Back home there is a poker room that has a BBJ approaching $1M. Apparently it got that high because of some odd rules. I'm curious to know what the chances to hit are.

- NLHE
- Both cards must play.
- Quads over quads ONLY.

I am notoriously shaky on these types of questions, but I'll take a crack at it.

Assume 9-handed NLHE. Board must be two pair and a fifth card of another rank. Two of the nine players each must hold one of those exact two pair. All hands go to showdown (or at least every player stays in the hand if they have any chance of the BBJ).

Then I get the number of deals that qualify to be:

= C(13,2)*C(4,2)*C(4,2)*C(11,1)*C(4,1)*C(9,2)*C(2,2) *C(2,2)*C(43,14)*[1*3*5*7*9*11*13]

And the total number of deals is:

= C(52,23)*C(23,5)*[1*3*5*7*9*11*13*15*17]

The probability is the ratio (in simplified form):

= 2,592 / 22,511,825

= 0.0001151

which is 1 in 8,685 deals.

Others should confirm or deny.

Oh my!

I just programmed a quick-and-dirty simulation and got nowhere near the answer I posted above.

I will not post my simulation results, or attempt to identify/correct any mistakes in my analytical derivation, for fear of unduly influencing others.

For the time being I will just leave this out there (for amusement purposes if nothing else). Eventually, in the absence of other posts, I may return to rectify my errors.

Oh, I think I may have discovered my error.

The numerator should not have the C(9,2) factor since the way I wrote the denominator is wholly oblivious to who has what hands. That is, all the permutations of {H1,H2,H3,H4,H5,H6,H7,H8,H9} in the denominator are counted as a single "deal".

Thus, it is incorrect to treat {Q1,Q2,H3,H4,H5,H6,H7,H8,H9} as C(9,2)=36 separate deals in which two opponents each have quads. If someone could confirm or deny this, I'd be grateful.

Anyway, presuming that is an error (and the only error) in the above analytical derivation, the correct probability becomes:

= 72 / 22,511,825

= 0.000003198

= 1 in 312,664 deals

P.S. I warned you that I was shaky on these types of questions.

Is it that quads naturally block other quads that makes this BBJ so much less likely to happen than a quads vs straight flush?

And...

Is this what happens when poker rooms are run by mathematically challenged individuals?

What about a boat board where a person holds the 4th card of the board's trips and a higher kicker than the board pair the other has the board's matching pair.

If this is Motor City Casino in Detroit, they also require both players to have a pocket pair. Quads with trips on the board does not qualify.

For completeness of the math, suppose that the above scenario does indeed qualify for the BBJ.

Since I botched the easier of the two cases of quads over quads where two players use both cards (two separate 2+2 quads), let me try the harder case described above.

Assume 9-handed NLHE. Board has a full house (three of one rank and two of another rank). One player has pair of the board pair. Another player has the remaining card of the board trips and his other hole card is higher than the rank of the board pair (so both players make quads with both hole cards playing).

All hands go to showdown (or at least every player stays in the hand if they have any chance of the BBJ).

Then I get the number of deals that qualify to be:

= C(13,3)*3*C(4,3)*C(4,2)*C(2,2)*C(1,1)*C(4,1)*C(43, 14)*[1*3*5*7*9*11*13]

The wrinkle is the " *3 " which comes from only half of the 3! (6) orderings of three different ranks have the "third" rank higher than the "second" rank. I think this is the correct approach here.

And the total number of deals is:

= C(52,23)*C(23,5)*[1*3*5*7*9*11*13*15*17]

The probability is the ratio (in simplified form):

= 48 / 22,511,825

Adding the probabilities of these two cases together gives the overall prob of quads over quads (both cards play) yields:

= (72+48) / 22,511,825

= 120 / 22,511,825

= 0.000005331

which is 1 in 187,599 deals.

Usual caveats apply.

whosnext, what does the bracket notation ([...]) mean?

13C2*(4C2)^2*44 two pair boards.

9P2 ways to deal 2 players quads.

43C14*14!/(2^7) ways to deal the remaining 7 hands

13C2*(4C2)^2*44*9P2*43C14*(14!/2^7) ways to deal quads over quads

52C23*(23!/(2^9*5!)) total possible deals.

I get 1 in 312,664 deals.

Don't know why I used brackets. It is simply parentheses.

7!! = 1*3*5*7*9*11*13

= 135,135

is the number of ways to deal 14 cards into 7 2-card hands where the order of the players and the order of the cards in each hand does not matter.

As you know this can also be written as:

= 14! / [(2^7) * 7!]

= 135,135.

This has also been true for every other place I've seen with a quads vs. quads qualifier.

But that is for a "super" jackpot (or whatever a particular place calls it) and they'll still have a regular jackpot with an aces full of tens or better beaten qualifier.