Probability at least one card is hit [Texas hold'em]

Question

The cards in hand = [9][9]. The question is as follows:

What is the probability we hit at least one 9 on the flop, turn or river (so, on the next 5 cards)?

There are (50, 5) = 2.118,760 board combinations and two outs [9][9] left.

P(at least one 9) means the following:

(i) P(hitting exactly one 9);
(ii) P(hitting exactly two 9's);

And combine those probabilities.

ad (i):
p(How many boards have exactly one 9)?
(49, 5) / (50, 5) = 0.9
1 - 0.9 = 10% 
ad (ii):
p(How many boards have exactly two 9's)?
(48, 5) / (50, 5) = 0.8
1 - 0.8 = 0.19% 

Combining the probabilities

p(one 9) + p(exactly two 9's) * 1 - p(one 9) 
= .10 + .19 * (.10)
= .119%

So the probability of hitting at least one 9 is 11.9%. Is this correct?

Answer 1

With a pair of 9 in hands, the probability that a 9 not occurs on the flop, turn and river, assuming that we only know these two cards in hands, is given by

$$\frac{48}{50}\frac{47}{49}\frac{46}{48}\frac{45}{47}\frac{44}{46}$$

thus the probability that we hit at least one 9 on the flop, turn or river is

$$1-\frac{48}{50}\frac{47}{49}\frac{46}{48}\frac{45}{47}\frac{44}{46}\approx 19.2 \%$$

With a direct calculation we have that

• the number of boards with one 9 is given by $2\binom{48}{4}$
• the number of boards with two 9 is given by $\binom{48}{3}$

thus the probability that we hit at least one 9 on the flop, turn or river is

$$\frac{2\binom{48}{4}+\binom{48}{3}}{\binom{50}{5}}\approx 19.2 \%$$