Players A and B each have a well shuffled standard pack of cards, with no jokers. The players deal their cards one at a time, from the top of the deck, checking for an exact match. Player A wins if, once the packs are fully dealt, no matches are found. Player B wins if at least one match occurs. What is the probability that player A wins?
Since player A is dealing from a shuffled (randomized) pack, the probability that A wins is independent of the order in which B's cards are dealt. So, without loss of generality, we can assume B's cards are dealt in order: 1, 2, 3, ... , 52. Therefore the probability that player A wins is the fraction of permutations of (a1, a2, ... , a52) for which ai
i, for all i from 1 to 52. Such permutations are known as derangements.
Let d(n) be the number of derangements of n elements. Then, by the Inclusion-Exclusion Principle,
| d(n) = | (total number of ways to deal n cards) |
| − sum over i (number of deals for which ai = i) | |
| + sum over distinct i, j (number of deals for which ai = i and aj = j) | |
| − sum over distinct i, j, k (number of deals for which ai = i, aj = j and ak = k) | |
| ± number of ways in which ai = i, for all i from 1 to n | |
| (with the final sign dependent on the parity of n) |
Here, we start with n! deals, subtract those with one matching card, then add back the number with two matching cards (we just counted these combinations twice), and so on.
| So d(n) | = n! − nC1 · (n−1)! + nC2 · (n−2)! + ... ± nCn · (n−n)! |
| = n! − n!/1! + n!/2! + ... ± (−1)n |
Therefore, the probability that A wins is d(52) / 52! = 1 − 1/1! + 1/2! − 1/3! + ... + 1/52!
Note that this expression is the first 53 terms of the Maclaurin series for e−1. The series converges very rapidly to 1/e; the above probability is within 1/53!
2.34×10−70 of 1/e. Therefore, somewhat surprisingly, for any reasonably large number of cards, say, 10 or more, the probability that A wins is almost independent of the number of cards in the decks.
Source: Traditional