N holdem hands (just 2 cards from a standard 52 cards deck) are dealt to N players
How to compute the probability that:
-exactly k players have a pair ( Two cards of the same value e.g.: 7, 7).
-at least k players have a pair
The case $n=1$ is simple. The probability that the (one and onl player has a pair is $3/51=0.0588\dots$.
For more than one player, one method of computation is simulation. The following Mathematica code simulates $10^8$ hands and counts the number of hands with $k$ pairs, $0\le k\le n$, for $1\le n\le10$.
The problem seems simpler that I thought.The probability that any player has a pair is constant (3/51=0.0588). Hence,if I am right: