Permutation without fixed points.

Question

We have a deck of 52 playing cards. We can assume an initial order of them; when i mix it, I ask if is more likely that P: "none card remains in the original position" or Q:"not(P)". And if we consider a deck of 40 cards?

Thanks.

Answer 1

If no card is in its original position, the permutation is a derangement of the deck. Let $d_n$ be the number of derangements of a deck of $n$ cards; then

$$d_n=n!\sum_{k=0}^n\frac{(-1)^k}{k!}\;,$$

which for $n\ge 1$ is equal to $\left\lfloor\frac{n!}e+\frac12\right\rfloor$, the integer nearest to $\frac{n!}e$. The probability that a randomly chosen permutation is a derangement is therefore

$$\frac{d_n}{n!}\approx\frac1e\approx0.36788$$

and is certainly less than $\frac12$ for decks of $40$ and $52$ cards. In fact $\frac{d_n}{n!}<\frac12$ for all $n\ge 3$.