Originally Posted by OmahaDonk
Let's assume all players are playing all quads and all pots are checked down to river. What are the odds two players have quads and their two ranks make the nut straight on river?
Assuming 9 players, I get about 1 in 1.1 Billion.
The hard part of that is counting the number of boards where someone can make a nut straight. For that, I can leverage off of the work I used in this post
where I found the probability that the nuts at the river is a set. In that post I counted the number of ways to choose 5 ranks for the board so that no straight is possible. There are 79. That means that there are C(13,5) - 79 = 1208 ways to choose the 5 ranks so that a straight would be possible. In these cases someone can have the nut straight when a flush is not possible. The other post also counted the number of ways to choose the suits so that a flush isn't possible. That was
[C(4,2)*C(2,1)*C(5,2)*C(3,2) + C(4,1)*C(5,2)*3!].
We then just need the probability that the 2 players each have 4 of one of the nut ranks. Assuming 9 players this would be
Putting all that together gives
[C(4,2)*C(2,1)*C(5,2)*C(3,2) + C(4,1)*C(5,2)*3!] / C(52,5) *
=~ 1 in 1.1 Billion