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timtyler comments on Beauty quips, "I'd shut up and multiply!" - Less Wrong
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Re: "P(monday and heads)=1/2. P(monday and tails)=1/4. P(tuesday and tails)=1/4. Remember, these have to add to 1."
Yes, but those Ps are wrong - they should all be 1/3.
My assumptions and use of probability laws are clearly stated above. Tell me where I made a mistake, otherwise just saying "you're wrong" is not going to move things forward.
Well, the correct sum is this one:
"Suppose this experiment were repeated 1,000 times. We would expect to get 500 heads and 500 tails. So Beauty would be awoken 500 times after heads on Monday, 500 times after tails on Monday, and 500 times after tails on Tuesday. In other words, only in a third of the cases would heads precede her awakening. So the right answer for her to give is 1/3. This is the correct answer from Beauty's perspective."
That gives:
P(monday and heads)=500/1500. P(monday and tails)=500/1500. P(tuesday and tails)=500/1500.
You appear to have gone wrong by giving a different answer - based on a misinterpretation of the meaning of the interview question, it appears.
So you are not willing to tell me where I made a mistake?
P(heads)=1/2, p(monday | heads)=1. Which one of these is wrong?
You're using expected frequencies to estimate a probability, apparently. But you're counting the wrong thing. What you are calling P(monday and heads) is not that. There is a problem with your denominator. Think about it. Your numerator has a maximum value of 1000 (if the experiment was repeated 1000 times). Your denominator has a maximum value of 2000. If the maximum possible values of the numerator and denominator do not match, there is a problem. You have an outcome-dependent denominator. Try taking expectation of that. You won't get what you think you'll get.
Re: "If the maximum possible values of the numerator and denominator do not match, there is a problem.
The total possible number of awakenings is 2000.
That represents all tails - e.g.:
P(monday and heads) = 0/2000; P(monday and tails) = 1000/2000; P(tuesday and tails) = 1000/2000;
These values add up to 1 - i.e. the total numerators add up to the commonn denominator. That is the actual constraint. The maximum possible value of the numerator in each individual fraction is permitted to be smaller than the common denominator - that is not indicative of a problem.
Oh, it is a huge problem. It proves that your ratio isn't of the form # of events divided by # of trials. Your ratio is something else. The burden is on you to prove that it actually converges to a probability as the number of trials goes to infinity.
Using cell counts and taking a ratio leads to a probability as the number of trials goes to infinity if you have independent draws. You don't. You have a strange dependence in there that messes things up. Standard theory doesn't hold. Your thing there is estimating something, you just don't know what it is
The total number of events (statements by Beauty) adds up to the total number of trials (interviews).
You should not expect the number of statements by beauty on Monday to add up to the total number of interviews alltogether. It adds up to the number of interviews on Monday. This is not very complicated.
Do you have to make a condescending remark every time you respond? You told me things that I already know, and then said "This is not very complicated." Great, but nothing accomplished.
You are using an estimator that is valid when you have counts from independent trials. Coin flips are independent here, but interviews are not. You need to take that into account.
Or they all should be 1/2.
Impossible - if they are to add up to 1.
For Jack's bookie, I agree, you have to use 1/3 – but if you want to calculate a distribution on how much cash Beauty has after the experiment given different betting behavior, it no longer works to treat Monday and Tuesday as mutually exclusive.