In 1997, a problem called The Absent Minded Driver appeared in a paper published in Games and Economic Behaviour (Ref. 1). Its authors, Michele Piccione and Ariel Rubinstein, had invented the problem in 1994 with the intention of illustrating how beliefs could be determined in situations of imperfect recall. In March 1999, it was posted, with minor alterations, in the newsgroup rec.puzzles by Jamie Dreier who, a few days later, posted a similar problem called The Sleeping Beauty Problem. The interpretation of this latter problem 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?
Illustrator : Warwick Goble

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).

 

KNOWLEDGE ACQUIRED DURING THE EXPERIMENT. There is no way of SB knowing which thread she is following, the single H awakening or the double 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 be indistinguishable from the single one of the H thread.

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. She also knows that if a tail is thrown she will be awoken twice. However, she has no way of deciding which of these two awakenings is occurring.

ASSIGNING PROBABILITIES TO UNKNOWNS. Whatever Beauty does not know she must assign a probability value to (see diagram below). At any time she is awoken, she does not know whether the coin has fallen a head or tail. If it were just a simple question of a coin toss, with no falling asleep, she would give a probability of 1/2 to each outcome. Now let us put her to sleep. Suppose that the coin falls a tail and she is informed on being awoken that the coin is a tail. She still does not know whether she is in the first (B) or second (C) awakening. In fact, the probability that she could guess her correct awakening of these two given this supposition is 1/2 (this probability being based on the equal division of a two-day time interval). If the coin fell as a head and she were told of this fact there is only one awakening which occurs with probability 1 (outcome A).

It is easy to compute the composite probabilities available to SB. The probability that the coin falls a head and she knows which awakening she is at is P(H)P(A)=1/2x1=1/2. The probability the coin falls a tail and she is at the first awakening is P(T)P(B)=1/2x1/2=1/4. The probability the coin falls a tail and she is at the second awakening is P(T)P(C)=1/2x1/2=1/4. So the probability of a head is the first alternative 1/2 and the probability of a tail is the sum of the second and third ones 1/2.

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 and here state of knowledge has not changed since then.


CONCLUSION. The only knowledge available to SB is that a fair coin falls H or T with probability 1/2 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, these awakenings have a different weight, 1/4 each, to the single H awakening of 1/2. This fact refutes the thirder position. So her credence for the probability that the coin landed H must be 1/2.
REFERENCES

 

I am grateful to Tim Edmonds for pointing out this problem to me.