I'm curious because of a promotion that pays only a certain set of quads each day. As in today you must get quad 7's to win. The only rule is you must have a pocket pair. So, what are the odds that any one of 9 players at a table will hit a specific set of quads? Also, how would you calculate the odds of any player hitting the specific quads over 1,000 hands? The question is for any of the 9 players to hit the quads and not just one specific player.

I'll give it a try. There are several ways to derive this probability. Here is one (they should all wind up with the same probability, of course).

I'll do the example with Aces, but the prob is the same for any other rank under the standard assumption that all hands go to showdown on every deal.

You need the following:
- one player is dealt pocket aces
- no other player is dealt an ace
- the remaining two aces appear on the five-card board.

Then the prob of this occurring on any random deal is given by:

= [C(4,2)*C(48,16)*15!!*C(32,3)*C(2,2)] / [C(52,23)*C(23,5)*17!!]

= 18 / 54,145

= 0.000332441

which is approx once in every 3,008 deals.

As per usual, I encourage others to post their derivations (be they confirming or refuting) or any other comments.

Agree. Did it a bit differently.

Pr(1 player is dealt 4 aces) = C(4,2)/C*52,2) * [C(2,2)*C(48,3)/C(50,5)] = 0.0000369

Since only one player out of 9 can get 4 aces, multiply this result by 9 to get whosnext result.

You guys are forgetting that a player can also have any pocket pair and have the four 7's (or whatever) on the board...making quads for everyone.

(Rule just says player most hold a pocket pair - not that he must hold two of the cards needed for the quads)

I think that was implied.

I agree.

Plus winning the jackpot in that manner would be very rare. And it would definitely be a lot of work to derive that probability! Assuming that that is a valid way to win the jackpot, let's see if we can ballpark the new prob.

Prob of quads of specific rank appearing on board:

= [C(48,1) * C(4,4)] / C(52,5)

= 1 / 54,145

And I would hazard to guess that the prob of the winner of the deal with that specific quads on board having a pocket pair is less than 50%. Perhaps far less than 50%

Ballpark that prob at 25% just to see what that does. Then the overall prob of the jackpot becomes:

= 18.25 / 54,145

= .000337058

which is approx once in every 2,967 deals.

4/52 times 3/51 times 2/50 times 1/49 times 10 times 9.

9 * C(5,2) / C(52,4)

Just as a side note, even if the jackpot requires the quads and pocket pair to be the same rank (which seems very likely to be the case), the "other" case where a player wins the hand with a pocket pair and a different rank quads are on board is sort of interesting to me.

I coded up a simulation of this case and let it run over night. The simulation pre-supposes quads of a specific rank on board and tallies how often the hand is won (or split) by a player having a different pocket pair.

9,613,082 deals out of 100,000,000 are won this way. Of course, that is approx 9.61%.

I don't know how easy or difficult it would be to derive this probability analytically. If anybody is looking for something to do, they may want to try to tackle this problem.

Take P(quads on board) and multiply by P(at least one pocket pair | quads on board). For the latter, refer to the calculations in the sticky (specifically this post of mine) and adjust the numbers to reflect the removed cards.

Yes, but there is more to the story (and that can be a bit tricky).

As you undoubtedly realize, if board is AAAA5 then somebody with pocket Tens (say) loses to another player with K2.

So you're saying one needs to win the hand, and you want the probability of that as well? Then it becomes a question of real-life ranges and tendencies rather than a pure probability question. But in real life, the pocket pair requirement definitely refers to a PP of the same rank as the quads.

Sorry, I am not following you.

Someone above said (or appeared to say) that OP's "win with quads while holding a pocket pair" jackpot could be won not only with a pocket pair making quads of that same rank but also a player winning the pot while holding a pocket pair when quads are on board. As I have said repeatedly, I find it very hard to believe that that is the correct interpretation of the jackpot rules.

But leaving that aside, it is a somewhat interesting question of how you would derive that probability. And, it seems to me, it has nothing to do with ranges, player tendencies, etc. Same assumptions apply. Every player goes to showdown on every deal.

I simulated 100 million deals and found that when quads appear on board the winning player has a pocket pair approx 9.61% of the time. I further wondered if this probability would be amenable to analytical derivation.

If I have some time down the road I may try to tackle this myself.

Ah ok, well it's a tedious problem.

P(quads on board with kicker x) * P(exactly n players have PP's <= y rank | quads on board) * P(highest PP wins/ties)

If exactly one person was dealt 88 and the board is JJJJQ, P(tie) is a little less than C(37,8) / C(45,8)

The winner must have a pocket pair and make quads with the remaining 2 cards on the board to win. I assume this post to be correct for that, but 1:3,008 seems a bit low to me. I honestly would've guessed it to be closer to 1:30,000. Thank you very much!!!!