Last week I went to a talk based on the following paper (http://philsci-archive.pitt.edu/8556/). I didn’t really follow the argument that a quantum many-worlds version made the argument clearer, but it did get me thinking about the problem.

Beauty agrees to the following arrangement. After she goes to sleep on Sunday a coin is tossed. When she wakes up she is asked how likely it is that the coin landed heads – with the following proviso. If it landed heads she is just woken up on Monday, but if it is tails then after being woken on Monday she is hypnotised so she forgets that she has been woken up. She is then woken again on Tuesday with the same question. In no case does she know what day it is when she is woken up.

The paradox is that on Sunday she believes that the coin will land heads with a probability of 50%. She does not gain any new information when she is woken up, but if she makes a bet then the rational thing to do is to bet that is has landed heads with a probability of 33.3%.

When I came out of the talk I was convinced that the ‘correct’ probability had to be 50%. But then I began to construct an argument for this and I changed my mind – I now think that it is 33.3%

Suppose there were two ‘sleepers’ Peter and Paul. In fact they don’t have to sleep, you just arrange as follows. If the coin lands tails you ask each of them separately for the probability (so that the other doesn’t know about it), but if it lands heads then you just ask one of them at random. In this case a probability of 33.3% is certainly correct, but the person being asked has got some new information – the fact that he is being asked.

Suppose now that there is just one person, but the arrangement is that you hypnotise him to make him think that he is Peter or Paul – if the coin lands tails you ask each persona, if heads then just one at random. This seems equivalent to there being two people. Maybe you could just show him his ‘name’ on a card when he is woken up – but what difference could knowing the name make to the probability of the coin landing heads. So this must be just the same as with Beauty. Hence her correct choice of probability is 33.3%.

It seems to me that although in one sense Beauty does not gain any new information, in another she does, just as Peter or Paul gain new information when they are woken up and asked for the probability

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JeffJoThe fallacy in the halfer argument is that SB does, indeed, gain new information. On Sunday, she sees three possible wakings {H-Mon, T-Mon, T-Tue}. Each has a probability of 1/2. They add up to more than 1 because T-Mon and T-Tue represent the same *future*. The universal event is represented by only {H-Mon, T-Mon}.

But during a waking, when she must view the wakings in the *present*, the three possibilities are disjoint. The probability must be updated to reflect the new universal event {H-Mon, T-Mon, T-Tue}: P'(H-Mon) = P(H-Mon)/[P(H-Mon)+P(T-Mon)+P(T-Tue)] = (1/2)/[1/2+1/2+1/2] = 1/3.