neq1 comments on Beauty quips, "I'd shut up and multiply!" - Less Wrong

6 Post author: neq1 07 May 2010 02:34PM

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Comment author: rwallace 07 May 2010 04:39:35PM 7 points [-]

This is one of those cases where we need to disentangle the dispute over definitions (1), forget about the notion of subjective anticipation (2), list the well-defined questions and ask which we mean.

If by the probability we mean the fraction of waking moments, the answer is 1/3.

If by the probability we mean the fraction of branches, the answer is 1/2.

Comment author: neq1 comment score below threshold [+] (8 children)

Comment author: PhilGoetz 07 May 2010 06:06:45PM * 1 point [-]

On a random waking moment, Monday preceded by heads is equally likely as Monday preceded by tails.

I think you're thinking of a similar problem that we discussed last year, which involves a forgetful driver who is driving past 1 to n intersections, and needs to turn left at at least one of them. That problem is different, because it's asking about the probability of turning left at least once over the course of his drive.

Comment author: Vladimir_Nesov 07 May 2010 09:21:57PM 1 point [-]

The absent-minded driver is essentially the same problem, but it's easier to analyze because explicit payoff specification prompts you to estimate expected value of possible strategies. In estimating those strategies, we use the same probability model that would say "1/2" in the Beauty problem.

Comment author: neq1 comment score below threshold [+] (5 children)

Comment author: timtyler 09 May 2010 02:19:24PM * 1 point [-]

How come P(monday and heads) and P(monday and tails) are not the same? This is an ordinary unbiased coin, yes?

How come P(monday and tails) and P(tuesday and tails) are not the same. Nothing happens in the interim, yes?

Comment author: JGWeissman 07 May 2010 07:13:04PM 1 point [-]

Before she wakes, the probabilities SB would assign if she were conscious are P(monday and heads) = P(monday and tails) = p(tuesday and heads) = p(tuesday and tails) = 1/4.

After waking, she would update to p(tuesday and heads) = 0 and P(monday and heads) = P(monday and tails) = p(tuesday and tails) = 1/3, since p(tuesday and heads | wakes up) = 0 and p(monday and heads | wakes up) = p(monday and tails | wakes up) = p(tuesday and tails | wakes up) = 1.

Comment author: neq1 07 May 2010 08:47:15PM -1 points [-]

Ugh. That makes no sense. Can you explain why she would update in such a manner?

Comment author: JGWeissman 07 May 2010 08:57:49PM 2 points [-]

Odds(monday and heads : monday and tails : tuesday and heads : tuesday and tails | wakes up)
= Odds(monday and heads : monday and tails : tuesday and heads : tuesday and tails) * Likelihood(wakes up | monday and heads : monday and tails : tuesday and heads : tuesday and tails)
= (1 : 1 : 1 : 1) * (1 : 1 : 0 : 1)
= (1 : 1 : 0 : 1)
= (1/3 : 1/3 : 0 : 1/3)

SB starts out with four equaly likely possibilities. On observing that she wakes up, she eliminates one of them, but does not distinguish between the remaining possiblities. Renormalizing the probabilities gives probability 1/3 to the remaining possibilities.

neq1 comments on Beauty quips, "I'd shut up and multiply!" - Less Wrong

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