We turn cards simultaneously from two well shuffled decks. Exact match means same card from each deck. Find probability of at least one exact match.

Question

Problem Suppose we turn over cards simultaneously from two well shuffled decks of ordinary playing cards. We say we obtain an exact match on a particular turn if the same card appears from each deck; for example, the queen of spades against the queen of spades. Let $p_M$ equal the probability of at least one exact match.

• Show that $$p_M=1-\frac{1}{2!}+\frac{1}{3!}-\frac{1}{4!}+...-\frac{1}{52!}$$ Hint: Let $C_i$ denote the event of an exact match on the $i^{th}$ turn. Then $p_M = P(C_1 \cup C_2 \ldots C_{52})$ Now use the the general inclusion-exclusion formula. Note that $P(C_i) = \frac1{52}$. And, hence $p_1=52(1/52)$.

• Show that $p_m$ is approximately equal to $1-e^{-1}=0.632$

I do not understand why $p_1=52(1/52)$, could someone explain to me please? Thank you very much.

Answer 1

I think $$p_1 = \sum\limits_{i=1}^{52} \mathbb{P}(C_i) = \sum\limits_{i=1}^{52} \frac{1}{52} = 52 \cdot \frac{1}{52} = 1$$

Answer 2

Here is a different way of thinking about it.

A derangement is an arrangement with no cards in the same position. If we don't get a derangement then at least 1 card matches.

And the probability we get a derangement is the number of derangement over the number of arrangements $\frac {!52}{52!}$

And the probability that at we don't get a derangement is $1 - \frac {!52}{52!}$

How do count derangements?

First we count all of the arrangements the deck. Then we subtract all of the arrangements that fix at least 1 card $-{52\choose 1}(51!)$ But we have counted arrangements that fix two cards twice, so we must add one set back in $+{52\choose 2} 50!$ this is "exclusion-inclusion," and we must keep passing this along for all numbers of fixed cards.

$!52 = 52! - {52\choose 1}51! + {52\choose 2} 50! + \cdots + (-1)^k{52\choose k}(52-k)! + \cdots = \sum_\limits{k=0}^{52} (-1)^k\frac {52!}{(52-k)!}$

$\frac {!52}{52!} = \sum_\limits{k=0}^{52} (-1)^k\frac {1}{k!} \approx e^{-1}$

$1-\frac {!52}{52!} = 1-\sum_\limits{k=0}^{52} (-1)^k\frac {1}{k!} = \sum_\limits{k=1}^{52} (-1)^{k-1}\frac {1}{k!} \approx 1- e^{-1}$