Suppose we have a normal deck of $52$ cards. We shuffle them well and then turn over the first $13$ cards one-by-one. If the first card is one of the four aces we say that a match has occurred; similarly, if the second card is one of the twos; the third card is one of the threes, etc.; until the $13$th (one of the kings).

What is the expected number of matches?

my solution: $${\frac{4}{52}} + \frac{4}{51} + ...$$ and add until the $13$th draw.


2 Answers 2


One of the things that is difficult to grasp about expectations is that they are additive. Let $M_i$ be the number of matches in the position $i$ - so that $M_i$ is either $0$ or $1$.

Then $E(M_i)=\frac{4}{52}=\frac{1}{13}$.

Now, the total number of matches is $M=M_1+\cdots+M_{13}$. So, by additivity, $$E(M)=E(M_1+\cdots + M_{13}) = 13\cdot\frac{1}{13}=1$$

This additivity property is often confusing, because it seems to miss the conditional probabilities. It doesn't, but it takes some effort to understand why.


By Linearity of Expectation the expected number of matches is the sum of the expectation of getting a match in each position.

$$\dfrac{4}{52} \times 13 = 1$$

The counterintuitive fact of the Linearity of Expectation is that it does not require independence of the summed random variables.   The expectation operator is linear.

$$\mathsf E(X+Y) = \mathsf E(X) + \mathsf E(Y) \\[2ex] \mathsf E(\sum_i X_i) = \sum_i \mathsf E(X_i)$$


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