Probability of Three of a Kind in a 5-Card Hand
A standard 52-card deck has 13 ranks (Ace through King) and 4 suits. A five-card hand is a three-of-a-kind if it contains exactly 3 cards of one rank and exactly 1 card each of two other distinct ranks (not a full house, not four-of-a-kind).
To the nearest ten-thousandth, what is the probability that a randomly dealt 5-card hand is a three-of-a-kind?
Hints
- Break the count into four independent choices: the rank of the triple, the 3 suits for the triple, the 2 distinct kicker ranks, and the 1 suit per kicker.
- Use $\binom{4}{3} = 4$ for the triple's suits and $\binom{12}{2}$ (not $12 \times 11$) for the kicker ranks -- order does not matter among the kickers.
- The total is $13 \times 4 \times \binom{12}{2} \times 4^2 = 54{,}912$ favorable hands out of $\binom{52}{5} = 2{,}598{,}960$ total.
Worked Solution
How to Think About It: Count the number of valid three-of-a-kind hands and divide by the total number of 5-card hands. The counting breaks into three independent choices: (1) which rank appears three times, (2) which 3 of the 4 suits appear for that rank, and (3) which two other ranks and which suits fill the remaining two slots. The only subtlety is making sure the two 'kicker' cards are from distinct ranks -- otherwise you would be counting full houses.
Quick Estimate: There are 13 ranks and for each we choose 3 suits from 4: $\binom{4}{3} = 4$ ways. Then we need 2 kicker ranks from the remaining 12, and each kicker has 4 suit choices: $\binom{12}{2} \times 4^2 = 66 \times 16 = 1056$. Total three-of-a-kind hands $\approx 13 \times 4 \times 1056 = 54{,}912$. Over $\binom{52}{5} = 2{,}598{,}960$ total hands, that is about $2.1\%$.
Formal Solution:
Total 5-card hands: $$\binom{52}{5} = \frac{52!}{5! \cdot 47!} = 2{,}598{,}960$$
Counting three-of-a-kind hands:
Step 1 -- Choose the rank for the triple: 13 ways.
Step 2 -- Choose 3 suits for the triple: $\binom{4}{3} = 4$ ways.
Step 3 -- Choose 2 kicker ranks (from the remaining 12 ranks, must be distinct to avoid full house): $\binom{12}{2} = 66$ ways.
Step 4 -- Choose the suit of each kicker (4 choices each, independently): $4 \times 4 = 16$ ways.
Total three-of-a-kind hands: $$13 \times \binom{4}{3} \times \binom{12}{2} \times 4^2 = 13 \times 4 \times 66 \times 16 = 54{,}912$$
Probability: $$P(\text{three-of-a-kind}) = \frac{54{,}912}{2{,}598{,}960} \approx 0.0211$$
Answer: $\approx \mathbf{0.0211}$ (to the nearest ten-thousandth).
Intuition
Counting card hands is a prototypical combinatorics problem because each step of the construction is independent and multiplicative. The key discipline is identifying exactly what makes a hand belong to a category and ensuring your counting choices map precisely onto that definition. Here, the two kicker ranks must be distinct (hence $\binom{12}{2}$, not $12 \times 11$) to avoid double-counting or including full houses.
Three-of-a-kind ranks below two pair and full house in hand frequency and above straights and flushes, which makes intuitive sense: getting three of a rank is harder than getting two pairs (less specific) but easier than also pairing the kickers (full house). The $\approx 2.1\%$ probability means you expect to see it roughly once every 47 hands. In general, the skill of breaking a combinatorial count into independent steps -- rank choice, suit choice, kicker rank choice, kicker suit choice -- is directly applicable to portfolio construction, combinatorial trading problems, and any counting problem where you are assembling objects with multiple independent attributes.