2
$\begingroup$

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.

$\endgroup$

2 Answers 2

8
$\begingroup$

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.

$\endgroup$
4
$\begingroup$

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.

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

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .