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Bayes and Coin Toss?

November 10th, 2020, 11:38 pm

Hi -

Suppose someone tossed 100 coins and then covered them up. They ask you how many heads you expect, and absent other info you say 50. Suppose now that this person uncovers 20 of the 100 coins and shows you 20 heads. Note that they don't randomly uncover 20 coins which happen to be all heads - they purposefully show you 20 heads. So it's as if we have a crystal ball that gives us new info which is, "there are at least 20 heads out of the original 100 coin tosses". Do you change your estimate of how many heads in total and how? I'm having some trouble constructing Bayes rule for this particular example - thanks for your help!
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Re: Bayes and Coin Toss?

November 11th, 2020, 3:07 am

Well, if your prior is that the coin tosses are iid with probability of a head is one half, then the probability of at least 20 heads are in the neighborhood of .9999999995, so you haven’t learnt much. So, up to rounding, 50 still seems like a good number.
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Re: Bayes and Coin Toss?

November 11th, 2020, 9:31 pm

The Bayes' rule was constructed in the 18th century and still applies. No need to construct a new one.
If the person cherry-picked the 20 heads, you get bearish's answer in the following way:
A - probability of tossing 50 heads exactly
B - probability of tossing at least 20 coins
P(A|B) = P(A&B) / P(B) = P(A) / P(B)

P(A|B) - P(A) =  4.440717e-11 according to an R console i've found in the Internet.