While we are here, let's post the results for all standard number of players at a poker table. Here is the probability of anyone flopping a set (or quads), pocket pair required, with N players at the table (NLHE assumed).
N=1
This has been derived many times (including in this thread above) and in many different ways. This prob is:
= 179,712 / 25,989,600
= 144 / 20,825
= 0.006914766
= 0.6914766%
N=2
Derivation is very straightforward (can use the approach given above in this thread). Prob is:
= 193,763,232 / 14,047,378,800
= 310,518 / 22,511,825
= 0.013793551
= 1.3793551%
N=3
Derivation is very straightforward (can use the approach given above in this thread). Prob is:
= 95,663,101,248 / 4,635,635,004,000
= 2,322,822 / 112,559,125
= 0.0206364611
= 2.06364611%
N=4 thru 10
The derivation can follow the approach given above. The algebra becomes a bit messy, and the resulting formula is a bit clunky (and requires a "simplification process" to evaluate on some calculators), but here it is. [Note, the notation X!! refers to the double-factorial expression and for X an odd integer is the product of the odd integers from 1 to X.]
Prob of anyone flopping a set (or quads), pocket pair required, for N = 4 to 10 is:
= [155,664*(2N-3)!!*C(2N,2) - 2,430*(2N-5)!!*C(2N,4) + 36*(2N-7)!!*C(2N,6)] / [22,511,825*(2N-1)!!]
Summary Table
Here is a table that summarizes our results.
Number of Players | Prob of Anyone Flopping a Set* |
---|
1 | 0.6915% |
2 | 1.3794% |
3 | 2.0636% |
4 | 2.7444% |
5 | 3.4215% |
6 | 4.0951% |
7 | 4.7651% |
8 | 5.4317% |
9 | 6.0947% |
10 | 6.7541% |
* Prob is of at least one player flopping a set (or quads), pocket pair required.