Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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)


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

Am I thinking correct?

share|cite|improve this question
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
up vote 2 down vote accepted

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%).

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.