But not, I’m afraid, in a good way.

Less Wrong comes down on the wrong side of the “Sleeping Beauty” problem.

Sleeping Beauty volunteers to undergo the following experiment and is told all of the following details. On Sunday she is put to sleep. Afair coin is then tossed to determine which experimental procedure is undertaken. If the coin comes up heads, Beauty is awakened and interviewed on Monday, and then the experiment ends. If the coin comes up tails, she is awakened and interviewed on Monday and Tuesday. But when she is put to sleep again on Monday, she is given a dose of an amnesia-inducing drug that ensures she cannot remember her previous awakening. In this case, the experiment ends after she is interviewed on Tuesday.

Any time Sleeping beauty is awakened and interviewed, she is asked, “What is your credence now for the proposition that the coin landed heads?”

The intent of the problem is presumably that she NOT know what day it is until after giving her answer and since in a sequence of repetitions of the experiment there are twice as many expected wakeups for tails as for heads the probability of head on any given wakeup is 1/3.

(If she’s told what day it is the problem is not interesting. On Monday it’s just a normal coin toss with P(H)=.5,and on Tuesday it’s a coin toss where all the heads were discarded so P(H)=0.)

LessWrong refers to a couple of other arguments for p=1/3 including a gambling argument from Richard Neal and Nick Bostrom’s argument from the extreme case but ends coming down as a “halfer” on the basis of the following:

A probability tree can help with the intuition (this is a probability tree corresponding to an arbitrary wake up day):

If Beauty was told the coin came up heads, then she’d know it was Monday. If she was told the coin came up tails, then she’d think there is a 50% chance it’s Monday and a 50% chance it’s Tuesday. Of course, when Beauty is woken up she is not told the result of the flip, but she can calculate the probability of each.

When she is woken up, she’s somewhere on the second set of branches. We have the following joint probabilities: P(heads, Monday)=1/2; P(heads, not Monday)=0; P(tails, Monday)=1/4; P(tails, Tuesday)=1/4; P(tails, not Monday or Tuesday)=0. Thus, P(heads)=1/2.

The problem with that tree diagram is that it really just amounts to giving a long calculation to prove what she already knows (or at least has been told) – namely that the coin is fair.

If so then of course the P(H)=.5 but that question is even less interesting than the cases in which she is told what day it is.

The actual question we are asked is what credence she should attach to the statement that when she wakes up she will be shown a head, and both Richard Neal’s gambling approach and Nick Bostrom’s extreme case should make it clear that the answer is 1/3.

Or, if you really like tree diagrams, think of it this way:

p=(1/2) Heads
p=(2/3) – it’s Monday <
/ p=(1/2) Tails
Wakeup<
\ p=0 Heads
p=(1/3) – it's Tuesday <
p=1 Tails

This gives p(H)=(2/3)*(1/2)+(1/3)*0=1/3

and p(T)=(2/3)*(1/2)+(1/3)*1=2/3

Part of the difficulty in all this is that what we mean by asking for the “probability” of something is not well defined without a clear specification of the “game” or “experiment” involved. The 1/3 solution is the answer to the question “When you get woken up like this, in what fraction of the cases do you expect to see a head?” – which to my mind is the correct interpretation of “when you get woken up like this what will you take as the probability of seeing a head?”

That is very different from the question “when you get woken up like this what will you think was the probability of getting a head when the coin was tossed?”

Examples like this are why I have little trust in people who discuss probability but toss aside questions like “What is the actual game or experiment in terms of which you are saying that the probability of event E is p?”