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Here is a similar question: Question on the 'Hat check' problem

But the follow-up is, what is the standard deviation of number of people who leave with their own hat? I understand that expectation would be 1/n x n, but how to calculate variance and standard deviation?

RobPratt
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The other things
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    The question in the title ("What is the expected number of people who leave with their own hat?") does not match the question in the body ("What is the standard deviation of number of people who leave with their own hat?"). – Dan May 23 '23 at 05:05
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    The standard deviation is $1$. Here is a long proof. Note that in the link, the $X$ refers to the number of people who do not get their own hat. – Dan May 23 '23 at 05:27
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    @Dan FYI A slightly quicker way to calculate the SD is to use indicator variables: $Var(\sum I_i) = \sum Var(I_i) + \sum Cov (I_i, I_j) = 12 \times ( 11/12^2) + 132\times (1/(11*12^2) ) = 1. $ (for the specific case of $n = 12$). – Calvin Lin May 23 '23 at 15:09

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Let $X_i$ be the random variable which is 1 when person $i$ leaves with their own hat and 0 otherwise. Then $$E[X_i]= 1\cdot P(X_i=1)=\dfrac{1}{n}$$ and $$\mathrm{Var}[X_i]= 1^2\cdot P(X_i=1)-E[X_i]^2=\dfrac{1}{n}-\dfrac{1}{n^2}=\dfrac{n-1}{n^2},$$ while $X=\sum_{i=1}^nX_i$ is the random variable which counts the number of people who leave with their own hats.

Now the $X_i$ are not independent, but still $E[X]=\sum_{i=1}^n E[X_i] = 1$.

However, for $\mathrm{Var}[X]$ you need the result $$\mathrm{Var}[X]=\sum_{i=1}^n \mathrm{Var}[X_i]+ 2\sum_{i=1}^{n-1} \sum_{j=i+1}^n \mathrm{Cov}[X_i,X_j]= n\mathrm{Var}[X_1]+ n(n-1) \mathrm{Cov}[X_1,X_2].$$ and calculate $$\mathrm{Cov}[X_1,X_2]= 1^2\cdot P(X_1=1 \;\mathrm{\&}\; X_2=1)-E[X_1] E[X_2]=\dfrac{1}{n(n-1)}-\dfrac{1}{n^2}=\dfrac{1}{n^2(n-1)},$$ so finally $\mathrm{Var}[X]= \dfrac{n-1}{n}+ \dfrac{1}{n} =1$.

mcd
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