On Sunday, Sleeping Beauty is put to sleep and then a fair coin is tossed. On Monday, Sleeping Beauty is awakened and interviewed. She is then put to sleep again with an amnesia-inducing drug that makes her forget that awakening. On Tuesday, if the coin landed tails she is awakened, interviewed and put to sleep once more. Otherwise, if the coin landed heads, she is left asleep. On Wednesday, she is awakened and the experiment ends.
When she is awakened and interviewed she does not know which day it is or whether she has been awakened before. During the interview, Beauty is asked, "What is the probability that the coin landed heads?" The experiment is illustrated below.
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| Sleeping Beauty problem (illustration by Stuart Armstrong) |
There are two popular but opposing solutions that are commonly given by philosophers. The solution known as the thirder position is that Sleeping Beauty should report that the probability of heads is 1/3. [1] The intuition is that Sleeping Beauty cannot distinguish between the three awake states, so she should assign the same probability to each state. Also, if the experiment is run many times, Sleeping Beauty is awakened on average in a state where the coin has landed heads 1/3 of the time and in a state where the coin has landed tails 2/3 of the time.
The halfer position is that that Sleeping Beauty should report that the probability of heads is 1/2. [2] The intuition here is that Sleeping Beauty seems to learn no new information when she is awakened. So she should not update on the probability that she held on the Sunday before the experiment began. That is, it is the coin toss that is the relevant event, not how many times she is awakened. [3]
So which position is correct?
Before the experiment starts, Sleeping Beauty knows that during the experiment she can be in one of the four following states with equal probability.
She also knows that she will not be awakened on Tuesday if the coin lands heads, so the probability for that state is 0 when she is awakened. The question then is how the other probabilities should be updated.
Monday Tuesday Heads 1/4 1/4 Tails 1/4 1/4
The halfers distribute the probability to the remaining heads state, as follows:
From the table:
Monday Tuesday Heads 1/2 0 Tails 1/4 1/4
P(Heads and Monday) = 1/2
P(Tails and Monday) = 1/4
P(Monday) = P(Heads and Monday) + P(Tails and Monday)
= 1/2 + 1/4 = 3/4
Using the formula for conditional probability: [4][5]
P(Heads|Monday) = P(Heads and Monday) / P(Monday)
= 1/2 / 3/4 = 2/3
This, I think, reveals the fatal flaw with the halfer position. If Sleeping Beauty is told it is Monday, then she will think the probability for the coin landing heads is 2/3!
The thirders distribute the probability evenly to all awake states, as follows:
From the table:
Monday Tuesday Heads 1/3 0 Tails 1/3 1/3
P(Heads and Monday) = 1/3
P(Tails and Monday) = 1/3
P(Monday) = P(Heads and Monday) + P(Tails and Monday)
= 1/3 + 1/3 = 2/3
Using the formula for conditional probability: [6]
P(Heads|Monday) = P(Heads and Monday) / P(Monday)
= 1/3 / 2/3 = 1/2
So, if Sleeping Beauty is told it is Monday, then she will think the probability for the coin landing heads is 1/2, which intuitively seems correct.
But the thirder position seems to imply that Beauty has learnt new information since Sunday (when the probability for heads was 1/2) which causes her to update her probability for the coin landing heads to 1/3. What could that new information be?
That it is both Tuesday and that the coin landed heads is a possible state for Beauty to be in when she is asleep. So she learns that she is not in that state when she wakes up. As a consequence, she updates by excluding that state and normalizing the probabilities for the remaining awake states.
One further test for the thirder position is to consider the probabilities when Sleeping Beauty is awakened on, say, 1000 consecutive days when the coin lands tails (instead of two consecutive days). [7] The probability of the coin landing heads when Beauty is awakened is then 1/1001. This gives:
P(Heads) = 1/1001
P(Tails) = 1000/1001
P(Heads and Monday) = 1/1001
P(Tails and Monday) = 1/1001
P(Monday) = P(Heads and Monday) + P(Tails and Monday)
= 1/1001 + 1/1001 = 2/1001
P(Heads|Monday) = P(Heads and Monday) / P(Monday)
= 1/1001 / 2/1001 = 1/2
As with the original experiment, if Sleeping Beauty is told that it is Monday, then she will think the probability for the coin landing heads is 1/2 which (as before) intuitively seems correct. Note that there are still 1000 states that Beauty can be in as a result of the coin landing heads. But she will just not be awake when she is in 999 of those states so they are excluded from her awake state calculations.
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[1] The thirder position was initially introduced by Adam Elga in this paper.
[2] The halfer position was defended by David Lewis in his reply to Elga.
[3] The thirders interpret the problem as an experiment about awakening events (i.e., which branch Beauty will find herself on) whereas the halfers interpret the problem as an experiment about a coin toss event (i.e., which branch Beauty will be placed on). But both sides agree about what odds should be accepted if Beauty is offered a bet that the coin landed heads. If she is offered a bet whenever she is awakened, she should accept 1/3 odds. If she is only offered a bet once, she should accept 1/2 odds.
P(A|B) = P(A and B) / P(B)[5] The probability calculations for Sleeping Beauty when awake according to halfers:
P(Heads) = 1/2
P(Tails) = 1/2
P(Heads and Monday) = 1/2
P(Heads and Tuesday) = 0
P(Tails and Monday) = 1/4
P(Tails and Tuesday) = 1/4
P(Monday|Heads) = P(Heads and Monday) / P(Heads) = 1/2 / 1/2 = 1
P(Tuesday|Heads) = P(Heads and Tuesday) / P(Heads) = 0 / 1/2 = 0
P(Monday|Tails) = P(Tails and Monday) / P(Tails) = 1/4 / 1/2 = 1/2
P(Tuesday|Tails) = P(Tails and Tuesday) / P(Tails) = 1/4 / 1/2 = 1/2
P(Monday) = P(Heads and Monday) + P(Tails and Monday) = 1/2 + 1/4 = 3/4
P(Tuesday) = P(Heads and Tuesday) + P(Tails and Tuesday) = 0 + 1/4 = 1/4
P(Heads|Monday) = P(Heads and Monday) / P(Monday) = 1/2 / 3/4 = 2/3
P(Heads|Tuesday) = P(Heads and Tuesday) / P(Tuesday) = 0 / 1/4 = 0
P(Tails|Monday) = P(Tails and Monday) / P(Monday) = 1/4 / 3/4 = 1/3
P(Tails|Tuesday) = P(Tails and Tuesday) / P(Tuesday) = 1/4 / 1/4 = 1
[6] The probability calculations for Sleeping Beauty when awake according to thirders:
P(Heads) = 1/3
P(Tails) = 2/3
P(Heads and Monday) = 1/3
P(Heads and Tuesday) = 0
P(Tails and Monday) = 1/3
P(Tails and Tuesday) = 1/3
P(Monday|Heads) = P(Heads and Monday) / P(Heads) = 1/3 / 1/3 = 1
P(Tuesday|Heads) = P(Heads and Tuesday) / P(Heads) = 0 / 1/3 = 0
P(Monday|Tails) = P(Tails and Monday) / P(Tails) = 1/3 / 2/3 = 1/2
P(Tuesday|Tails) = P(Tails and Tuesday) / P(Tails) = 1/3 / 2/3 = 1/2
P(Monday) = P(Heads and Monday) + P(Tails and Monday) = 1/3 + 1/3 = 2/3
P(Tuesday) = P(Heads and Tuesday) + P(Tails and Tuesday) = 0 + 1/3 = 1/3
P(Heads|Monday) = P(Heads and Monday) / P(Monday) = 1/3 / 2/3 = 1/2
P(Heads|Tuesday) = P(Heads and Tuesday) / P(Tuesday) = 0 / 1/3 = 0
P(Tails|Monday) = P(Tails and Monday) / P(Monday) = 1/3 / 2/3 = 1/2
P(Tails|Tuesday) = P(Tails and Tuesday) / P(Tuesday) = 1/3 / 1/3 = 1
[7] A similar example is used by Nick Bostrom in this paper intending to show that the thirder position is counter-intuitive. I disagree, for the reasons I give in my conclusion. Bostrom also makes several other arguments against both the halfer and thirder positions and defends a hybrid position.
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