# What is the probability of matching numbered letters to numbered envelopes?

I have been given the following problem:

There are $$n$$ envelopes and $$n$$ letters, randomly assigned. Let $$A_i =$$ the event where the $$i$$th letter goes to the $$i$$th envelope (a match). Then $$P(A_i)=\frac{(n-1)!}{n!}, P(A_i \cap A_j)= \frac{(n-2)!}{n!},...,P(A_i \cap ... A_n)=\frac{(n-n)!}{n!}=\frac{1}{n!}$$

1. Find the probability of at least one match

2. Find the probability of no matches

To get a feel for the problem, I let $$n=10$$ and started evaluating probabilities. For example, when $$n=10$$, $$P(A_i)=\frac{(10-1)!}{10!}=\frac{9!}{10!}=\frac{1}{10}$$. Substituting $$n$$ back in for $$10$$ gives $$P(A_i)=\frac{1}{n}$$. Sure enough, as I increased the number of matches while doing this I confirmed that $$P(A_i \cap ... A_n)=\frac{(n-n)!}{n!}=\frac{1}{n!}$$.

Now, I know that $$P$$(no matches) = $$1-P($$at least one match) so answering either of these questions will make answering the other one trivial. My question though, is this: is it correct to think of $$P(A_i)=\frac{1}{n}$$ as the probability that at least one envelope matches or is $$P(A_i)=\frac{1}{n}$$ the probability that exactly one envelope matches? If it is the latter, how do I go about answering question 1?

No, $$P(A_i)=\frac 1n$$ is the probability that a given envelope matches its letter. Permutations where no items are fixed are called derangements. For large $$n$$ the probability is $$\frac 1e$$. $$10$$ is rather large, so you will be close. Even for smaller $$n$$, if you round $$\frac {n!}e$$ you get the right answer.