Probability of hitting trips [Texas hold'em]

Question

Let's say we have [K][9] in hand, what is the probability we hit at least (or better) trips on the flop, turn and river? Trips meaning that [9][9] will come on the board.

We have 3 outs left [9][9][9], and (50, 5) possible board combinations.

All possible board combinations without the 9's are (47, 5).

Logically I think the following calculation can be derived:

1 - ((47, 5)/(50, 5)) = 27%

of hitting trips. I find this probability of hitting trips way to high so the question is: what is the correct probability of hitting trips on the flop, turn and river?

Answer 1

With a 9 in hands, the probability that at least two 9 occur on the flop, turn and river, assuming that we only know these two cards in hands, can be calculated noting that

• the number of boards with two 9 is given by $3\binom{47}{3}$
• the number of boards with exactly three 9 is given by $\binom{47}{2}$

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

$$\frac{3\binom{47}{3}+\binom{47}{2}}{\binom{50}{5}}\approx 2.35 \%$$