There are n sets of twins (so in total 2n people) attending a party. Each twin wears an identical hat to his/her twin sibling, and there are n different kinds of hats. Each person gives there hat to the hat clerk at the beginning of a party and at the end of the party they are randomly given a hat back. If x is the random variable denoting the number of people who have their own hats back, what is pr{X=2n} and E(X)
The Hat problem with Twins
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probability
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1By "their ow$n$ ha$t$s", you mean "$t$heir own kind of ha$t$"? – 2012-11-02
1 Answers
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Each person will have their own kind of hat back if there are only swaps between twins. Each pair of twins can either swap or not, for a total of $2^n$ possibilities, whereas there are $(2n)!$ possibilities in all, so $\mathbb P(X=2n)=2^n/(2n)!$. Each person has a probability $1/n$ of getting their kind of hat, so by linearity of expectation $\mathbb E(X)=2n/n=2$.
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0@max: Yes, that's what I meant by "their own kind of hat" -- theirs or their twin's. Otherwise the probability would trivially be $1/(2n)!$. The variance is a bit trickier to calculate; I might do that later when I find the time. – 2012-11-02