Given two identical groups with X distinct entities, and randomly selecting Y entities. What are the chances each selection contains at least 1 match? My wife asked a version of this question because she was curious about the probability of getting matching villagers in animal crossing. I was curious to know the generalized method of solving this problem so I can answer her question. In Animal Crossing, there are 397 villagers, but you are only allowed to have 10 at a time. So, knowing that she's limited to 10/397, what's the chances that at least 1/10 is matched with one of my 10?
To give a much smaller example that can be brute forced by hand, consider a pool of 5 Villagers and you can only have 2 on your island. There are 10 combinations of 2 villagers:
AB / AC / AD / AE / BC / BD / BE / CD / CE / DE
From this we can determine that there's a 70% chance that at least one of my 2 would match with one of her 2.
Say I have villagers "B" and "E", I have at least 1 match with 7 of the permutations (and this is true with any group of 2):
AB / AE / BC / BD / BE / CE / DE
 A: As Gribouillis says in the comments the key to doing this computation is to express it as $1$ minus the probability that there are no matches. If you fix your wife's set of $10$ villagers without loss of generality, you can have ${397 \choose 10}$ possible sets of villagers, and in order to have no matches you have to choose your villagers from the ones your wife doesn't have, which can be done in ${387 \choose 10}$ ways. So the probability is
$$1 - \frac{ {387 \choose 10} }{ {397 \choose 10} } \approx 0.227 \dots $$
or about $\boxed{ 23 \%}$.
An easier and rougher version of this calculation which doesn't tell you the exact probability but gives you an upper bound on it, and which you can more easily do in your head on the fly, is the following. There are $10^2$ different pairs (your wife's villager, your villager), and each of them has a probability of $\frac{1}{397}$ of being the same villager. This means the expected number of same villagers is
$$\frac{10^2}{397} \approx 0.252 \dots$$
which is both an upper bound and not that far off from the probability.
In general if there are $n$ possible villagers and you pick $k$ of them the exact answer is
$$1 - \frac{ {n-k \choose k} }{ {n \choose k} } = 1 - \prod_{i=0}^{k-1} \left( 1 -  \frac{k}{n-i} \right)$$
which is a little tricky to analyze in detail, although it can be done. But the expected value computation easily gives $\frac{k^2}{n}$, which is an upper bound, and can be shown to be a pretty acccurate upper bound as long as $k$ isn't too big compared to $n$.
