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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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    @PeterTaylor Yes exclude, because that would be two pairs not 1 pair.2012-02-14

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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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    @PeterTaylor - Thank you so much for the response! Agree completely, and in thinking about it again _technically_ a pair could exist within the flush, so perhaps this question should be asked of the OP. In any event, thank you again, your answer helped me so much in my own calculations.2018-05-27