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During a period of $n$ days, two persons drink beers, one each. There are $n$ different beers $B_1, B_2, B_3,\ldots, B_n$, where $n\geqslant 1$ is an integer. t Person1 drinks the beers in order, i.e., on the $i$-th day, she drinks beer $B_i$. Person2 takes a uniformly random permutation $a_1, a_2, \ldots, a_n$ of the set $\{1, 2,\ldots, n\}$ and drinks beer $B_{a_i}$ on the $i$-th day. Let $X$ be the random variable whose value is the number of days during which person 1 and person 2 drink the same beer. Determine the expected value $\mathbb E(X)$ of $X$. I know that I should let $X$ be the number of days during which both persons drink the same beer. And $$ X_i = \left\{\begin{array}{rc} 1,&\text{if they drink the same beer on the }i\text{-th day} \\ 0,&\text{if they drink different beers on the }i\text{-th day}\end{array}\right. $$

Then $$\begin{align} \mathbb E[X_i] &= 1\cdot P(X=1) = P(\text{Person 1 and Person 2 drink beer }i) \\&= P(\text{Person1 drinks beer }i)P(\text{Person 2 drinks beer }i) \\ &= 1\cdot P(\text{Person 2 drinks beer }i) = \frac{1}{(n-i)!} \end{align}$$

So, is my approach right? If it's will the next step be: $\mathbb E(X) = n\cdot\frac{1}{(n-i)!}$?

ki3i
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user2847624
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1 Answers1

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The beginning is fine. Let $X_i=1$ if on day $i$ they both drink the same kind of beer, in other words if on Day $i$ Person 2 drinks Beer $i$. Let $X_i=0$ otherwise. Then the number $X$ of days that they drink the same beer is $X_1+\cdots+X_n$. By the linearity of expectation, $E(X)=E(X_1)+\cdots +E(X_n)$.

The probability that on Day $i$ Person 2 drinks Beer $i$ is $\frac{1}{n}$. For Beer $i$ is equally likely to be drunk by her on any one of the days. Thus $E(X_i)=\frac{1}{n}$.

It follows that $E(X)=n\cdot\frac{1}{n}=1$.

Remark: For much more information about this problem, in a less thirst-quenching setting, please see the Wikipedia article on derangements.

André Nicolas
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