A deck of playing cards is laid out on the table (for example, in a row). On top of each card, a card from another deck is placed. Some cards may have matched. Find:

a) the expected value of the number of matches;

b) the variance of the number of matches.

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a) Let's number the pairs from 1 to $N$ (we don't know how many there are) in the order they lie on the table. Let

the indicator $I_{k}$ be 1 if the two cards in the $k$-th pair are the same, and 0 if the cards in the $k$-th pair are different.

Obviously, $\mathrm{P}\left(I_{k}=1\right)=1 / N$. Therefore, $\mathrm{E}_{k}=1 / N$.

Let $S$ be the number of pairs with a match. It is equal to $I_{1}+I_{2}+\ldots+I_{N}$. Therefore, $\mathrm{E} S=\mathrm{E}I_{1}+\mathrm{E}I_{2}+\ldots+\mathrm{E}I_{N}=N \cdot{ }^{1} / N$ $=1$.

b) $S^{2}=\left(I_{1}+I_{2}+\ldots+I_{N}\right)^{2}=I_{1}^{2}+I_{2}^{2}+\ldots+I_{N}^{2}+2\left(I_{1} I_{2}+I_{1} I_{3}+\ldots+I_{j} I_{k}+\ldots+I_{N-1} I_{N}\right)$, where the sum of all possible different indicator products is in the parentheses.

Obviously, $I_{k}^{2}=I_{k}$, so $\mathrm{E} I_{k}^{2}=\frac{1}{N}$.

Now consider the quantity $I_{j} I_{k}$. It is 1 if the cards match in both the $j$-th and $k$-th pairs. The cards in the $j$-th pair match with probability $1 / N$. If this happens, the probability of the cards in the $k$-th pair matching is ${ }^{1}{ }_{N-1}$.

Therefore, $\mathrm{P}\left(I_{j} I_{k}=1\right)=1 / N(N-1)$. In all other cases, $I_{j} I_{k}=0$. Therefore, $\mathrm{E}\left(I_{j} I_{k}\right)=1 / N(N-1)$.

Thus, $\mathrm{E}S^{2}=N \cdot 1 /{ }_{N}+N(N-1) \cdot{ }^{1} / N(N-1)=1+1=2$, and therefore, $\mathrm{D}S=\mathrm{E}S^{2}-(\mathrm{E}S)^{2}=2-1=1$.

## Answer

a) 1 ; b) 1.

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