Thursday, March 20, 2008

Another argument for thirding in Sleeping Beauty

As usual, a fair coin is flipped on Sunday, without you seeing the result, and then you go to sleep.

Experiment 1 (standard Sleeping Beauty):
Tails: You get woken up Monday and Tuesday. Your memory is erased each time, and you don't know whether it's Monday or Tuesday when you wake up.
Heads: You get woken up Monday but not Tuesday.
Question: What should your credence in heads be when you wake up?

Experiment 2:
As soon as you have fallen asleep, a second coin is tossed. If it is heads, "Monday" is written down on a hidden blackboard in the experimenter's office, and if it is tails, "Tuesday" is written down on that board. You never see that board.
Tails: You get woken up Monday and Tuesday. Your memory is erased each time as in Experiment 1.
Heads: You get woken up on the day whose name is written in the experimenter's office, but not on the other day.
Question: What should your credence in the first coin's being heads be when you wake up?

I now claim (i) in Experiment 2, the answer is 1/3 regardless of how biased the second coin is, and (ii) it follows from (i) that the answer is 1/3 in Experiment 1.

Claim (ii) is intuitively clear. It shouldn't matter whether the heads wakeup day is Monday or Tuesday.

The harder to argue for claim is (i). Here goes. I am now awake. I give a new rigidly-designating name to today. Maybe the way I do it is I pick a bunch of letters at random to form the name (I neglect the probability that on multiple wakeups I'll choose the same name). So, let's say I have named this day "Xhfure". Let A be the following event: The name of the day written on the experimenter's blackboard refers to Xhfure. Note that A is a contingent event and has prior probability 1/2. Let H and T be the events of the first coin being heads or tails respectively. What is the most specific evidence I now have? I submit it is the following: H or (T and A). Let this evidence be E.

So, now I ask: What is P(H|E)? This is an easy calculation. P(H and E) = P(H) = 1/2. P(E) = P(H) + P(T)P(A) = (1/2) + (1/2)(1/2) = 3/4. Thus, P(H|E) = (1/2)/(3/4) = 1/3.

1 comment:

Alexander R Pruss said...

This argument may suffer from this problem.