The goal is to solve the probability of a Texas Holdem hand occurring where at the river one player has quad $9$s and the other player has exactly a $9$-high straight flush.
(For non poker players - in holdem each player has $2$ cards and the board has 5 cards - players make the best hand of $5$ by any combination using their hole cards and the board)
This requires that:
- (A) Player 1 holds 2 nines but not the nine of diamonds
- (B) Player 2 holds any two of the $5,6,7$ and $8$ of diamonds
- (C) The board by the river contains the other two diamonds not held by player 2 AND must not have the 10 of diamonds on the board, plus must have the last 9.
I know the final probability is A * B * C
A = 3/52 * 2/51
B = 4/50 * 3/49 (since we know A we can lower the card count right?)
How do we solve C?
I think it's something like: (4/48 + 3/47 + 2/46 + 1/45 + 1/44) - (odds of 10d) but ... I'm not sure if that math is right or how to calculate the odds of just the 10d.
(Edit - forgot the 4th 9 needed to be on the board)
The chances for $C$ are $43/{48\choose 5}=43*120/(44*45*46*47*48)$ There are $48\choose5$ possible tables. It needs the other two diamonds, the other two nines, and one other card which is not the ten. So there are 43 possible tables that work.
Overall, chances of it happening with diamonds is 43/154521166800; happening at all is 43/38630291700, slightly better than one in a billion. Double that if player 1 and player 2 can swap places.