This story begins on the river in a limped omaha pot, 6 ways, $30. Checked down to river. Board shows 9J46T with no flush, and 2 players got exactly what they wanted.

Player 1 with KKKK bets 30. Player 2 with QQQQ raises 120. Player 1 pots to 480, then Player 2 tanks and shoves for $1700. Player 1 realizes, "the only way he knows is cause he has all the blockers" and calls.

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?

Thanks,
OD

They don't have quads.

Right, what he's getting at is something different. Both guys know the nut straight isn't possible because they're holding all the cards you'd need to get them. So when QQQQ shoves KKKK knows he must have all the Qs because he shouldn't shove a non-nut straight there (QQQQ is shoving because he knows the other guy can't have the nuts)

The back story is pointless to the question, which I guess is just how often 2 people can get dealt quads and get to showdown.

Yeah, it only works though when the guy holding KKKK knows that the guy holding QQQQ is smart enough not to shove with the idiot end of the straight draw (or something worse like trips).

It also seems to me that another possible hand for the QQQQ hand is 8QQQ, knowing he has 3 of the 4 blockers, and depending on how smart the QQQQ player is...but maybe that is why I never really played much omaha.

We are assuming both players know they can't raise in limped pot on river without nuts. And they know the other player knows that.

Also, the QQQ8 can just call the 120 and has no need to rebluff, eliminating 2 levels of deliciousness.

Holy ****! I need to get into this game!

And the odds aren't good OP. The odds for a single player making quads is 13/270,725 = 1 in 20,825

(Note: Choosing 4 cards from 52 is C(52,4) = 52*51*50*49/(4*3*2*1) = 270,725.

The odds of two players even being dealt quads is roughly:

13*12*C(9,2)/C(52,4)/C(48,4) = 1 in 9.58 million

Then having a non paired board with no flush draws possible...omg...I don't even want to think about it.

As for one (or two) players limping in with quads, I reiterate:

Holy ****! I need to get into this game!

This doesn't discount the factor that the quads have to be close enough together that they make the nut straight

Well of course it doesn't. But it is too late for me to answer this now fully as asked. Having a few beers doesn't help my cause either. If I had to ballpark, it's definitely over 1 in 100 million! Calling BruceZ! Come in, BruceZ!

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

C(9,2)*2/C(47,4)*1/C(43,4).

Putting all that together gives

1208*
[C(4,2)*C(2,1)*C(5,2)*C(3,2) + C(4,1)*C(5,2)*3!] / C(52,5) *
C(9,2)*2/C(47,4)*1/C(42,4)

=~ 1 in 1.1 Billion