### The Sleeping Beauty Problem

Rather than explain it I'll point you here. It looks a bit like the Monty Hall problem at first. But the Monty Hall problem is simple once you use the right method - and almost everyone agrees when it's pointed out to them. But philosophers can't actually agree on the solution to the Monty Hall problem.

There isn't really any mathematical content. But it does say something interesting about the *interpretation* of mathematics. In particular, what exactly do we mean when we say the probability of an event X is p? Turns out that the meaning becomes quite subtle when you know your memory is unreliable. Anyway, follow the links to find out more.

Are you a halfer or a thirder?

Labels: mathematics

## 3 Comments:

Wow - I hadn't realized so much had been written outside all the philosophy literature! I've got a friend who's been working on an argument that should generalize to other probabilities for sentences like "today is Monday", and only a couple of the papers he's cited are listed on that other site.

Years ago, when I was a student, I was worried about sleeping through the alarm clock. It had a snooze button which allowed you to easily reset the alarm to go off a few minutes later up to three times. But if you were really tired in the morning you might wake, hit the snooze button, and immediately fall back to sleep and forget you had done so. At the time I realized there were some paradoxical probability questions you could ask under such circumstances but I never really made the effort to work out the details. I'm really surprised to find that this type of problem is actually taken quite seriously by so many people! I'll have to see if I can find a nice reformulation of the problem in terms of snooze buttons that seems less contrived than the sleeping beauty version.

A new take on an old problem - to take my mind off work.

Lets we look at the cost of making the wrong answer. If P(heads) less than P(tails) then this cost should be minimised by always choosing tails.

Suppose that every time you wake up you give a value p between 0 and 1. If the coin came up heads you get p £1000. Otherwise you get (1 - p) £1000. The expected return is then 1 - .5p which is minimised by always giving p=0 ie always guessing tails.

Heres the catch though - if they only give you this deal the first time you are woken up then the expected return is one.

So

P(heads) = 1/2

P(heads | first time ive woken up) = 1/2

P(heads | ive woken up) < 1/2 ie = 1/3

But for some reason this answer still bothers me.

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