Expected Number of Same-Rank Pairs in a Hand Dealt Without Replacement

A deck contains $M$ ranks with $N$ cards of each rank, so it holds $MN$ cards in total. You are dealt $K$ cards from the shuffled deck, without replacement. What is the expected number of pairs in your hand? Count a pair as any two cards in the hand that share a rank, so three cards of one rank contain three pairs. Give an expression in $M$, $N$ and $K$, then evaluate it for the values you are given; the specific numbers are generated randomly. Also say how the answer changes if a pair instead means a rank that appears exactly twice in the hand. ```hint Skip the full distribution You do not need the probability of every possible hand. Look for small events whose indicators add up to the number of pairs. ``` ### Constraints and Clarifications - $M$, $N$ and $K$ are positive integers, and $K$ is at most $MN$. - Every set of $K$ cards is equally likely to be dealt. ### Clarifying Questions - Does three of a kind count as three pairs, as one pair, or as no pair? - Should four cards of one rank count as six pairs, as two disjoint pairs, or as no pair? - Should the answer be exact or rounded to two decimal places? ### What a Strong Answer Covers - A precise definition of "pair", settled before computing - Linearity of expectation over well-chosen indicators, with no independence assumption needed - The correct probability that two cards drawn without replacement share a rank - How the answer changes under the other definitions of a pair, and a sanity check on a small or familiar deck ### Follow-up Questions - What is the expected number of ranks that appear exactly twice in the hand? - What is the probability that the hand contains no pair at all? - How does the answer change if the cards are dealt with replacement?