Discrete Math Question

Question

Suppose you are playing 5-card draw poker- the deck has the usual 52 cards and you are dealt 5. What is the chance that you are dealt three aces (but not 4 and not an additional pair)? Suppose you are dealt three aces and in order to improve your hand you discard the other two cards and draw two more in the hope of improving your hand-you might wind up with either four aces or three aces and a pair. What is the chance of that happening?

Answer 1

The denominator is the total number of hands you could have: ₅₂C₅

The numerator is the number of ways you could have exactly 3 aces: ₄C₃ ways to pick your three aces times the other cards: ₁₂C₂ for the number of ways to pick their ranks (Ranks are A,K,Q, etc. There are 13 ranks in total, so 12 ranks that are not aces, you pick two of them, since you're not allowed to pick another ace or a pair) times 4 for each one. So: (₄C₃*₁₂C₂*4^2)/(₅₂C₅).

The odds for improving the hand by replacing the unmatched cards gets tricky.