# Probability of getting a pair in Texas Hold'em poker

I want to know the probability of getting a pair after all 5 cards are dealt on the table and assuming that I haven't got a pocket pair.

Let's assume that I hold an ace and a king and 50 cards are left. So the probability of getting a pair is: P(getting an ace) + P(getting a king) + P(5 table cards having a pair of anything else than ace or king)

Then:

$P(pair) = \frac {3}{50} + \frac{3}{50} + \frac{44 \times 3 \times 40 \times 36 \times 32 \times {5 \choose 2}}{50 \times 49 \times 48 \times 47 \times 46} = 0.3592 = 35.92\%$

Am I thinking correct?

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Do you want the probability of getting exactly one pair or the probability of getting at least a pair (ignoring the possibility of getting flushes and straights)? –  Peter Taylor Feb 14 '12 at 14:59
@PeterTaylor Just the probability of getting 1 pair. –  Cobold Feb 14 '12 at 15:00
That doesn't answer the question. Let me rephrase: do you want to exclude the case where you get AKxxx on the table? –  Peter Taylor Feb 14 '12 at 15:01
@PeterTaylor Yes exclude, because that would be two pairs not 1 pair. –  Cobold Feb 14 '12 at 15:04

Almost. You actually need

P(getting exactly one ace, no kings, and no pair) +
P(getting exactly one king, no aces, and no pair) +
P(getting no aces or kings but exactly one pair)


The number of combinations for the first is $3 \binom{11}{4} 4^4$ where the $3$ is the number of remaining aces, the $\binom{11}{4}$ accounts for the combinations of possible values for the other cards, and the $4^4$ accounts for their suits. The number of combinations for the second is identical.

For the third, there are 11 possible values for the pair, with $\binom{4}{2}$ possible suit combinations, 10 values left which are neither the pair nor AK, so we have $11 \binom{4}{2} \binom{10}{3} 4^3$.

This gives a final result of $$P(\text{exactly one pair}) = \frac{6\binom{11}{4}4^4 + 11\binom{4}{2}\binom{10}{3}4^3}{\binom{50}{5}}$$

(The numerical result is $\frac{25344}{52969}$ or about 47.85%).

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Thanks, but I don't understand why suits are to the 4th and 3rd power. The suits shouldn't matter, only the ranks matters. –  Cobold Feb 14 '12 at 15:26
Four cards and three cards respectively. –  Peter Taylor Feb 14 '12 at 15:54