Probability of Four of a Kind in a 5-Card Poker Hand
A standard deck has 52 cards: 4 suits and 13 ranks (Ace through King). You are dealt a 5-card hand uniformly at random. What is the probability that your hand contains four of a kind -- exactly four cards sharing the same rank, plus one card of a different rank?
Hints
- The hand has a fixed structure: four cards of one rank plus one card of a different rank. Count by choosing the quad rank, then the kicker.
- Once you fix the quad rank, there is exactly 1 way to select all 4 cards of that rank; the kicker can be any of the remaining 48 cards.
- Divide by $\binom{52}{5} = 2{,}598{,}960$ to convert the count to a probability.
Worked Solution
How to Think About It: This is a counting problem. The probability is (number of favorable hands) / (total number of 5-card hands). Count favorable hands by choosing: (1) the rank that appears four times, (2) all four cards of that rank, and (3) the fifth card from the remaining deck. Everything is straightforward once you set up the structure.
Quick Estimate: There are 13 ranks and each rank gives $48$ fifth-card choices. So roughly $13 \times 48 = 624$ favorable hands. Total hands: $\binom{52}{5} \approx 2.6 \times 10^6$. Probability $\approx 624 / 2{,}598{,}960 \approx 0.00024$, or about 1 in 4{,}165$.
Formal Solution:
Total 5-card hands:
$$\binom{52}{5} = \frac{52!}{5! \cdot 47!} = 2{,}598{,}960.$$
Favorable hands: A four-of-a-kind hand has the structure $\{A, A, A, A, B\}$ where all four $A$'s share one rank and $B$ is any card of a different rank.
- Choose the four-of-a-kind rank: 13 ways.
- Take all 4 cards of that rank: 1 way (there is only one way to take all four).
- Choose the fifth card (the "kicker"): must be from the remaining $52 - 4 = 48$ cards (any rank except the four-of-a-kind rank, any suit). That is 48 ways.
Favorable hands: $13 \times 1 \times 48 = 624$.
Probability:
$$P(\text{four of a kind}) = \frac{624}{2{,}598{,}960} = \frac{1}{4{,}165} \approx 0.0240\%.$$
Alternative count (as a sanity check): Instead of picking the quad rank first, pick the kicker card first: - 52 choices for the kicker. - The quad must be one of the 12 remaining ranks (not the kicker's rank): 12 ways. - Take all 4 cards of that rank: 1 way.
This counts each hand once (the kicker is uniquely identified), giving $52 \times 12 = 624$. Same answer.
Answer: $P(\text{four of a kind}) = \dfrac{624}{2{,}598{,}960} \approx 0.0240\%$ (roughly 1 in 4{,}165).
Intuition
Poker hand probability problems are a staple interview question because they test structured counting, not formula memorization. The key skill is setting up the right combinatorial framework: identify the structure of the target hand, count each component independently, multiply, then divide by the total. The two-way count (fix the quad rank vs. fix the kicker) is a good self-check -- if both routes give 624, you have not double-counted.
The deeper lesson is about choosing the right anchor for your count. In the quad-rank-first approach, the five cards are naturally partitioned into a group of four and a singleton, which maps cleanly to the hand structure. In more complex hands (full house, flush), finding the right partition is the crux of the problem. Getting comfortable with structured counting under combinatorial constraints is directly applicable to quant work: from counting valid order book states, to enumerating scenarios in a Monte Carlo, to calculating the number of ways a portfolio can satisfy multiple constraints simultaneously.