0
$\begingroup$

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)

  • 1
    By "their own hats", you mean "their own kind of hat"?2012-11-02

1 Answers 1

0

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 $\mathbb E(X)=2n/n=2$.

  • 0
    Joriki, For this problem, I think that a twin receiving it's partner's hat back is still counted as the twin receiving his hat back. Does your answer take this into account?2012-11-02
  • 0
    Also, Joriki what is the variance?2012-11-02
  • 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