Cards from an ordinary deck of 52 playing cards
are turned face up one at a time. If the 1st card
is an ace, or the 2nd a deuce, or the 3rd a three,
or . . ., or the 13th a king, or the 14 an ace, and so
on, we say that a match occurs. Note that we do
not require that the (13n + 1)th card be any particular
ace for a match to occur but only that it be
an ace. Compute the expected number of matches
that occur.
My attempt is:
Let $X_i$ denote the indicator variable for the event that the i-th card is matched, 0 otherwise.
$N=\sum_{i=1}^{52} X_i$=number of matches that occur
$$\begin{align}E[N]=E[\sum_{i=1}^{52} X_i]&=\sum_{i=1}^{52} E[X_i]\\&=E[X_1]+E[X_2]+…E[X_{52}]\\&=\frac{4}{52}+…\frac{4}{52}+\frac{3}{52}+….\frac{3}{52}+\frac{2}{52}+…+\frac{2}{52}+\frac{1}{52}+…\frac{1}{52}\\&=\frac{4}{52}*13+\frac{3}{52}*13+\frac{2}{52}*13$+\frac{1}{52}*13\\&=\frac{10}{4}\\&=2,5\end{align}$$
because the probability of having a match is $\frac{4}{52}$ for the first 13 cards, $\frac{3}{52}$ for the second 13 cards, $\frac{2}{52}$ for the third 13 cards, $\frac{1}{52}$ for the last 13 cards.
But in my book the final result is 4 and the probability is $\frac{1}{52}$ for all the 52 cards.
Why?
Best Answer
Imagine all the 52 cards arranged in a row.
As you have said,
let $X_i$ denote the indicator variable for the event that the i-th card is matched, $0$ otherwise.
By linearity of expectation, we have $E\Sigma (X_i) = \Sigma E( X_i)$
The expectation of an indicator random variable is just the probability of the event it indicates, so $E(X_i) = P(i_{th}\; card\; matches) = \frac4{52}$,
and $\Sigma E(X_i) = \frac4{52}\cdot 52 = 4$