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.


Alexander R Pruss said...

This argument may suffer from this problem.

JeffJo said...

There's a better version of this argument. Forget the second coin, just use four volunteers. They know the following details:

On Sunday, the four are isolated from the others in four different rooms, chosen randomly. Each is told she will be wakened at least once, and maybe twice, as indicated on her door. One room has "You will be wakened unless it is Tuesday, after Heads was flipped" written on the *inside* of the door where the volunteer can see it. The other three possible schedules (i.e., {Tue,Tails}, {Mon,Heads}, and {Mon,Tails}) are each written on one of the other doors.

Regardless of the coin flip, on each day three of these volunteers will be wakened and asked "What should your credence in the coin result written on your door?" Since each knows that three are awake, and that only one is awake in a room where the door matches the coin flip, her answer can only be 1/3.

And it doesn't matter if they see the door, or not.