[PairTheBoard]

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Here is a cooler problem, imo: show that the expected value of the # of people who get their hat back is 1, independent of n.

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Hmm, how about n(1/n)=1!
uA

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Ok, mr genius, show that the outcomes are independent. I know this is "intuitively obvious" but actually providing a proof is an altogether different matter.


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Why do we care about whether or not the outcomes are independent? The question is about expected values, and the expected value of the sum is the sum of the expected values regardless of whether or not the variables are independent.

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n(1/n) = 1/n + 1/n + .... + 1/n (n terms in the sum) where the ith term represents the probability that the ith person receives his hat. If the events were not independent, the 2nd through nth terms could be different, yielding a different sum.

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Actually, the events are not "independent" as the term is normally used. To be "independent" would mean that the probality of me getting my hat stays the same regardless of whether Joe gets his hat or not. But it does change. If Joe gets his hat my chance improves to 1/(n-1) instead of 1/n.

So the concept being talked about is not "independence" as the term is used in probability. I'm not sure it has a name. But it's the same concept as the one we invoke when we say that the unseen cards in other players' hands do not affect my probability of hitting my flush card on the river. With the hats, each person can say ahead of time that he will have 1/n chance of getting his own hat regardless of what place in line he is in when they are handed out. That defines n dependent indicator functions each of whose EV can be added despite their not being independent.

What is the correct term for the concept that unseen cards don't affect my probabilities regardless of whether they are in other players' hands or at the bottom of the deck? I'm not sure. I don't think it's something we ask to be proved all the time though.



PairTheBoard