% letter at the first place, you won't
receive the 2nd letter.
But if you receive a 20% letter at first, you will
definitely receive a 2nd letter.
so the probability that you will receive the 2nd 20% off
letter is heavily depends on whether you receive a 20% off
letter at first, which has a probabilty of 0.5 because the
salesman decide this on a fair coin flip.
>
> With the given probabilities, it is twice as likely to be a 20% letter
> as it is a 40% letter.
>
> If Nick kept looking in his rubbish bin until he found a letter from
> Spam, then 20% and 40% letters would be equally likely. This,
The top letter Nick picked out of various mails is from Spam
already. That's a given condition.(if I read Nick's question
correctly). In Nick's question, he didn't keep looking into
the bin.
Patrick A. O'Donnell
> Suppose that you receive his first letter on the 'x-th' day (instead of
> assuming that this is when he sent the letter). Also, assume that you would
> receive the second letter on the 'xth + afew' day. When you pick a letter
> off of the top, it could only be from Spam Insurance if the day is the 'x-th'
> or 'xth + afew'. If it is any other day, then you would get junk mail from
> someone else. The three possibilities of (40%, 'xth day'), (20%, 'xth day'),
> and (20%, 'xth + afew day') are equally likely. Thus P[40%] = 1/3.
It's no the case that the probability of it being a 40% letter is 1/3
because the three possibilities are equally likely. The three cases
are equally likely because their probabilities all happen to be 1/3.
What is P[a 40% letter is on top] if the Spam Insurance agent sends
the 40% letter with probability 99%?
Matt McLelland
> It's not the case that the probability of it being a 40% letter is 1/3
> because the three possibilities are equally likely. The three cases
> are equally likely because their probabilities all happen to be 1/3.
Well, either fact implies the other. There isn't a causal relationship here. I
see a very simple reason why they should be equally likely. It is just as likely
that you picked the first 20% letter he sent as it is that you picked the 40%
letter, since the coin was fair. On the other hand, each 20% letter is obviously
just as likely as every other 20% letter. Conclusion: they are all equally
likely, and P[40% on top]=1/3.
> What is P[a 40% letter is on top] if the Spam Insurance agent sends
> the 40% letter with probability 99%?
Well, then they wouldn't be equally likely, would they!
Matt McLelland
[...]
Lets take your position to an extreme so that you can see it is wrong. Suppose that on
heads he sends you one single letter with a 40% discount inside. On tails, he sends you
a letter everyday for the rest of your life, each containing a 20% discount. If you go
to your mail bin and pick up a letter, and it just happens to be from Spam Insurance,
don't you think it is much more likely that you get such letters every day? In this
case, would you still think that the odds of 40% are 1/2?
Yuen, Tsz (EXCHANGE:WWP:1M42)
>
> > Suppose that you receive his first letter on the 'x-th' day (instead of
> > assuming that this is when he sent the letter). Also, assume that you would
> > receive the second letter on the 'xth + afew' day. When you pick a letter
> > off of the top, it could only be from Spam Insurance if the day is the 'x-th'
> > or 'xth + afew'. If it is any other day, then you would get junk mail from
> > someone else. The three possibilities of (40%, 'xth day'), (20%, 'xth day'),
> > and (20%, 'xth + afew day') are equally likely. Thus P[40%] = 1/3.
I agree with Matt that there are three possibilities. But
they are not equally likely.
P(40%, 'xth day') and P(20%, 'xth day') are equally likely.
But P(20%, 'xth + afew day') = 0 if you receive a 40% off at
the first place. (I think I repeated myself here) why would
you say they are equally likely if one of the event (2nd 20%
off) is a dependent of another event(whether the first
letter is 20% or 40%)?
>
> It's no the case that the probability of it being a 40% letter is 1/3
> because the three possibilities are equally likely. The three cases
> are equally likely because their probabilities all happen to be 1/3.
> What is P[a 40% letter is on top] if the Spam Insurance agent sends
> the 40% letter with probability 99%?
>
Yuen, Tsz (EXCHANGE:WWP:1M42)
so you are telling me that P(40%) would be zero using your
example??
I disagree.
Let use your example, further extreme.
The salesman decide this is a one time deal and he flip his
coin(50-50 chance). let me quote you: "Suppose that on heads
he sends you one single letter with a 40% discount inside.
On tails, he sends you a letter everyday for the rest of
your life, each containing a 20% discount. " Either case, he
would not flip the coin and decide again what letter to send
you. (if you assume the salesman will send me 20% off letter
for the rest of your life, that's pretty much the same case
here)
Let's say the salesman flip a head: that means I will
receive ONE 40% letter, nothing more(certainly no 20% off
letter because the salesman flip a head), and I put that
into a one element stack. So the 'TOP' element is a 40% off
letter. No matter when I go and check the TOP of the stack,
it still a 40% off letter. Agree with me so far??
Then, let's say the salesman flip a tail: that means I will
receive infinity numbers of 20% off letter, and I put them
into a stack that has infinite number of elements. The 'TOP'
element is still a 20% off letter. Again, no matter when I
go and check the stack, no matter how many 20%off letters
are in the stack, the top of the stack is still a 20% off
letter. Agree with me here?
Now the question, if I read Nick's question
correctly(assuming we only consider Spam letter only, this
is a given condition), what is the probability of the TOP
element is a 40% off letter?
if the salesman flip a head, there is no chance you will see
a 20% off letter. Agree?
It still comes down to whether the salesman flipped a head
or tail, don't you think?
Patrick A. O'Donnell
> > span of time the awakenings occur over supports the Thirder camp. If
> > the number of awakenings is not important to the question, the Halfer
> > camp prevails.
> For the record, I have not encountered a single question in which the halfer
> camp prevails. Some cases have been troubling to varying degrees, but I
> think that the thirder position is consistently the correct one.
How about the one that you elided from my message: SB is scored on
whether she answers "was it 'heads'?" correctly at least once.
Multiple correct answers have no additional value.
I will grant that there do seem to be more questions where the number
of awakenings matter than not, and so that the Thirder position is
more often helpful. I do not grant, however, the Thirder argument
that the awakenings are equally likely, therefore, P(H|awake) = 1/3.
In the cases where the awakenings matter, they are equally likely
because they are all 1/3.
> You can rightly conditionalize on the event "She was awake" to reach
> P[heads]=1/3, but she cannot rightly conditionalize on the event "Patrick
> visited me" to change her probability assessment. Suppose that I visit her
> on one day, and you visit her on the other. Either of us who finds her awake
> will conditionalize to P[heads]=1/3. Look at it from her point of view.
> When she wakes up, she knows that one of us is going to arrive. If she can
> conclude that P[heads]=1/3 no matter which it is, then she must be able to
> conclude that P[heads]=1/3 before either of us arrive, or for that matter,
> even if we never do arrive.
If you change the question, it should be no surprise that the answer
may change. If only I visit her on a day selected at random, and she
knows this, then she can conclude P[heads | I visit] = 1/3. She
cannot carry this conclusion over to the case where she will be
visited every day whether by the same or by different people. In that
case, every awakening is like another, and she can conclude nothing
more than if she were not visited.
I'm not sure where you intended to go with the closing "even if we
never do arrive." That seems to lead to yet another scenario, and the
conditionalizations won't carry over to that one, either.
Patrick A. O'Donnell
> > because the three possibilities are equally likely. The three cases
> > are equally likely because their probabilities all happen to be 1/3.
>
Well, right, but that's only helpful so long as one can establish
either fact. I'm not completely persuaded by your argument (just
below) that they are equally likely. (I'm not saying it's wrong, just
that I'm not convinced by it with the short time I can spend pondering
it. Particularly, I'm not sure that it's _obvious_ that "each 20%
letter is just as likely as every other 20% letter." It's true in
this case, but not immediately obvious.)
> There isn't a causal relationship here. I
> see a very simple reason why they should be equally likely. It is just as likely
> that you picked the first 20% letter he sent as it is that you picked the 40%
> letter, since the coin was fair. On the other hand, each 20% letter is obviously
> just as likely as every other 20% letter. Conclusion: they are all equally
> likely, and P[40% on top]=1/3.
>
> > What is P[a 40% letter is on top] if the Spam Insurance agent sends
> > the 40% letter with probability 99%?
>
> Well, then they wouldn't be equally likely, would they!
Right. I guess my point is that the "equally likely" argument is
rather limited, and it's more useful in the general case to establish
the probabilities by other means. Maybe that's not true for all folk.
Yuen, Tsz (EXCHANGE:WWP:1M42)
>
> "Patrick A. O'Donnell" wrote:
>
> > It's not the case that the probability of it being a 40% letter is 1/3
> > because the three possibilities are equally likely. The three cases
> > are equally likely because their probabilities all happen to be 1/3.
>
> see a very simple reason why they should be equally likely. It is just as likely
> that you picked the first 20% letter he sent as it is that you picked the 40%
> letter, since the coin was fair. On the other hand, each 20% letter is obviously
> just as likely as every other 20% letter. Conclusion: they are all equally
will receive more 20% letter. if you are sure about
something, why would you say 'as like as'? 'as like as'
sure?
> likely, and P[40% on top]=1/3.
I think I am repeating myself. But if you receive the first
20% off letter, you are sure that you will receive the 2nd
one. Agree?
If you agree, then P(2nd 20%off | 1st 20%off) = 1. agree
with the equation? Remember those probability diagarms? let
me draw one here if I may.
-----------------------------------
| | |
| | |
| | |
| 1st 20%off | 1st 40%off | This is a 50-50 split
| | |
| | |
| | |
-----------------------------------
now, P(2nd 20%off | 1st 20%off) = 1, that implies that the
area of P(2nd 20%off) is the same with P(1st 20%off), but
they are overlapped with each other(P()=1). The diagram
becomes
-----------------------------------
| | |
| | |
| 1st 20%off | |
| | 1st 40%off | This is a 50-50 split
| 2nd 20%off | |
| | |
| | |
-----------------------------------
If you still disagree, I don't know how describe the
condition, may be you can help me there.
Matt McLelland
> Let's say the salesman flip a head: that means I will
> receive ONE 40% letter, nothing more(certainly no 20% off
> letter because the salesman flip a head), and I put that
> into a one element stack. So the 'TOP' element is a 40% off
> letter. No matter when I go and check the TOP of the stack,
> it still a 40% off letter. Agree with me so far??
NO! You get mail every day. If you check the top of the stack *on the day the 40% arrives*,
then it might be on the top. If you check it the next day, then OTHER JUNK MAIL will have
been put on top of it. In this scenario, you only find the 40% letter if you check your mail
*on the day it arrives*. However, since the 20% mail will come every day, it could be at the
top of the stack on any day.
Matt McLelland
> Particularly, I'm not sure that it's _obvious_ that "each 20%
> letter is just as likely as every other 20% letter." It's true in
> this case, but not immediately obvious.
You are checking your mail on a random day. We are assuming that you find the piece of
mail if and only if you check your mail on the day it arrives. We are given that the
two pieces of mail never arrive on the same day, and may assume that either is just as
likely to be on top on the day that it arrives. I don't see the problem.
Matt McLelland
> Matt McLelland <mat...@flash.net> writes:
> > "Patrick A. O'Donnell" wrote:
> > > Basically, any question which weights the number of awakenings or the
> > > span of time the awakenings occur over supports the Thirder camp. If
> > > the number of awakenings is not important to the question, the Halfer
> > > camp prevails.
> > For the record, I have not encountered a single question in which the halfer
> > camp prevails. Some cases have been troubling to varying degrees, but I
> > think that the thirder position is consistently the correct one.
>
> How about the one that you elided from my message: SB is scored on
> whether she answers "was it 'heads'?" correctly at least once.
> Multiple correct answers have no additional value.
Thirders would choose the correct strategy in this situation as well. The key
observation that the thirder would make is that, if this were the second wakening,
and "tails" had been guess on the previous wakening, then guessing "tails" again
would gain nothing. Hence, while tails may be more probable, a guess of tails is
likely to be worth less. The correct strategy is to guess "tails" randomly with
some probability. In this way, you have some chance of guessing both heads and
tails. To find out what probability you should use, suppose you guess "tails"
with probability P. Then the probability of you guessing correctly just once is:
1/2(1-P) + 1/2(P + (1-P)P)
Differentiate and equate to 0 to find that the maximum occurs at P=1/2.
> I will grant that there do seem to be more questions where the number
> of awakenings matter than not, and so that the Thirder position is
> more often helpful. I do not grant, however, the Thirder argument
> that the awakenings are equally likely, therefore, P(H|awake) = 1/3.
> In the cases where the awakenings matter, they are equally likely
> because they are all 1/3.
>
> > You can rightly conditionalize on the event "She was awake" to reach
> > P[heads]=1/3, but she cannot rightly conditionalize on the event "Patrick
> > visited me" to change her probability assessment. Suppose that I visit her
> > on one day, and you visit her on the other. Either of us who finds her awake
> > will conditionalize to P[heads]=1/3. Look at it from her point of view.
> > When she wakes up, she knows that one of us is going to arrive. If she can
> > conclude that P[heads]=1/3 no matter which it is, then she must be able to
> > conclude that P[heads]=1/3 before either of us arrive, or for that matter,
> > even if we never do arrive.
>
> If you change the question, it should be no surprise that the answer
> may change.
Some problems have different answers, some the same. It just depends on the
problems!
> If only I visit her on a day selected at random, and she
> knows this, then she can conclude P[heads | I visit] = 1/3.
No, she can't. Halfers would like to think that she could, but really she can't.
Consider her position. She wakes up. Whether or not you will come *today* is a
random event to her. It has a 50% chance of happening - even though she doesn't
know the day, she knows that you are picking a day at random to visit. Further, it
is clear that you visiting her *today* is independent of the coin toss.
> She cannot carry this conclusion over to the case where she will be
> visited every day whether by the same or by different people. In that
> case, every awakening is like another, and she can conclude nothing
> more than if she were not visited.
You're assertion here is correct - she cannot conclude anything about the cointoss
from anyone's arrival in this case. Of course, she couldn't have done it in the
previous case either. My purpose for bringing in a second person is that the
people who visit her *can* conditionalize when they find her up. So, how do you
plan on dealing with this dilemma? Again, lets say that you visit her on a random
day, and I visit her on the other day. One of us will find her up, lets say you.
Now, since you have found her up, you will think P[heads]=1/3. But by your own
admission, she cannot have learned anything by your arrival. Hence, if she is a
halfer, then she will still think that P[heads]=1/2. Which of you is right?
> I'm not sure where you intended to go with the closing "even if we
> never do arrive." That seems to lead to yet another scenario, and the
> conditionalizations won't carry over to that one, either.
My point is that if she is going to conclude that P[heads]=1/3 no matter which of
us arrives, then she might as well draw that conclusion before we get there.
Jack
Matt McLelland wrote:
Please recheck the original question. The TOP letter IS THE SPAM letter!!! If you care about
OTHER JUNK MAILS, I don't think your 1/3 answer is correct neither.
Jack
Matt McLelland wrote:
> "Yuen, Tsz (EXCHANGE:WWP:1M42)" wrote:
>
> > let me try this again:
> >
> > 1) Since you picked the letter already and it's from Spam,
> > your condition that you have a large amount of mails in your
> > bin does not matter.
>
> This is wrong. =)
would you care to explain why it's wrong?
Matt McLelland
> Matt McLelland wrote:
>
> > "Yuen, Tsz (EXCHANGE:WWP:1M42)" wrote:
> >
Yes it is. What we are talking about here are the periods of time in which the top letter would be
from Spam Insurance under the two different coin toss results. Here is what I am saying. The
event which you have witnessed (the spam letter is on top) could possibly occur everyday if the
coin landed one way, and could only occur on one random day if it landed the other way. Therefore,
upon finding the Spam letter on this random day, you should think it is more likely that there are
many Spam Letters delivered to your mail bin.
Agree now?
James Smith
> >Someone has suggested that the 'thirder' argument lies in the belief
> >that after SB removes one of the states the other three are
> >equi-probable.
> >But they are not. The probability of the removed state doesn't go away,
> >it accretes to the probability of one of the other states, and so the
> >three remaining states are not eqi-probable. That is the flaw in the
> >argument.
>
> You put this well. "Accrete" is a good word for it.
>
> I think I understand the halfer position. Of the four cases (H/Mon,
> H/Tue, T/Mon, T/Tue), we know that H/Tue is not possible. In my view it
> "goes away". In your view, it "accretes" to H/Mon.
>
> How do you decide that it accretes to H/Mon and not to T/Tue?
Because on the original coin toss, heads has a 50% probability and tails
has a 50% probability. You then split the tails event into two separate
events each of 25% probability. If you were to split the heads event
along the same lines then there would be two events each with 25%
probability and you would have your four state. But you don't split the
heads event and so it remains at 50%.
>>You are trying to argue that if you toss a coin there is a 1/3 chance
>>that
>>it will come down heads.
>I have never tried to argue that. I have argued that, from SB's
>standpoint, when the coin was tossed there was a 1/3 chance that it came
>down heads.
Sleeping Beauty has no information upon which to form a subjective
viewpoint other than the information available in the objective one.
So her viewpoint cannot differ from the objective one.
Jim
Jack
Matt McLelland wrote:
This is wrong. =)
Jamie Dreier
> Jamie Dreier wrote:
>
> > > Ok, no problem. Lets consider the variant in which Beauty is woken on a
> > > random day (either Monday or Tuesday) when the coin flips heads, and on
> > > both Monday and Tuesday on tails.
> > >
> > > If you go to visit Beauty on a random day (M or T) and find her
awake, then
> > > you will think P[heads| she is awake]=1/3. Now suppose that she asks you
> > > what she should think. Since you are a halfer, it looks to me like
you are
> > > torn. You think that P[heads]=1/3, but you think she should think that
> > > P[heads]=1/2. What do you tell her the probability is if she asks?
>
> > I would not be entirely sure what she means when she asks, but if I had to
> > answer the question as stated I would say that the probability is 1/3.
>
> But you do agree with the conflict as I have described it? That is, you
> acknowledge that in this situation, there will be two people who should not be
> able to reach a consensus on the probability of heads?
Hm, I am not sure about that.
What if I told her that I just picked Monday or Tuesday at random, and
found her awake? Maybe she should then change her assessment to 1/3. Does
that count as reaching consensus, or have I told her something you were
ruling out?
> Do you further agree
> that there is no piece of information which you could give Beauty [without
> knowledge of the coin toss] which should convince her that P[heads]=1/3?
I'm not sure, I don't know quite what you mean. I am not sure what the
restriction in the square brackets is.
Do you mean, there is nothing that Beauty could be told by someone who
doesn't actually know the outcome, that she could conditionalize on to
reach pr(Heads) = 1/3? I don't agree about that at all.
> > There are much more mundane cases like this.
>
> Hmmm. I'm not sure what you mean by this... could you elaborate, maybe
give an
> example?
Now I'm not sure because I don't think I understand the point of the
example. What I meant was that there are many mundane examples in which I
can sensibly have a different probability assessment from yours, and then
if you ask me what the probability is I might be unsure of whether I'm
supposed to express my own judgment or try to say what your rational
judgment should be.
-Jamie
--
SpamGard: For real return address replace "DOT" with "."
Matt McLelland
[...]
Ok. The situation that I was attempting to convey is the Objective Advisor argument
that I mentioned as one of my favorites. I will attempt to set the situation up
properly.
---------------------
One day, Sleeping Beauty visits the Institute of Memory Loss to be part of an
experiment. She will be put to sleep on Sunday night, and a coin will be tossed.
If the coin lands heads, then Beauty is woken either on Monday or Tuesday at
random. On tails, she is woken on both Monday and Tuesday, subject to usual memory
loss.
In this experiment, Dr. Blue and Dr. Green, the world's foremost authorities on
statistics, are hired to assist Beauty in her probability reasoning. If the coin
landed heads, then one of the Statisticians is chosen randomly to help assist her.
If the coin landed tails, then one of the PhDs gets her on one day, at random, and
the other gets her on the other day.
Now, using this setup, lets analyze the situation from Beauty's perspective in a
random wakening. Clearly, after she wakes up and before she finds out which doctor
she will visit, her probability assessments will be just as they are in the original
Sleeping Beauty situation. Now, at that point, she knows that she will either meet
with Dr. Blue or Dr.Green, but doesn't know which. Can her probability assessment
change when she finds out which will advise her? Clearly not. Both men are, by
symmetry, equivalent to her. One is not any more associated with heads or tails
than the other. Hence, her probability assessment of the coin toss should be the
same when she visits her advisor as it was when she woke.
So, what will her advisor say about the probability of heads? Well, if the coin had
landed heads, then he would only have had a 50% chance of meeting Beauty. On the
other side of the coin, his meeting her would have been a certainty. Since he does
get to meet her, he will believe that the probability of heads is only 1/3, when
Beauty and him discuss the probability of heads in the coin toss, he will be quite
insistent about this (you know how these Ph.D. types can be). As it stands, it is
possible that the Statistician has been able to more accurately assess the situation
with the piece of information that he has which she doesn't - he knows the day.
Well, lets have him tell Beauty the day. How can this affect her probability
assessment? Again, in this variant, it can't. By symmetry, neither Monday nor
Tuesday is more indicative of heads or tails than the other. After this disclosure,
Beauty and her advisor have exactly the same relevant information. It stands to
reason that if they don't agree on the probabilities at this point, then one of them
is wrong.
Patrick A. O'Donnell
> After this disclosure,
> Beauty and her advisor have exactly the same relevant information. It stands to
> reason that if they don't agree on the probabilities at this point, then one of them
> is wrong.
I don't agree that one of them has to be wrong. We seem to be in
agreement that if an advisor visits Beauty on all of her awakenings,
then Beauty gains no useful information. Each advisor (ignorant of
the actual result of the coin flip, and also ignorant of the day of
the week) may reasonably believe that the conditional probability of
heads is 1/3. Depending on Beauty's goals, her assessment should be
and remain either 1/2 or 1/3. One of the advisors is more likely to
be right in prefering tails (the Tuesday visitor), the other is more
likely to be wrong (the Monday visitor). (I'm too lazy right now to
work out the math on that -- I don't know if it really helps or not.)
If there is to be only one visitor on one day (day still unknown to
both Beauty and the visitor), and Beauty knows this, then the visit is
indeed enough to allow Beauty to adjust her assessment to 1/3, under
_either_ interpretation of the question!
Patrick A. O'Donnell
> Matt McLelland wrote:
> > Agree now?
>
> This is wrong. =)
Jack, I think your wit is misplaced, here. Matt is right.
Yes, the Spam Ins. letter is on top, by the problem statement. This
fact places the problem into one of three possible time periods, each
of which is (happens to be) equally likely. If there were no other
junk mail to be considered, then there would be no restriction on the
time period, and the answer would be 1/2. The problem _did_ say there
was other junk mail, and that there was a single bin. It did _not_
ask for the probability of a 40% Spam discount being on top at any
given day, and it did not ask for the probability that the highest
Spam discount letter in the bin was 40%. The probabilities were
conditionalized on the top letter being from Spam Ins, taking into
account the other junk mail.
Wei-Hwa Huang
>Wei-Hwa Huang wrote:
>Nevertheless, I still disagree with your claim that
>> >> But if you were going to dig down to the first Spam Insurance envelope
>> >> you saw (and it just happened to be the first) then the probability
>> >> becomes 1/2 again.
>Lets suppose that he has a bottomless rubbish bin which never gets emptied. He
>throws away all kinds of junk mail every day. On a random day he gets bored and
>decides to dig until he finds something from Spam Insurance. Let D be the depth at
>which he finds such a letter. What is P[ 40% | D=1]? It is 1/3.
Why disagree with my claim, then? I claimed that P[40%] = 1/2.
The probability changes depending on whether you're allowed to use the
D=1 information or not. In the passage you quoted above, I acknowledge
that D=1 but am not using that information.
--
Wei-Hwa Huang, whu...@ugcs.caltech.edu, http://www.ugcs.caltech.edu/~whuang/
---------------------------------------------------------------------------
Is there a metonym in this sentence?
Matt McLelland
> Matt McLelland <mat...@flash.net> writes:
> > After this disclosure,
> > Beauty and her advisor have exactly the same relevant information. It stands to
> > reason that if they don't agree on the probabilities at this point, then one of them
> > is wrong.
>
> I don't agree that one of them has to be wrong.
> We seem to be in agreement that if an advisor visits Beauty on all of her awakenings,
> then Beauty gains no useful information.
> Each advisor (ignorant of the actual result of the coin flip, and also ignorant of the
> day of
> the week) may reasonably believe that the conditional probability of
> heads is 1/3.
Let me make sure you understand this situation. In this variant Beauty, on heads, is
woken randomly on either Monday or Tuesday. So, she can safely learn the day without
learning anything about the coin toss. Thus, I have not arranged for the Advisor to be
ignorant of the day.
> Depending on Beauty's goals, her assessment should be
> and remain either 1/2 or 1/3. One of the advisors is more likely to
> be right in prefering tails (the Tuesday visitor), the other is more
> likely to be wrong (the Monday visitor). (I'm too lazy right now to
> work out the math on that -- I don't know if it really helps or not.)
No. Due to the random day on heads, mentioned above, neither advisors favors heads or
tails. Please reread my scenario. You have missed out a key fact in the situation, and
thus I don't think grasped argument.
> If there is to be only one visitor on one day (day still unknown to
> both Beauty and the visitor), and Beauty knows this, then the visit is
> indeed enough to allow Beauty to adjust her assessment to 1/3, under
> _either_ interpretation of the question!
This isn't true. Ask yourself this: when Beauty wakes, knowing that an advisor will
visit only one time, what should she think the odds of seeing him *today* are? There are
two possibilities for her: seeing the advisor, or not seeing the advisor. Which is an
indication to her of heads and which of tails?
Patrick A. O'Donnell
>
> No. Due to the random day on heads, mentioned above, neither
> advisors favors heads or tails. Please reread my scenario. You
> have missed out a key fact in the situation, and thus I don't think
> grasped argument.
No, I didn't miss the key fact. I merely glossed over it for brevity
in stating my position. Evidently that was a mistake. If you prefer,
you can retry reading my post substituting "awakening-on-heads-day"
for "Monday", "non-awakening-on-heads-day" for "Tuesday", and
"ignorant of whether this is the 'awakening-on-heads-day'" for
ignorant of the day of the week. Choosing the awakening day at random
is not a novel idea for me; I just didn't think it needed emphasising
as you had described it well enough.
> > If there is to be only one visitor on one day (day still unknown to
> > both Beauty and the visitor), and Beauty knows this, then the visit is
> > indeed enough to allow Beauty to adjust her assessment to 1/3, under
> > _either_ interpretation of the question!
>
> This isn't true. Ask yourself this: when Beauty wakes, knowing that an advisor will
> visit only one time, what should she think the odds of seeing him *today* are? There are
> two possibilities for her: seeing the advisor, or not seeing the advisor. Which is an
> indication to her of heads and which of tails?
>
Interesting question, what she should think the odds of seeing him
are. I'll have to ponder it for a bit. However, I don't see how it
is relevant to her being able to conditionalize her probability of
heads during the advisor's visit, once the visit starts and the
probability of seeing him *today* is 1.