Counting Poker Hands

"How many ways can you deal a five-card poker hand?" sounds like it wants you to start listing hands. It doesn't — with 2,598,960 possible five-card hands from a 52-card deck, listing is impossible, and the interviewer knows it. What they want is whether you reach for combinations, $\binom{n}{k}$, instead of trying to enumerate anything by hand. This is the standard warm-up for a longer line of questions ("how many are a flush," "how many are two pair") that build on the same technique.

The tool: choosing without order

A five-card hand is a set of cards — the order you're dealt them doesn't matter, and no card repeats. That's exactly what $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ counts: the number of ways to choose $k$ items from $n$ without regard to order. The total number of five-card hands from a 52-card deck is

$$\binom{52}{5} = \frac{52!}{5!\,47!} = 2{,}598{,}960 .$$

In plain English: there are about 2.6 million distinct five-card hands, and every poker probability question is really "how many of those 2.6 million hands have property $X$, divided by 2,598,960."

Worked example 1: counting flushes

A flush is five cards of the same suit (ignoring straight flushes for now, and yes, real poker excludes straight flushes from "flush" — a good interview move is to name that ambiguity and pick a convention). Pick the suit: 4 ways. Given a suit, pick 5 of its 13 cards: $\binom{13}{5} = 1{,}287$ ways. So the count of same-suit hands is

$$4 \times \binom{13}{5} = 4 \times 1{,}287 = 5{,}148 .$$

The probability of any five-card hand being a flush (including straight flushes) is $5{,}148 / 2{,}598{,}960 \approx 0.198\%$ — about 1 in 505 hands. This is the general pattern: break the hand into independent choices (suit, then ranks within that suit), count each choice with a combination, and multiply.

Worked example 2: counting one pair

"Exactly one pair" (two cards of one rank, three cards of three different other ranks, no further coincidences) needs more care because you must avoid double-counting and avoid accidentally including three-of-a-kind or two-pair hands. Break it into steps: choose the paired rank, $\binom{13}{1} = 13$ ways; choose 2 of the 4 suits for that rank, $\binom{4}{2} = 6$ ways; choose the 3 other (distinct) ranks from the remaining 12, $\binom{12}{3} = 220$ ways; choose 1 suit out of 4 for each of those 3 cards independently, $4^3 = 64$ ways. Multiply:

$$13 \times 6 \times 220 \times 64 = 1{,}098{,}240 .$$

That's about $1{,}098{,}240 / 2{,}598{,}960 \approx 42.3\%$ of all hands — one pair is by far the most common non-trivial hand, which matches poker intuition. The technique — decompose into a sequence of independent choose-steps, multiply the counts, and only then divide by the total — generalizes to every "count hands with property $X$" question you'll be asked.