0

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?

Anne
  • 2,931
  • Use the law of total probability to define each $\Pr[X_i]$. You need the sum of the conditioned probabilities, an you get that $\Pr[X_i]=1/52$. Check this. – Masacroso Jan 22 '17 at 17:11
  • Sorry, I mean each $E[X_i]$, not each $\Pr[X_i]$. – Masacroso Jan 22 '17 at 17:20
  • by saying ``3/52 for the second" you are assuming you got a match on the first. – Daniel Xiang Jan 22 '17 at 17:22
  • @Masacroso to have a match I need an ace as the first card and there are four aces in the deck. For this reason I have a probability of 4/52 to have a match and so on – Anne Jan 22 '17 at 17:32
  • @Anne, yes, I think you are right, probably the result of the book is wrong or incomplete. Anyway the other expectations are wrong. Observe that the $X_i$ are not independent variables. The expectations are $E[X_i]=4/52$, hence it addition is $4$. – Masacroso Jan 22 '17 at 17:57

2 Answers2

5

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 $\Sigma E(X_i) = \frac4{52}\cdot 52 = 4$

true blue anil
  • 41,295
0

Here is where you went wrong in your reasoning, you assumed that in order to have a match you need an ace in the first position and you have 4, so 4/52 chance, which is correct; however, on the next possible match position you will have 39 cards not 52 so there is a 3/39 chance of another match, and so on. There is an inherent symmetry here, exactly like the hat problem--if you are familiar with it.

Ben Cross
  • 1