In 5-card poker, the odds of specific hands reveal the beautiful combinatorics of a 52-card deck.

Problem:
What is the probability of being dealt each of the following 5-card hands?

  1. Four-of-a-kind
  2. Full house
  3. Two pairs

Total Possible Hands

There are:

\[ \binom{52}{5} = 2,!598,!960 \text{ possible 5-card hands} \]

We’ll calculate the number of favorable hands for each case and divide by this total.

1. Four-of-a-Kind

  • Choose the rank for the four-of-a-kind: \(13\) choices
  • Choose 4 suits for that rank: exactly 1 way
  • Choose the fifth card: any of the remaining \(52 - 4 = 48\) cards (but not the same rank)

\[ \text{Favorable hands} = 13 \times 48 = 624 \] \[ \text{Probability} = \frac{624}{2,!598,!960} \approx \boxed{0.000240} \]

2. Full House

  • Choose the rank for the three-of-a-kind: \(13\) choices
  • Choose 3 suits from 4: \(\binom{4}{3} = 4\)
  • Choose a different rank for the pair: \(12\) choices
  • Choose 2 suits from 4: \(\binom{4}{2} = 6\)

\[ \text{Favorable hands} = 13 \times 4 \times 12 \times 6 = 3,!744 \] \[ \text{Probability} = \frac{3,!744}{2,!598,!960} \approx \boxed{0.001440} \]

3. Two Pairs

  • Choose 2 distinct ranks for the pairs: \(\binom{13}{2} = 78\)
  • Choose 2 suits for each pair: \(\binom{4}{2} \times \binom{4}{2} = 6 \times 6\)
  • Choose a third rank (not used above) for the kicker: \(13 - 2 = 11\)
  • Choose any 1 suit for that card: 4

\[ \text{Favorable hands} = 78 \times 6 \times 6 \times 11 \times 4 = 123,!552 \] \[ \text{Probability} = \frac{123,!552}{2,!598,!960} \approx \boxed{0.047539} \]

Final Answers

Hand Count Probability
Four-of-a-kind 624 0.000240
Full House 3,744 0.001440
Two Pairs 123,552 0.047539

Reference