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I came across the following thought experiment, and I would like to understand whether the controversy around it is justified.

Imagine an experiment in which a mathematician is put to sleep with some kind of drug. He is located in a room that is designed in such a way as to keep him completely isolated from any kind of external information. The researchers have a sleep inducing drug that is able to put you to sleep and make you forget it was even administered. After the researchers have put the mathematician to sleep with this drug, they toss a fair coin. If it comes up heads they will wake the mathematician up once and administer the drug again. If it comes up tail they will wake him up twice, each time administering the drug again.

Whenever the mathematician is awoken during the experiment, they will ask him for his best guess regarding the result of the coin toss. Eventually the experiment ends, and the researchers will awaken the mathematician a final time and tell him the experiment has ended. During the experiment, what answer should the mathematician give as his best guess for the result of the coin toss ?

I think he should say that odds are the coin came up tail, but I am very curious what other people make of it, and whether there are any grounds for dissent at all.

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5 suggests that the problem is philosophical in nature and has yet to be fully resolved. – Qiaochu Yuan Jan 28 '12 at 19:56
thank you ! ideally I m looking for a more rigorous defence of the equal odds position (or refutation more likely). I don t find equal odds argument in that article convincing ! – Beltrame Jan 28 '12 at 20:00
See also here – Count Iblis Nov 18 '14 at 22:13

Give him a flower if he is awoken after heads come up and a beer every time he is awoken after tail came up. If you ask him before he gets his gift, he should always guess that he will get a beer (with $p=\frac{2}{3}$ he will get one).

If the coin has been tossed 1000 times, he will have approximately 500 flowers and 1000 beer.

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This argument is tempting but is debunked (well, at least attacked) in Nick Bostrom's paper ( on the subject. – Qiaochu Yuan Jan 28 '12 at 20:15
Hm, clearly we don't have to tell the mathematician that we give him beer and flowers. If best guess means anything at all, then we should count the number of right guesses in the long run? Otherwise it's pure psychology, and the mathematician should be asked "how sure are you that..." – Blah Jan 28 '12 at 20:24
Have you read the paper? The discussion of the long-run argument begins on page 12 and is subtle so I don't want to summarize it too briefly here. – Qiaochu Yuan Jan 28 '12 at 20:26
Quite ununderstandable, the question there is: When she awakes on Monday, what should her credence be that the coin will fall heads? – Blah Jan 28 '12 at 20:30

What follows is perhaps only a non-mathematical rambling, but let me expose some thoughts I just have had.

Assume you are betting on a soccer match: Barcelona versus Madrid. For all you can tell, Barcelona and Madrid each can win the match with probability 1/2. However, you encounter a very peculiar broker, who offers you the following bet:

  • if Madrid wins, you get 1 euro;
  • if Barcelona wins, you get 2 euros.

What is your best guess ? Meh, both teams have the same chance. What bet is the best ? Definitely Barcelona.

Perhaps that you can just replace Madrid by "heads" and Barcelona by "tails". Both outcomes have the same probability, but choosing one of the bet can maximize the expected number of good answers. So, basically, all boils down to what you consider as your "best guess" or "credence": the probability of each event, or the guess which gives you more good answers.

I have the feeling that the paper of Nick Bostrom plays on this ambiguity, but I need to read it more carefully to be sure (or not). [Edit : actually he raises the exact same point]

I am sorry if this anwer does not fit the requirements of math.stackexchange. If deemed so, I'll delete it (but this is quite a fascinating problem - I should at least find somebody to discuss it with).

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The solution to the question as worded is 50%. I will now explain why there are two common answers to this problem.

We know that the coin will be heads 50% of the time and tails 50% of the time. We know that in the event of heads, the man is interviewed 1 time. In the event of tails, the man is interviewed 2 times.

We arrive at two different solutions depending on how we define an experimental success. If an experiment is defined as a single interview, then one should guess tails, because 2/3 of interviews will be when the coin has landed on tails.

If however we allocate a single experiment to each coin flip, then guessing tails in two consecutive interviews only counts as 1 success. Therefore, despite guessing correctly in 2/3 of the interviews, you are only successful in 50% of the experiments.

The way the question is worded above, it indicates that an experiment is for a single coin toss, and therefore it matters not whether one guesses heads or tails.

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It's amazing how many probability "paradoxes" fall apart if one pins down exactly what constitutes an "experiment". The "correct" answer to the Monty Hall problem, for example, depends upon whether an correct initial guess will always, never, or sometimes result in the contestant being given a chance to switch, and likewise for an incorrect initial guess. If those things are unstated, the probability that switching is the correct choice of action may be anywhere from 0% to 100%. – supercat Feb 22 '14 at 18:17

I'm addressing the question "I would like to understand whether the controversy around it is justified." I'll do it by introducing a (backwards) analogy.

If you roll a pair of standard dice, what is the probability of getting a sum of 7? (This is meant to be easy: 1/6).

If you roll the dice twice, what is the probability of getting 7 on at least one roll? (Not as easy, but still straight forward: 1/6 + 1/6 - 1/36 = 11/36. The subtraction is because 7's on both rolls are counted twice otherwise.) The important point is that it is not 1/6.

But what if you roll the dice one-and-a-half times? While that seems to be impossible, it is exactly what happens with Sleeping Beauty. She is wakened an expected one-and-a-half times. If she rolls the dice when she is wakened (backwards from the SB problem, where the random selection is done once), what is the probability she will roll 7 (A) this time, or (B) over the course of the experiment? The (A) answer is still 1/6, but the (B) answer is more complicated. I'm not going to address it because the value is irrelevant. The only point of using the analogy is that the answer is not 1/6.

Going back to the question, if you believe that there can be only one logical answer, and everybody should arrive at that logical answer, then no controversy is ever justified. But to answer unorthodox riddles, you need to use unorthodox methods. Controversies can justifiably arise when the unorthodoxy is not clear, as it is in this problem. But the existence of the controversy means you need to address its cause, which almost nobody does.

Some try to answer question (B) for Sleeping Beauty, and get 1/2 because the probability over the course of the experiment is 1/2. But an awake SB can only address question (A). The answer to (A) is 1/3, because it applies to her making her evaluation one-and-a-half times. And you can't claim to have solved the problem unless you address what that means.

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Here is a write up. The answer is $\frac{1}{3}$ with the heads/tails case reversed. There is also an interesting "day" angle to it.

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Each time that Sleeping Beauty is asked what the probability is that the coin came up heads, we may write that probability P(Heads | Asked, Background). That is, what is the probability, given the fact that she is asked the question, plus any other background information she has, that the coin came up heads? The problem is constructed so that Asked and Background are the same in each possible case. Sleeping Beauty should give the same answer in each case, and a straightforward application of Bayes Theorem will give the answer.

Bostrom rejects the application of Bayes Theorem on two possible grounds, I think. (I have not bothered to wade through all his reasoning. ;)) First is that it is possible to reject the idea that one's credence (if numerical) will tell you how to bet (assuming that you must always bet). That is true, but then we have veered away from probability as it is normally understood. Second is that the application of Bayes Theorem gives the same answer as the Self Indication Assumption in anthropic reasoning. Well, so what? Beauty is not duplicated in the problem as given.

He then goes into questions about agent parts and so on, which he believes are relevant. These would fall into the Background category, but I do not see how he gets different backgrounds for the different cases. Anyway, based upon such questions, he thinks that the application of Bayes Theorem to this problem has undesirable consequences. To get to where he is you have to agree that such questions are relevant, and that the consequences he sees are undesirable. Moi, I do not agree that those questions are relevant, and if the consequences of assuming that they are are undesirable, then don't assume that they are.

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