The Sleeping Beauty Problem

This probability problem was presented to the newsgroup rec.puzzles in the spring of 1999 by Jamie Dreier. It originated from a male graduate student at MIT who was considering a similar problem called The Case of the Absent-minded Driver. It’s interpretation strikes at the very heart of the meaning of mutually exclusive events and those who commit to a view on the problem usually fall into one of two groups : halfers or thirders, corresponding to the probability answer they arrive at. The original problem, for which this article shall argue the halfer case, is as follows.

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 about the above details of our experiment. What credence should she state in answer to our question?

THE RELATIVITY OF PROBABILITY. Probability is person-related in that estimations of the probability of an event might be different for different people, depending on their state of knowledge. To illustrate this, consider the following scenario. Two friends, A and B, are shown three identical boxes whose lids are covered in dust. They are blindfolded and told that a coin is to be placed in one of the boxes. Once this is done, the blindfolds are removed and they are each asked to estimate the probability that the coin is in Box 1. Person A states one third. However, person B notices that the dust on the lids of Boxes 1 and 2 has been disturbed whereas that on the lid of Box 3 has not, indicating to him that the lid of Box 3 has not been raised. This narrows his choices to two boxes so he states one half. One might ask, who has estimated correctly? The answer is that both have answered correctly according to their knowledge at the time. Our interest is in what probability estimation is available to Sleeping Beauty (SB).

WHAT SLEEPING BEAUTY KNOWS. SB knows that the coin is fair and that as a result the probability of Heads (H) or Tails (T) is one half each. This probability perspective is that of the experimenters before they toss the coin. Once the coin is tossed, the experimenters can assign a probability of zero or one to the H and T. SB, however, must persist with her original perspective unless she acquires additional knowledge to change it. There are two points to consider : (a) Does the structure of the experiment give SB additional knowledge? (b) Can she acquire additional knowledge about the result during the experiment?

KNOWLEDGE FROM THE STRUCTURE OF THE EXPERIMENT. Shown (below) is the probability tree diagram for the experiment which SB knows about. There are two outcomes of the coin throw H or T with probability ½ each and these are shown by the two branches of the tree diagram. If a H is thrown, the Monday awakening A certainly follows with probability 1. If a T is thrown then the Monday awakening B has probability 1 as does the Tuesday awakening C which must follow B.

SB could reason that “Each of the three awakenings is equally likely to occur and the coin will be T for two of them. This gives the probability of H as 1/3.” These, however, are not mutually exclusive events. If the first awakening B occurs then C must follow, and if awakening C occurs then B must have preceded it. The condition for mutually exclusive events of having one without the other is not met here. The two consecutive awakenings are actually ONE composite consequence of throwing T, albeit a larger one (in the temporal sense) than the single H awakening, and so T does not have more than one way of occurring.

There are actually only two mutually exclusive events : the single awakening (A) H thread and the composite awakening (B and C) T thread and they are equally probable. SB can only admit the following from the probability tree diagram : “It doesn’t matter whether I am at B or C, the tree diagram leads back to the same probability ½ for T and so H must be ½ ”. So SB can derive no additional knowledge about the coin throw from the way the experiment is constructed.

KNOWLEDGE ACQUIRED DURING THE EXPERIMENT. There is no way of SB knowing which thread she is following, the single H awakening or the composite T awakening. She knows about the details of the experiment so her memory loss is known to her. This means that on awakening she is aware that she could be at any of the three awakenings. Without this knowledge she could only believe that she was at the awakening for H or the first of the two for T. Either way, there is no way for her to determine which thread she is on. Her memory loss also prohibits her from temporal awareness so any of the two awakenings of the T thread would seem identical to the single one of the H thread.

ERRONEOUS ARGUMENT AGAINST THE HALFER POSITION. Consider the following argument which I shall show to be in error. “Assume that the probability that the coin lands H is ½. Repeat the experiment a million times. On each awakening, Beauty bets one dollar, even money, that the coin was tails. If the probability was 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 million dollars ahead. So the assumption that the probability is one half is not correct.” In other words, the probability of T must be different to ½ if the expected return is not zero. This, however, is a bogus argument because the size of the bets are determined by the result and, unknown to SB, every time the desired outcome T occurs then the larger amount is placed. If the amount bet is x, the expected return based on ½ probabilities is then -0.5x + 0.5(2x) = x and cannot be zero. More importantly, though, the results of these bets would not be available to SB.

ERRONEOUS ARGUMENT THAT PROBABILITY CANNOT BE ESTIMATED. Some analysts maintain that since there is no definition as to how the probability may be measured then a solution cannot be given. They argue that it is not clear whether the probability is based on (a) the number of runs of the experiment where the coin appears tails or on (b) the percentage of interrogations for which the coin is tails. However, a probability cannot be found for (b) because it relies on the misapprehension that two awakenings each result in one tail whereas, in reality, one tail results in two awakenings. There is a crucial difference, because the former, which leads to the thirder result, claims that one can manufacture a tail in two ways, whereas the latter tells us that the probability of a tail or head being thrown is already determined before the awakenings and is therefore a half. This is clearly a causal distinction and appears to contain the essence of the problem. Let us illustrate this with the following scenarios. The first leads to the thirder result and shows how the causal condition must be met that the awakenings must determine the event outcomes. Suppose SB was chemically put to sleep as before and woken up once a day over three consecutive days. On two of these days, a red mark is made on a sheet of paper and on the other day a blue one is made. The matching of colours and days can be decided at random beforehand. Here the choice of the day of the awakening decides the colour. SB is aroused on one particular day and asked to estimate the probability that a blue mark is made on the paper that day. Since there are two ways a red can be caused and one way for a blue she can reliably answer one 1/3. Now consider instead the scenario that a blue or red mark, randomly chosen, is to be made on a piece of paper on any particular day, the choice of colour being "the event". If a blue is made, SB is to be aroused once, and if a red is made she is to be aroused twice. Here the event outcomes determine the number of awakenings. What answer can SB now have for the probability of a blue mark? None other than 1/2 because this result is determined before the awakenings occur.

CONCLUSION. The only knowledge available to SB is that a fair coin falls H or T with probability ½ each . She has no opportunity to modify what she knows either from knowledge of the experiment construction or from impressions during the experiment. The fact is, she knows that there is a probability of one half (for T) that she will repeatedly be awoken and a probability of one half (for H) that she will be awoken once. If she elects to answer T on each awakening she knows that two awakenings out of three she will be saying the correct result, however, this is a repeated consequence of an event with probability ½ and has no influence on the probability that it happened. So her credence for the probability that the coin landed H must be ½.