Some "Sleeping Beauty" postings

In spring 1999, the newsgroup rec.puzzles carried several threads discussing versions of the "Sleeping Beauty" problem. This page was originally created to record some of the more interesting postings on the problem. It now also contains some other related material.

I would like to thank Jamie Dreier for introducing this problem to the newsgroup; and Matt McLelland for his many convincing contributions (formerly on the halfer side, latterly on the thirder) to the discussion.

The Sleeping Beauty problem, and the older Paradox of the Absentminded Driver, were first posed by Piccione and Rubinstein, in 1994. See below for references.

The Case of the Absentminded Motorist

The lead-up to the sleeping beauty problem was "The case of the absentminded motorist":

TitleThe case of the absentminded motorist
Posted byJamie Dreier
Ralph is in a bar late at night and he's planning to drive home. Though he's sober enough to drive, he's a little woozy.

The problem is that to get home, he has to take the second highway exit, and the exits are not marked. If he gets off at the first exit by mistake, he will be in a horrible neighborhood (his utility for getting off at the first exit is 0). If he gets off at the second exit, he'll get home ok (utility = 4). If he misses both exits, he'll have to drive for miles and end up spending the night at a motel (utility = 1). But, woozy as he is, when he arrives at an exit he won't be able to tell whether it's the first or the second, and he will not remember whether he has passed an exit already. After thinking about it for a while, he concludes that he cannot make a plan that will get him home tonight. For he can only [see note on mixed strategies below] plan to take an unknown exit when he gets to it, in which case he ends up in the horrible neighborhood, or plan to drive by an unknown exit when he gets to it, in which case he spends the night in a hotel. He decides to adopt the drive-by plan. This plan has utility 0 (obviously).

So he's driving along and just ahead he sees an exit. Although he has already made his plan, now he reconsiders. The chance that this is the first exit is 1/2, and the chance that it's the second is 1/2. So the expected utility of taking this exit is (1/2)*0 + (1/2)*4 = 2. This is higher than the utility of his plan, so he decides to exit.

Ralph got *no* new information, and yet his optimal plan has changed. Which plan should he follow???

Note on Mixed Strategies:

When I said 'he can only plan...', I was ignoring mixed strategies. The mixed strategies are also paradoxical in just about the same way. Making a plan in the bar, Ralph will conclude that the optimal mixed strategy is to exit with probability 1/3. This yields expected utility

  (1/3)*0 + (2/3)*(1/3)*4 + (2/3)*(2/3)*1 = 4/3.

But if Ralph adopts this plan, then when he arrives at some unknown exit he will surmise that (on the assumption that he is actually following his own plan) the chance that this is the first exit is 3/5, the chance that it is the second exit is 2/5. The expected utility of exiting is thus 8/5, which is higher than the utility of his original plan. Should he therefore decide to exit?


This problem is due to Ariel Rubenstein and Michele Piccione, and is discussed formally in their GAMES AND ECONOMIC BEHAVIOR 20 (1997), 3-24. Any arithmetic errors in this posting are certainly mine and not theirs. Any conceptual confusions are also extremely likely to be mine and not theirs.

The Case of Sleeping Beauty

The central problem was "The case of the sleeping beauty". It is essentially the same problem as the case of the Absentminded Motorist, but expressed more clearly.

TitleThe case of sleeping beauty
Posted byJamie Dreier
We plan to put Beauty to sleep by chemical means, and then we'll flip a (fair) coin. If the coin lands Heads, we will awaken Beauty on Monday afternoon and interview her. If it lands Tails, we will awaken her Monday afternoon, interview her, put her back to sleep, and then awaken her again on Tuesday afternoon and interview her again.

The (each?) interview is to consist of the one question: what is your credence now for the proposition that our coin landed Heads?

When awakened (and during the interview) Beauty will not be able to tell which day it is, nor will she remember whether she has been awakened before.

She knows the above details of our experiment.

What credence should she state in answer to our question?


p.s. Don't worry, we will awaken Beauty afterward and she'll suffer no ill effects.

p.p.s. This puzzle/problem is, as far as I know, due to a graduate student at MIT. Unfortunately I don't know his name (I do know it's a man). The problem apparently arose out of some consideration of the Case of the Absentminded Driver.

p.p.p.s. Once again, I have no very confident 'solution' of my own; I will eventually post the author's solution, but I am not entirely happy with that one either.

Some people, "halfers", consider that the answer to this question is obviously "1/2". Others, "thirders", consider that it is obviously "1/3".

Matt McLelland's Get Rich Clinic

TitleRe: The Case of Sleeping Beauty
Posted byMatt McLelland

Poor? Work all of the time and barely make ends meat? Feel like ending it all? Come to me first!

At my clinic, we will buy you a lottery ticket. Now everyone knows that your odds of winning our state lottery are only 1 in a million, but the $20 million dollar jackpot this time can almost certainly be yours if you just call me now! When you arrive you will be given a warm bath and fed a four star meal of your choice. When you are done being pampered, you just lie down and go to sleep. While you are sleeping, our trained personnel will watch the TV to see if your lottery ticket is a winner. If it isn't, then you just pay our small $250 service fee and go on about your day. If it is, then you will be repeatedly awakened into lavish surroundings and by beautiful women every day for the rest of your life. How you ask?!? Well, at the end of every 30 minute interval, we will wipe your memory clean up to the time you arrived! The mathematics behind this miracle are complicated, so I won't bore you, but if we suppose that you will live 50 more years, then your odds of awakening to find that you have won are amazingly over 50%! Wouldn't you pay $250 to have a 50% chance of winning the lottery?!? The number is 1800-IMAFOOL. Operators are standing by.

Andy and Bob

TitleNice new paradox
Posted byMatt McLelland
The following paradox just occurred to me. It stemmed from the Beauty debate. It is a little too complicated to be a great paradox, but have fun trying to resolve it.

Sometime in the future we devise a procedure for cloning objects by replicating their atomic structure. We are even able to clone humans. One day, Andy shows up at a clinic and makes arrangements to be cloned. He has an unusual request. He asks that after he is put under anesthesia, a coin be flipped and that he only be cloned if it should come up tails. Furthermore, he asks that before he is woken he be placed randomly into one of two identical rooms. His clone, if one exists, should be placed into the other room.

When Andy awakens, he realizes that he doesn't know whether or not he has been cloned or even if he is the clone. Soon, before he learns anything about the toss or whether he was cloned, his friend Bob shows up. Bob knows all about what Andy was going to do, and that he is in one of the two identical rooms. He randomly picks one and goes in, to find Andy there in bed. After shooting the breeze for a bit, they have the following conversation:

Andy: "Before I was put under anesthesia, I asked myself 'what is probability that I will ever wake up due to heads?'. Since it would happen if and only if the coin flipped heads, I knew then the odds of that were 1/2. Now I have woken just like I knew I would, and I still know nothing about the coin toss result. The odds of heads must still be 1/2."

Bob: "Wait. I know more than that. I picked a door at random and walked in. If you were cloned, I would have found you for sure. If you weren't I would have had only a 50% chance of finding you. Since the event that I would find you was twice as likely if the coin landed tails, the probability now that the coin did actually land tails is twice that of heads given that I actually did find you. That is, heads has probability 1/3."

Andy: "No. Look at it from my point of view. When you randomly picked a room, there were only two possibilities for me: either you would find me or you wouldn't. Since you picked at random, not only are these possibilities equally likely, but the event that you would find me is completely independent of the coin toss. Hence, the conditional probability of heads given that you found me must be the same as I would have assessed it before you came - 1/2."

After arguing for a bit, they are unable to reach an agreement on the probabilities. They finally settle on a wager. Bob will give Andy $4 if the coin is heads, and Andy will give Bob $3 dollars if it is tails. Now each man has exactly the same information as the other, and yet each man has calculates this wager to be favorable for himself. Obviously it can't be favorable for both of them – which man has reasoned incorrectly?

The Baron

This was an attempt to remove the "memory-loss" element which may have contibuted to misunderstanding in some other formulations of the problem.

TitleRe: Nice new paradox
Posted byNick Wedd
In article <>, Matt McLelland <> writes

>Please resolve the following paradox:
>Beauty has a large purse full of money. She hasn't counted the
>money in a long and really has no idea how much she has. Before
>she goes into the experiment, her friend Sally tells her that
>each and every time she wakes up Sally will offer the following
> If the coin was tails, then Beauty pays Sally 2 dollar.
> If the coin was heads, then Sally pays Beauty 5 dollars.
>Sally says she will offer it every time, even if she already knows
>she will lose from the previous time. When beauty wakes up she
>thinks the odds that the coin was tails are 2/3. Now, if the coin
>was heads then I will win 5 dollars. On the other hand, if the
>coin was tails then I will pay her 2 dollars. But wait, since I am a
>rational person, I will end up paying her twice! In reality, I am
>going to end up paying her 4 dollars if the coin came up tails.
>Working out the math, Beauty estimates that her expected loss if she
>takes the bet is 4*2/3 - 5*1/3 = 1 dollar. Clearly she should have
>accepted, what happened?

This is the most clearly perplexing formulation of the problem that I have seen yet.

SB can think
"There are three equiprobable cases.
(1.) The coin was heads, and I have an option of winning $5.
(2.) The coin was tails, this is my first awakening, and I have an option of losing $4.
(3.) The coin was tails, this is my first awakening, and I have an option of losing $4.
Therefore I reject the bet."
As you say, this is clearly wrong.

Here is another version of it. This one is not confused by memory loss, lack of consciousness, or non-conservation of persons.

You are a loyal agent of the baron. You know that all of his agents are as intelligent as you are. You are suddenly summoned to the king's court, where a courtier tells you

"The king decided to play a game. He tossed a coin. If it came up heads, he put the gift of five manors in an envelope, and had a representative of your baron summoned, to be offered the envelope. If it came up tails, he put two separate documents, each ordering the confiscation of two manors, in two separate envelopes, and had two representatives of your baron simultaneously and separately summoned, each to be offered an envelope."

"You are such a representative. Here is an envelope. Do you want it?"

The answer is obviously "accept the envelope". Yet the logic which I have used to defend "1/3" for the Sleeping Beauty problem says otherwise.

Nick, a thirder, assumed when he posted this that his difficulty in answering was caused by the attraction of the halfer view. He realised later that he had in fact been tempted by an extreme and erroneous form of thirdism.

Sleeping Beauty's Fatal Disease

TitleRe: The Case of Sleeping Beauty
Posted byMatt McLelland
David A Karr wrote:

> I'm not convinced by it either. In fact I retract what I said earlier
> about 1/2 being as good an answer as 1/3 due to an "ambiguity" in the
> problem. I now think the question in the original problem can most
> reasonably be taken as to estimate the conditional probability of
> heads *given* the knowledge that Beauty has when she wakes up on
> Monday or Tuesday.

This conditional probability bit is a sham. She could have conditioned on "I will wake up on Monday or Tuesday" before she went to sleep. People seem to have more confidence in the Beauty problem than they do in my cloning problem. I don't see the difference, so I am going to go ahead and rephrase the cloning problem that I just posted in terms of Beauty. Consider this:

Beauty is rushed to the hospital after experiencing a sharp pain in her head. When she arrives, doctors administer preliminary tests which are very pessimistic indeed. They tell her that she will live at least 23 more hours. Then, there is a 1 in 2 chance that she will drop dead in the next hour. However, they tell her that if she does live past 24 hours, then all danger from this condition will have passed. Beauty is horrified at the news. Her daughter is touring Europe, and would most likely only be able to make it back to see her mother for a short few minutes before she dies. Nevertheless, steps are taken to bring her daughter back. Seeing her daughter before she dies is the most important thing on Beauty's mind.

Over the next hour, the doctors perform more extensive tests which conclusively determine if Beauty will be dead at the end of the day. Beauty, a mathematician, doesn't want to know. Instead, she wants to engage in the same procedure that we have seen repeatedly on this thread. She is to be put to sleep. If the test says she will die, then she is awakened once. She isn't told of the test results, and is just left to live out the rest of her day. If the test says she will live, then for the next ten years Beauty will have her memory cleared every other day. (She will get to experience two days, each time, before having her memory wiped).

Beauty is put to sleep. When she wakens she feels great piece of mind knowing that her odds of dying by the end of the day are now less than 1 in 1750. Wow she thinks, I really don't have much to worry about with that whole disease thing. This procedure, while not actually having done anything to decrease her odds of dying, has put her mind at rest. The day seems to pass quickly, just like any other day, with her mind at ease. She is told that in 15 minutes, she will be possibly be in the hour period where she risks death. Beauty laughs as she thinks about how improbable that is. "Oh wait, my daughter should be here by now", thinks Beauty. "Oh well, I'm tired. I'll just see her in the morning." And with that thought, Beauty turns out her light and goes to sleep.

Wouldn't you have done the same thing?

The Million Clones of Sleeping Beauty

TitleRe: The Case of Sleeping Beauty
Posted byMatt McLelland
Tim Firman wrote:


> Everyone allowed a decision is acting rationally, but the one forced to
> play (the third party / the new clone) is getting screwed over. I don't
> see a paradox here.

True, but this isn't where the paradox lives. The paradox (to me) lies in this situation:

A coin is flipped 10 times in a row.
If (and only if) it comes up tails each time, then you are cloned a million times.

When you are wakened, after the possible cloning, you (and clones of you) are asked "Were they all tails?"
If you answer correctly, you live happily.
If you answer incorrectly, you are tortured and killed.

The paradox is that thirders would answer yes to this question. (I wouldn't - but can't justify my decision)

Beauty and Betty

TitleAmnesia [Final]
Posted byMatt McLelland

You wake in a cell with total amnesia.
On a paper in front of you is the following:

Yesterday, Beauty and Betty were placed in cells just like this one. A coin was tossed. If it landed heads, then Beauty alone was given a memory wiping drug. If it landed tails, then both Beauty and Betty were drugged. A copy of this note was placed in both of their cells. You are either Betty or Beauty.

What are the odds that you are Beauty?
What are the odds that Betty got drugged?
If you learn that you are Beauty, what are the odds that Betty was drugged?

Sleeping Beauty Goes Parachuting

TitleSleeping Beauty Goes Parachuting [was SB probability thread]
Posted byGerry Quinn
Last weekend, Beauty decided that she needed some healthy outdoor exercise. As always, she combined it with a bizarre experiment in probability.

She hired a helicopter and went to a local farm in which there were two large fields. One field was known to contain a single daisy, and one field was known to contain two.

Climbing into the helicopter, she instructed the pilot thus:
"Secretly toss a coin, and if it comes up Heads fly to the 1-daisy field. If it is Tails go to the 2-daisy field. I will jump out and land on a random spot. I will then look around me for a daisy. Pick me up again five minutes later, and we will repeat the experiment until I land near a daisy."

After a time, Beauty landed near a daisy. What credence did she then give to the statement "The coin came up heads"?

Was Sleeping Beauty Wrongly Convicted?

TitleWas Sleeping Beauty wrongly convicted?
Posted byNick Wedd
There are two forms of the SB problem currently being discussed, one 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.

Michael Will pointed out that this problem is stated badly. The mistake can be fixed by specifying
"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."

Spam Insurance

TitleRe: Was Sleeping Beauty wrongly convicted?
Posted byNick Wedd
In article <>, James Smith 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%?

Language note: by "bin" Nick meant "trash can", not "mail bin".


This problem has been included because it has the delightful property that the thirder answer is "a half", and the halfer answer is "a third".

Posted byJoe Keane

You visit Dr. Evil's lab for an experiment. The following protocol is explained to you and you remember it at all times.

On Sunday you're put to sleep with drugs. That night, Geddy Lee flips a fair coin and rolls a fair die.

On Monday and Tuesday, according to the conditions below, you may be woken up by Nurse Chapel. After this, Will Smith pulls out his penlight and pushes the history eraser button, and you forget being woken up.

On Monday, you're woken up if and only if the die shows 1.

On Tuesday, you're woken up if and only if: the coin shows heads and the die shows 1; or, the coin shows tails and the die shows 2.

On Wednesday, Dr. Franklin shoots you up and sends you on your way.


You're woken up by Nurse Chapel.

What's the chance the coin is heads?

There was now a long interval without Sleeping Beauty postings.

When they resumed early in 2000, one of the more interesting postings (which I have not been able to find) was by Derek Holt. He pointed out that if, during an awakening, Sleeping Beauty accidentally learns that it is Monday, she must then think that the probability that the coin landed heads is 1/2. But halfers believe that she was thinking that already, without the accidental information. It seems odd that accidentally learning that it is Tuesday causes her to lower her estimate of the probability of heads, but accidentally learning that it is Monday does not cause her to raise it.

Two new views were presented.

Jamie Dreier, formerly a thirder, became a halfer, but with the view that "today is Monday" is not a proposition. Therefore statements about P(Monday) or P(Heads | Monday) are ill-formed. This disposes of the problem raised by Derek Holt. This may be called the "rejectionist halfer" view.

The other new view was that the question is ambiguous. Those who hold this say that "What is the probability that the coin now shows heads?" should be answered "1/3", but "what is the probability that the coin landed heads?" should be answered "1/2".

Sleeping Beauty and the bed shortage

This was posted Nick Wedd 2000/02/02.
The director of the hospital hosting the Sleeping Beauty experiment has decided to reduce non-emergency bed use by 25%. Therefore the original Sleeping Beauty experiment is modified as follows.

If the coin is heads, the director draws a card from a pack. If it is diamonds, the experiment ends immediately and there are no awakenings. Otherwise it proceeds as usual for heads and there is one awakening.

If the coin is tails, the director draws a card from a pack. If it is diamonds or hearts, the two awakenings are reduced to one. Otherwise it proceeds as usual for tails and there are two awakenings.

Of course, SB is not aware of what card was drawn.

When (if) SB wakens in the experiment, what is her probability that the coin is heads?

The thirder answer is "1/3". I would like to know how halfers answer.

Jamie answered 3/7.

SB variant

This variant problem was originally posted by Jonathan Perkins on 2000/02/13. Next day he posted again with modifications to it. I have taken the liberty of amalgamating these into a single statement.
Imagine you are a subject of a SB-like experiment without the memory loss. You are not told anything about the mechanics of the experiment.

There is a box beside your bed with glue on the floor and a hole in the top. Each morning you wake up, look in the box, and see there is a coin stuck to its floor. Over the course of many thousands of wakings you determine that Heads shows up 1/3 of the time and Tails shows up 2/3 of the times. The tester asks you "what is the probability that tomorrow's coin will show Heads?". IMHO the rational answer would be 1/3.

Then you think a little more carefully about your observations and realize that Tails always came in pairs (even numbers). If today's coin showed Heads you answer 1/2. If today's coin was the first Tail in a pair (odd number) then you answer 0. If todays card was the second Tail in a pair (even number) then you answer 1/2.

The tester then gives you a little pill that makes you lose track of the last few results (though you still remember the long term observations and logic). What is the rational answer to "What is the probability that tomorrow's coin will say Heads?". I think that 1/3 is a reasonable answer because the memory loss pill has ruined your ability to use the pattern and that is equivalent to setting you back to the state before you figured out the pattern in the observances of A's and B's. I guess that makes me a thirder.

Another way to get the 1/3 answer is to say that you are dropped randomly into the sequence and there is a 2/3 chance that the right answer is 1/2 and a 1/3 chance that the right answer is 0 and (2/3)*(1/2)+(1/3)*(0) = 1/3

My main qualm with this line of argument is that it relies on the sequence of results and the pure SB experiment is a single-shot. But then again, my gut feeling for the definition of probability relies on repetitions of trials and I think that probabilities are ambiguous for experiments that by definiton cannot be repeated.

A halfer intuition?

This was posted Jamie Dreier on 2000/02/14.
Before Beauty is put to sleep, she is given a ticket. The ticket is either of kind A or of kind B. A tickets are worth $5 iff the coin lands heads. B tickets are worth $5 if the coin lands tails. Tickets are cashed only once, of course, not once per awakening.

We then proceed in the usual way.

Upon awakening, Beauty cannot remember which kind of ticket she got. Does she hope that she got an A ticket, or a B ticket, or is she indifferent between them?


This was posted Joe Keane 2000/02/15.
Sally is taking part in an experiment.

On Saturday, she is led into a room and put to sleep with a drug.

On Sunday, two doctors, Mike and Tim, each flip a fair coin. They do not know the result of each other's flip.

However, they both show their results to a third doctor, Bill. Bill has a box, and he puts a new-style $100 bill in the box if the two coin-flip results are the same, otherwise the box is empty.

On Monday, if Mike's coin flip came up tails, he wakes up Sally, introduces himself, and asks her a question.

On Tuesday, if Tim's coin flip came up tails, he wakes up Sally, introduces himself, and asks her a question.

In both cases, the question is this:

``What is your credence that the box has money in it?''

The effects of the drug ensure that if Sally is woken up on Tuesday, she doesn't know if she was woken up on Monday, and vice versa.

On Wednesday, Sally is woken up, given the contents of the box, if any, and sent on her way.

Sally knows this whole procedure. She is very smart and she understands conditional probability at least as well as David Moews.

What does she answer, if Mike asks?

What does she answer, if Tim asks?

Thirders answer "1/2". Halfer opinions vary.

A Halfer Argument by Michael Hochster

This was emailed to me by Michael Hochster on 1999-08-02.
Here is the statement of the halfer position that seem most plausible to me.

When Beauty awakes, her experience is the same regardless of whether the coin is heads or tails. Therefore, her awakening cannot provide any new information about whether the coin came up heads or tails. Since her credence is 1/2 before the experiment, it must be 1/2 during the experiment.

[Since I am not a halfer, I must disagree with this somewhere. My disagreement starts at the word "therefore."]

This is similar to Jim Ferry's initial response to the original Beauty posting. Jamie Dreier has also posted in an apparently sophisticated way in favor of halfism, but I never really understood his argument.

And finally: there is a version of halfism which is not usually made explicit by halfers which says that all probability statements (including Beauty's statements of credence) must refer to the randomness *in the coin* only. Using this probability space, the only three possibilities for credence are 0, 1/2, and 1, so the answer is clearly 1/2. This position appears odd, but it is actually in accord with much of modern statistical practice.


A Halfer Argument by Aaron Bergman

TitleRe: Request: Original statement of Sleeping Beauty puzzle
Posted byAaron Bergman
In article <>, Nick Wedd wrote:
>This archive is still lacking a convincing statement of the halfer
>position. I tried to write one myself, but failed dismally. So I would
>appreciate it if anyone would send me one. (It need not convince me,
>but it has to be accepted by halfers.)

Here's one argument:

Given no information about the coin toss, the probability is 1/2 by definition. In order for this probability to change, sleeping beauty has to be given information that distinguishes the cases in any way. However, due to the parameters of the problem, she cannot distinguish the heads case from the tails case by simply being woken up. Therefore, she does not have any information as to the result of the coin toss, so the probability she must state is 1/2.

(while sleeping beauty being woken multiple time may allow others to gain information from the waking, her memory loss does not allow her to do so)


A Halfer Argument by Bill Vanyo

I received this by email from Bill Vanyo on 1999-10-07.

A slight variation on the original puzzle:

After the experiment is over (Monday afternoon if it was heads, Tuesday afternoon if it was tails) the experimenters will tell SB "The experiment is now over" and then ask her "What is the probability the coin landed heads?" She doesn't yet know if it's Monday or Tuesday at that point. Even the 1/3'ers agree she answers 1/2 to this question. There are 2 equally likely possibilities – it's Monday, or it's Tuesday, corresponding (given that the experiment is over) to heads or tails.

SB knows she will be asked this question when the experiment is over, and she knows throughout the experiment that her answer will be 1/2. So, when they wake her up, she knows that later (when the experiment is over) she will be answering 1/2. And she knows that she will then have no less information then than she does now. If she knows that she will absolutely be answering 1/2 later, when she has as much information or more information than she has now, how can she answer 1/3 now?

Some people argue that, when they wake her up, she does not yet have the information that the experiment is over, so she can't answer 1/2 yet. But she knows that she will definitely get that information, and that when she does she will conclude 1/2.

It's one thing to say that later I will know the answer, and another to say that later I will know that the answer is 1/2. If she knows that later she will know the answer is 1/2, then she knows it is 1/2 now.

Tell me what you think.

Bill Vanyo

A Halfer Argument by Charles Bryant

TitleRe: Was SB ever solved?
Posted byCharles Bryant
In article , Nick Wedd wrote:
>Here it is, in the original words of Jamie Dreier (I don't suppose he
>will thank me for reviving it):
>We plan to put Beauty to sleep by chemical means, and then we'll flip a
>(fair) coin. If the coin lands Heads, we will awaken Beauty on Monday
>afternoon and interview her. If it lands Tails, we will awaken her
>Monday afternoon, interview her, put her back to sleep, and then awaken
>her again on Tuesday afternoon and interview her again.
>The (each?) interview is to consist of the one question: what is your
>credence now for the proposition that our coin landed Heads?
>When awakened (and during the interview) Beauty will not be able to tell
>which day it is, nor will she remember whether she has been awakened
>She knows the above details of our experiment.
>What credence should she state in answer to our question?

I think I can see why it's so difficult to agree on the solution. There are two reasonable interpretations and people answer based on one interpretation only, leading to confusion and controversy. Here are the two interpretations:

1) If she is offered an evens bet on the state of the coin how should she assess this. The answer here is to regard the probability of tails as 2/3. This is because she sees tails twice but heads once.

2) If she is supposed to produce an objective assessment of the probability that the coin landed tails, she should regard it as 1/2. In this scenario she may be asked the same question again, but obviously the same question deserves the same answer.

Interestingly enough, the second scenario occurs in real life (or, rather, a more realistic equivalent). First, we assume that SB uses deterministic reasoning to produce her answer, so whatever it is she can record it in advance and we needn't bother waking her at all, we can simply consult the answer once or twice as approprate.

Now consider a meteorologist on days when she decides the chances of rain tomorrow are 1/2. Assume this is a perfect predicition. She uses this information to make the weather forecast she presents on TV after the main evening news. Obviously she tells the public that the probability is 1/2. But suppose the broadcast is recorded and the TV station has a policy that every day there is rain they play the previous day's forecast as part of a programme discussing the accuracy of weather forecasts. Should the meterorologist now change forecasts to say 2/3 chance of rain, on the grounds that when it does rain the broadcast gets twice the exposure?

Eppur si muove

A Halfer Argument by Greg Detre

This was told to me by Niall Cardin on 2000/04/27.

A fair coin is to be tossed without your seeing it, and you will be asked for your credence that it has landed tails. If it lands heads, you will be asked once; tails, you will be asked twice, but the second time you will be compelled to give the same answer as the first time. You know all this in advance.

The answer is obviously 1/2.

This is effectively the same as SB's scenario.

Three Thirder Arguments by Matt McLelland

TitleRe: Sleeping Beauty Archives?
Posted byMatt McLelland
Solomon Wright wrote:

> Sorry, I missed most of the Sleeping Beaty thread, and most of the stuff I
> did catch seemed a bit complicated without the foundations. Can anyone
> point me in the right direction for decent, clear arguments for both the
> halfist and thirdist positions? (I've already seen the questions on Nick
> Wedd's page, what I want now is some answers to check my ideas against)

Here is something I posted a while back, explaining the (IMHO) three most convincing arguments for 1/3. Attached at the end are some details for the objective advisor setup. [From the postings of mine on Nick's page, you could probably tell that I was a halfer. I am now a thirder.]

Conditionalization Argument
Summary: After waking, if Beauty learns that the day is Monday, then she should think P[heads]=1/2. (There is a simple argument for this based on the fact that knowledge of the day was exactly the piece of information she lost by the drug, and she thought P[heads] was 1/2 before she lost that information). On the other hand, if she learns that the day is Tuesday, then she should obviously think P[heads]=0. If she doesn't know the day, then the probability must lie somewhere in-between.

Objective Advisor Argument
Summary: We can arrange for Beauty to meet with an objective advisor to discuss the probability of the coin toss. We can do it in such a way that the advisors presence can't influence Beauty's assessment of the coin toss, while the advisor, through Beauty being awake, does gain information supporting a particular outcome of the toss. Through standard probability techniques the advisor will compute that the probability of heads is only 1/3, and advise Beauty as such. If the situation is setup correctly, this can be the case even if Beauty and the advisor disclose every piece of information they have to each other. *For the record, all attempts to get Beauty to meet with an advisor who thinks that P[heads]=1/2 have resulted in Beauty getting new information. (and that information always allows her to agree with P[heads]=1/2)

New Information Argument
Summary: The situation can be modified so that Beauty does seem to receive new information. Instead of waking her more times when the coin lands tails, instead wake her the same number of times regardless of the toss outcome, always without memory of any other wakening. Clearly, by symmetry, she should think that heads and tails are just as likely upon waking. So, suppose that we wait five minutes and then put her back to sleep unless the coin landed tails or it is her first day up. When she finds that she has not been put back to sleep she has new information which leads her to P[heads]=1/3. Further, the cases in which she is awake after five minutes are exactly the cases in the original problem.

-- Details for the Objective Advisor argument --

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.

A Thirder Argument by Wei-Hwa Huang

TitleAnother Beauty argument
Posted byWei-Hwa Huang

Suppose we didn't erase Beauty's memory, and she knows this. Everything else remains the same as in the previous problem -- on heads, she gets wakened on Monday; on tails, she gets wakened on Monday and Tuesday.

Obviously, when we ask her the question on a wakening, she will either think:

  1. "Well, I don't remember being wakened, so it must be Monday. This means that the probability of the coin being heads was 1/2."
  2. "Well, I remember being awakened yesterday, so today is Tuesday and the probability of the coin being heads is 0."
Now, suppose we DO erase her memory but don't tell her. Then she will always think it is Monday, and she will always answer 1/2.

And finally, if tell her we're going to erase her memory, we revert to the original problem (and note that it doesn't matter if we actually erase her memory or not): A halfer SB will think:

"My memory's been erased, so I don't know what today is for sure. But by such-and-such an argument, the probability is 1/2."
And a thirder will think:
"My memory's been erased, so I don't know what today is for sure. But by such-and-such an argument, the probability is 1/3."

Obviously the answer is different between the two problems because of the added memory wipe and SB's knowledge of it.

The halfer, then, thinks that Sleeping Beauty's knowledge of the erasure is irrelevant, while the thirder seems to think it is.

A Thirder Argument by Nick Wedd

I wrote this myself on 1999-07-23. I was trying to state the halfer position as clearly as I could manage, despite being a thirder. The result was that I made my thirdism very clear to myself.

Halfer: When Beauty goes to sleep (for the first time), she knows that the coin is tails with p=1/2. She also knows that when she next wakes she will not know what day it is, and will not have gained or lost any information.
Thirder: Yes she will. She will have lost the information that it is Monday.
Halfer: I meant, "She also knows that each time she wakes within the experiment, she will not know what day it is, and will not have gained or lost any information."
Thirder: "Each time she wakes within the experiment", the coin will be tails with p=2/3. This is because each of the three awakenings is equally likely to occur, and the coin will be tails for two of them.

A Thirder Argument by Chuck Gaydos

I received this by email from Chuck Gaydos on 1999-10-07.

The probability that the coin landed heads cannot be one half.

Assume that it is one half. Repeat the experiment a million times. On each awakening, Beauty bets one dollar, even money, that the coin was tails. If the probability is one half then she'll break even in the long run. But she won't break even in the long run. On each heads she'll bet once and lose. On each tails she'll bet twice and win twice. In the long run she'll be about a half million dollars ahead. So the assumption that the probability is one half is not correct.

A Thirder Argument by Tom Slater

I received this by email via Niall Cardin on 2006-06-05

First of all, I'm going to make it so that if it's a Head, then Sleeping Beauty is awoken on either Monday OR Tuesday with equal probability, but definitely not both. I can't see how anyone could object to that (but please do if you think it's wrong). It's somewhat necessary to make everything neat and symmetrical.

Next we'll examine a further different problem, as above, but where you are told the day when you are woken up. Let's see what Sleeping Beauty thinks.

Case 1: It's a Monday.

One of two things has happened. It was a tail (original probability 50%), or it was a head and Monday was selected (original probability 25%).

Case 2: It's a Tuesday.

One of two things has happened. It was a tail (original probability 50%), or it was a head and Tuesday was selected (original probability 25%).

So in our new problem Sleeping Beauty concludes that it was a tail 2/3 of the time, and a head 1/3 of the time, whatever she is told about the day. Now if we step back to the modified problem, she wakes up but is not told the day. She thinks to herself... "If I had been told the day, I would give the answer 1/3 WHATEVER I was told. Here I haven't been told, but should that bother me?"

A Thirder Argument by Bill Spight

I received this by email on 2013-02-21

Let's simplify the problem so that if the coin comes up heads, Beauty is not asked what is the probability (for her) that it came up heads, but if it came up tails, she is. The correct answer is obvious, but it is also obvious how to derive it via Bayes Theorem. You just do the same with the original problem. :)  Beauty's amnesia means that the conditions of the probability are the same each time she is asked, so her answer is always the same.

A Thirder Argument by Jeff Jordan

I received this by email on 2019-02-21

There is a simple transformation that allows the Sleeping Beauty Problem to be solved without ambiguity. Run the experiment with four volunteers instead of one. The same drugs are used to induce sleep and amnesia. Each volunteer will be wakened once, or twice, based on the same coin flip and a schedule selected specifically for her. Each will be asked for her assessment of the probability, or her confidence, or whatever you want to call it (they are all the same thing) that she will be wakened exactly one time during the experiment.

SB1 will be wakened each day, except Tuesday if the coin landed on Heads. This is the same schedule as in the original problem. And since “the coin landed Heads” and “SB1 will be wakened exactly one time during the experiment” represent the same event, no matter how you look at it, the question is the same as well.

SB2 will be wakened each day, except Tuesday if the coin landed on Tails.

SB3 will be wakened each day, except Monday if the coin landed on Heads.

SB4 will be wakened each day, except Monday if the coin landed on Tails. Obviously, the same answer is correct for each of the volunteers. And just as obviously, the correct answer here is correct for the original problem. But now there is a new methodology for finding it.

No matter which volunteer I am, if I am awake, I know that there are two others who are also awake. And one who is asleep. Of the three who are awake, two will be wakened on both days, and one will trade (or has traded) places with the one who is currently asleep. But none have any information that would allow her to say she is more, or less, likely to be in either category that the other two who are awake now. So each has a 1/3 chance to be the one who will be wakened once, and a 2/3 chance to be one of the two who will be wakened twice.

The answer is 1/3.

PS: It doesn’t matter if you tell the volunteers which schedule they have, or not. It also doesn’t matter if you bring the three who are awake together, as long as they are forbidden to reveal their own schedule to the others.

A Thirder Argument by Dennis Rager

I received this by email on 2021-04-06.

First, consider a case where she only wakes up on Monday and regardless of the coin outcome she is not put back to sleep. Her credence would obviously be 1/2 as there are only two possible outcomes: {heads, Monday} and {tails, Monday}.

Now let's return to the original case, keeping in mind that she doesn't know what day it is, only that she could possibly be answering this question on Monday or Tuesday. If she were to be answering on Monday, there would only be the original two possible outcomes, meaning the probability would be 1/2. However if she were answering on Tuesday, the probability would become 1/3 as there are now three possible outcomes {heads, Monday}, {tails, Monday}, and {tails, Tuesday}. While she may not be aware of how much time has passed, she is aware that time is passing and that the amount of time that has passed can impact her answer. Because of this knowledge, she would answer according to which day she is most likely to be asked upon. As there are two outcomes where she is asked on Monday, she would give her credence to be the answer that would be correct on Monday, that is 1/2.

Even if you transform the original question into one where, in the case of tails, she is put to sleep again each day and asked again each day repeated indefinitely, the outcome would not change. This is because on Wednesday the answer would be 1/4, on Thursday it would 1/5, and so on and so on. Each of those days would have a unique answer while 1/2 would be true in both outcomes on the original monday. Meaning that responding with 1/2 would have 2/(x +1) odds of being correct while 1/3 and every other possible answer would only have 1/(x+1) odds where x equals the number of days that the experiment is allowed to continue. This means that responding with 1/2 would always give Sleeping Beauty the greatest odds of being correct.

An Argument for Ambiguity by Saar Wilf

This was emailed to me by Saar Wilf on 2002/01/14.

Defining the Sleeping Beauty Paradox

Saar Wilf

The sleeping beauty paradox was first presented as described above.

Some people feel the answer is 1/2 ('halfers') and some think it's 1/3 ('thirders'). Despite much debate, no one yet provided a clear-cut answer, which is quite strange considering that this seems like a problem in the mathematical domain.

Some people believe the question is ambiguous or doesn't define accurately what SB's goal is in answering the question. A statement of this view is given on the Grey Labyrinth site. Nevertheless, like the halfer and thirder view, this view has not been proven, and the paradox is still considered unsolved.

I will attempt to clearly prove below that the question is indeed ambiguous, and hopefully put the Sleeping Beauty Paradox to sleep.

In order to expose the ambiguity, let's consider the question SB is asked:

What is your credence now for the proposition that our coin landed Heads?

The key word is 'credence', and we should accurately define it. A person's credence in an event is his estimate of the probability that the event has occurred (or will occur). So we now have:

What is your estimate now of the probability that our coin landed Heads?

Next we should define probability. The relevant definition in Merriam-Webster is:

The ratio of the number of outcomes in an exhaustive set of equally likely outcomes that produce a given event to the total number of possible outcomes

In more professional texts it's simply described as "the long-run relative frequency of the event". So what we're actually asking SB is:

If we run this test repeatedly, what is your estimate of the relative frequency of Heads from among all possible coin outcomes?

Replacing 'relative frequency' with a more clear wording:

If we run this test repeatedly, what is your estimate of the fraction of times the coin would have landed Heads from among all possible coin outcomes?

Now that we clearly defined the question the ambiguity becomes clear: the term 'times' is not well defined. It can either be 'times you are awakened' (leading to the answer of 1/3) or 'times we run this test' (leading to 1/2).

This is the heart of the paradox – some people consider the whole experiment as the basic unit, while others feel more comfortable considering each awakening separately. While some may feel that one view is superior to the other, this is a matter of philosophy and not mathematics. A definite answer cannot be reached.

The ambiguity is also exposed by our inability to agree on a definitive test for SB. While some suggest replacing the question 'what is your credence...?' with a bet on every awakening, others offer to provide her with a ticket that wins $5 on tails and ask her how much the ticket is worth on every awakening. In any normal situation, every reasonable test would lead to the same result (and the problem doesn't remain unsolved for years...). We need a very unique setting involving memory-erasing drugs to distort our intuition and leave such an ambiguity unnoticed.

Appendix – The Absentminded Motorist Paradox

The Absentminded Motorist was the first version of the SB paradox and is considered to be logically equivalent to it. As I explain below, this is simply incorrect and DM belongs to a completely different family of paradoxes.

This paradox is stated above.

In fact, AM is a recursive reasoning paradox (my term). It creates an infinite reasoning loop that doesn't seem to reach a definite solution. It is similar to Newcomb's paradox where you try to make a decision after an all-knowing being tried to predict it.

To understand how AM is similar we simply need to go further with AM's reasoning: After he decides he should exit, he must immediately rethink this decision considering that he knows himself to reason in that way: he is inclined to enter at the first exit, so this can't be the second exit (he never reaches it). But now, if this is the first exit then he should drive on to the next exit – and we're back to square one.

It is clear that the problem here is different and cannot be solved for different reasons than the SB paradox.

An Argument for Ambiguity by James Fingas

This was emailed to me by James Fingas on 2005-05-12.

Sleeping Beauty: time of decision clarifies question

James Fingas

The key difference between the answers 1/2 and 1/3 is when the decision is made. Simply this:
  1. If SB were going to predict heads versus tails on Sunday, then remember her answer when she wakes up, then probability of her predicting correctly is 1/2, regardless of whether she a priori decides it will be H or T.
  2. This corresponds to picking the answer when waking up, and using the measure of correctness "getting the right answer at least once". The appropriate probability in this case is likewise 1/2.
  3. Contrast that to SB deciding heads versus tails only at the moment she wakes up. There are three waking points, each equally likely, so if she wants the highest degree of success at that instant, then the probability of having heads is indeed 1/3.
  4. The corresponds to the measure of correctness "getting it right as many times as possible". The betting calculations used for defending the 1/3 position are done in this way, and the answer is truly 1/3, maximizing the number of times that she will be correct.
The difference between these two answers is that in the 1/2 case, you reason that she will make the same decision both times, so she will be correct in her assertions in 1 out of every two experiments. In the 1/3 case, you reason that she can make decisions independently each time she wakes up, so she would be correct in asserting the coin heads in 1 out of every 3 wakings.

Notice that this is fundamentally the same argument that Saar Wilf makes. Correlate "getting it right at least once" to the frequency "per experiment", and "getting it right as many times as possible" to the frequency "per waking"

Here is a betting scenario that favours the halfers:

The bets are set up so that Sleeping Beauty gets $36 if she predicts the throw correctly at least once.

If she has deterministic thought processes, and must therefore answer the same in both cases, then the payout will be $18 regardless of which choice she makes (thirders take note).

However, if she can make truly random guesses at each waking (flip of a coin, perhaps!), then she will get the best payout by using the probability of 1/2, which gives her a payout of $22.50 (5/8 of $36). Using the probability of 1/3 would give her a payout of $22.00 (11/18 of 36).

So the "halfers" are just as justified as mathematically justified as the "thirders".

Rhetorically, is SB two different people the two times she wakes up? If she is correct in asserting "Tails" on Monday, does it help to parrot her answer "Tails" on Tuesday? Does success "stack"? Therein is the rub.

For the million clones scenario, consider whether it is more important to save one genetic copy of yourself, or more important to save as many genetic copies of yourself as possible, as these two considerations give different answers.

This page last updated significantly 2021-05-18

If you have any corrections or additions to make to this page, please email Nick. A convincing statement of the halfer case will be particularly welcome.

Nick's home page.