Lets see if some of you can figure this out.

Assume there are X number of players at a table. Each hand is forced to play until the river. A bad beat jackpot includes 2 hands. One where there is at the least a 4 of a kind with jacks or better which is beaten by an even stronger hand.

Both hole cards must be used in the winning and losing hand.


What is the chance of a bad beat jackpot happening on a single hand?

How many hands must be dealt on average for a bad beat jackpot to happen?

Solve the answer for X equaling 2 to 10.

So in a 10-handed example, every player is forced to play until the river?

Yes, they are forced to play until the river.

I tried solving this myself, but I'm pretty sure I made some mistakes somewhere along the way.

Here is my attempt. Actually this was all done with my daughter off-and-on during her extended school holiday break. I believe all of the numbers that follow have been "confirmed" via computer programs (but, of course, computer programs can have errors just like human counting methods so take that with a grain of salt). As is typical in these types of derivations, we assume that each player on each deal goes to showdown.

Introduction

A bad beat jackpot is when a player loses a hand of NLHE with quad jacks or better and MUST use both of his hole cards to make the best hand (both cards are required to make the best hand which I interpret to mean that if I have a king kicker hole card and there is another king kicker on board, that would not qualify for BBJ).

For our purposes, it is useful to classify hands into three categories: (i) straight flushes including royal flushes; (ii) quads with a pocket pair; (iii) quads without a pocket pair. Clearly, quads with a pocket pair requires the matching pocket pair to appear on the board. And quads without a pocket pair requires the board to have trips. (There is no way for a BBJ deal to have quads on board since no player would be able to use both of his hole cards to make his best hand.)


BBJ at 2-Player Tables

The table below shows the tally of BBJ deals at a 2-player table (the order of the hands does not matter, meaning that [QQ,JJ;QQJJ5] is considered the same as [JJ,QQ;QQJJ5]).

Hand 1	Hand 2	BBJ Count
Straight Flush	Straight Flush	38,272
Straight Flush	Quads (PP)	40,824
Straight Flush	Quads (no PP)	6,768
Quads (PP)	Quads (PP)	9,504
Quads (PP)	Quads (no PP)	1,152
Quads (no PP)	Quads (no PP)	0
------------	------------	----------
TOTAL	.	96,520

The total number of deals in 2-player NLHE is:

= C(52,9)*C(9,5)*(1*3)

= 1,390,690,501,200

This means the probability of a BBJ deal at a 2-player table is:

= 96,520 / 1,390,690,501,200

= 6.9404371 * 10^-8

which is equivalent to:

= Once in every 14,408,314 deals.


BBJ at 3-Player Tables (involving all three players)

Clearly, a BBJ deal will occur more frequently if more players are at the table. Just like the probability of any player being dealt a pair goes up as the number of the players at the table increases. As a first and crude approximation (as we have seen in previous threads), if we call the probability of one player being dealt a pocket pair P1, then the probability that any player at a table with N players is dealt a pocket pair is approximately equal to N*P1.

Again, as we have encountered in previous threads, the Principle of Inclusion-Exclusion (PIE) is the mathematical way to derive these exact probabilities involving more than one player at a table. PIE properly takes into account the "dependence" between the various players' probabilities (which can be thought of as the degree of "overlap" in the events in a Venn diagram).

In each BBJ scenario, we are dealing with two (or more) players, so the PIE machinery needs to track pairs of players (not just individual players). This makes the formulas more complex, but the underlying reasoning is the same.

To get the correct answer for the probability of a BBJ deal at tables with more than two players, we will need to know how often (if at all) a BBJ deal can actually involve more than two players. That is, is it possible for one player to win a BBJ deal in which there are two other players who each had quad jacks or better (both hole cards playing).

The table below shows the tally of BBJ deals involving exactly three players.

Hand 1	Hand 2	Hand 3	BBJ Count
Straight Flush	Straight Flush	Straight Flush	0
Straight Flush	Straight Flush	Quads (PP)	2,304
Straight Flush	Straight Flush	Quads (no PP)	264
Straight Flush	Quads (PP)	Quads (PP)	1,080
Straight Flush	Quads (PP)	Quads (no PP)	0
Straight Flush	Quads (no PP)	Quads (no PP)	0
Quads (PP)	Quads (PP)	Quads (PP)	0
Quads (PP)	Quads (PP)	Quads (no PP)	0
Quads (PP)	Quads (no PP)	Quads (no PP)	0
Quads (no PP)	Quads (no PP)	Quads (no PP)	0
------------	----------	------------	----------
TOTAL	.	.	3,648

The total number of deals in 3-player NLHE is:

= C(52,11)*C(11,5)*(1*3*5)

= 418,597,840,861,200

This means the probability of a BBJ deal at a 3-player table involving all 3 players is:

= 3,648 / 418,597,840,861,200

= 8.7148084 * 10^-12

which is equivalent to:

= Once in every 114,747,215,148 deals.


BBJ at 4-Player Tables (involving all four players)

The table below shows the tally of BBJ deals involving exactly four players.

Hand 1	Hand 2	Hand 3	Hand 4	BBJ Count
Straight Flush	Straight Flush	Straight Flush	Straight Flush	0
Straight Flush	Straight Flush	Straight Flush	Quads (PP)	0
Straight Flush	Straight Flush	Straight Flush	Quads (no PP)	0
Straight Flush	Straight Flush	Quads (PP)	Quads (PP)	36
Straight Flush	Straight Flush	Quads (PP)	Quads (no PP)	0
Straight Flush	Straight Flush	Quads (no PP)	Quads (no PP)	0
Straight Flush	Quads (PP)	Quads (PP)	Quads (PP)	0
Straight Flush	Quads (PP)	Quads (PP)	Quads (no PP)	0
Straight Flush	Quads (PP)	Quads (no PP)	Quads (no PP)	0
Straight Flush	Quads (no PP)	Quads (no PP)	Quads (no PP)	0
Quads (PP)	Quads (PP)	Quads (PP)	Quads (PP)	0
Quads (PP)	Quads (PP)	Quads (PP)	Quads (no PP)	0
Quads (PP)	Quads (PP)	Quads (no PP)	Quads (no PP)	0
Quads (PP)	Quads (no PP)	Quads (no PP)	Quads (no PP)	0
Quads (no PP)	Quads (no PP)	Quads (no PP)	Quads (no PP)	0
------------	----------	----------	------------	----------
TOTAL	.	.	.	36

The total number of deals in 4-player NLHE is:

= C(52,13)*C(13,5)*(1*3*5*7)

= 85,812,557,376,546,000

This means the probability of a BBJ deal at a 4-player table involving all 4 players is:

= 36 / 85,812,557,376,546,000

= 4.1951902 * 10^-16

which is equivalent to:

= Once in every 2,383,682,149,348,500 deals.

It turns out, and is probably obvious at this point, that there is no way for five players to ever all be involved in a BBJ deal. So we can stop our journey at four players.


Putting it all Together

We are now almost ready to derive the various probabilities of a BBJ deal occurring at tables of 2 players, 3 players, 4 players, 5 players, 6 players, 7 players, 8 players, 9 players, and 10 players.

The one final piece of the puzzle is the PIE formulas for the respective probabilities. Remember that since the BBJ "event" necessarily involves two players these are similar to, but slightly different from, the traditional PIE formulas involving only single players.

I have derived these formulas via a mixture of partitions, permutations, and computer programs.

Notation:

Let P(N) denote the desired probability of a BBJ deal at a table of N players (involving any 2 or more of the N players).

Let P(All_N) denote the probability of a BBJ deal at a table of N players involving all N players.

Then per the PIE methodology on pairwise events, we have:

P(2) = P(All_2)

P(3) = 3*P(All_2) - 2*P(All_3)

P(4) = 6*P(All_2) - 8*P(All_3) + 3*P(All_4)

P(5) = 10*P(All_2) - 20*P(All_3) + 15*P(All_4) - 4*P(All_5)

P(6) = 15*P(All_2) - 40*P(All_3) + 45*P(All_4) - 24*P(All_5) + 5*P(All_6)

P(7) = 21*P(All_2) - 70*P(All_3) + 105*P(All_4) - 84*P(All_5) + 35*P(All_6) - 6*P(All_7)

P(8) = 28*P(All_2) - 112*P(All_3) + 210*P(All_4) - 224*P(All_5) + 140*P(All_6) - 48*P(All_7) + 7*P(All_8)

P(9) = 36*P(All_2) - 168*P(All_3) + 378*P(All_4) - 504*P(All_5) + 420*P(All_6) - 216*P(All_7) + 63*P(All_8) - 8*P(All_9)

P(10) = 45*P(All_2) - 240*P(All_3) + 630*P(All_4) - 1008*P(All_5) + 1050*P(All_6) - 3240*P(All_7) + 2835*P(All_8) - 80*P(All_9) + 9*P(All_10)

Using the values for P(All_2), P(All_3), and P(All_4) derived above, and knowledge that all "higher-order" values are zero, the table below presents these P(N) values, the probability that a BBJ deal will occur at a table with N players.

Number of Players at Table	Prob of BBJ Deal	Once in Every How Many Deals
2	6.9404371 * 10^-8	14,408,314
3	2.0819568 * 10^-7	4,803,174
4	4.1635651 * 10^-7	2,401,788
5	6.9386942 * 10^-7	1,441,193
6	1.0407170 * 10^-6	960,876
7	1.4568818 * 10^-6	686,397
8	1.9423464 * 10^-6	514,841
9	2.4970934 * 10^-6	400,466
10	3.1211054 * 10^-6	320,399

I typically say at the end of these types of posts that comments and questions are welcome and encouraged. Those sentiments doubly apply here since, after working on this project off-and-on for two weeks, I honestly have no idea whether or not these figures are in the right ballpark. While I have tried to be careful at each stage of this derivation, mistakes have been known to creep in and, once in, are notoriously difficult to identify and eradicate.

So, as I say, comments and questions are welcome and encouraged.

Wow whosnext, This is really great work! I will definitely have to look this over. At the time being I have to take care of Christmas decorations, but I must say this is very impressive. Thank you for taking the time to do this.

This could be a pretty cool tool for deciding if certain BBJ tables are profitable. I've been toying with the idea of how high a BBJ would be in order to make a highly raked table profitable or even make hands like 32s profitable. There are also things like 5-bet shoving JJ against a tighter opponent being profitable due to the BBJ being high enough.

A lot of those involve the BBJ being really high in order to justify the play, but I've been watching some tables with BBJs and even a dormant ultra tight strategy can be profitable since some BBJs payout to anyone that is sitting at any table at the time of a BBJ event. So I could play a certain strategy with very little thinking while doing another monotonous task.

Anyway, this is really cool. I think it would be really cool to win a BBJ someday, seems pretty rare and most likely won't ever happen, but it is fun to know the chances. I may end up taking this a step further and eliminating hands that get folded along the way to make things even more accurate.

Bump

I'm not sure I understand why you are subtracting. For example: P(3) = 3*P(All_2) - 2*P(All_3)

I see that you multiply by 3 for the total combinations of bad beat jackpots (1 winner and 1 loser) which makes sense, but I don't get why we subtract 3 way bad beat jackpots twice.


I made a little calculation of my own. When you hold, say, JJ and you get it in against an overpair your equity is better due to the bad beat jackpot.
With JJ facing AA the odds of hitting a bad beat jackpot is:
(2/48) * (1/47) * (2/46) * (1/45) * 10 * 3
Which reduces to 120 / 4,669,920
Which further reduces to 1 / 38,916

Multiply that by the current jackpot at the site I am playing at which is roughly $200,000 and we get $5.14 which can make a difference, especially against a short stack. Everytime I get it in badly against an overpair I am gifted an additional $5.14 in equity.

Hopefully my math is right.

I will give a short explanation for the weird +/- terms above. Others can undoubtedly do a better job than I can.

In the example you cited above, P(3) = 3*P(All_2) - 2*P(All_3), the second term is required to account for the "overlaps" (double-counting) inherent in the first term. A Venn diagram of three overlapping circles is a good mental picture of what is going on.

Perhaps the easiest way to think about it is to consider ABC (a deal in which all three players at the table are involved in a BBJ). Then this deal is automatically covered by each of P(AB), P(AC), and P(BC) since each of the three pairs of players is clearly also part of the BBJ.

So each ABC deal is actually counted three times when it should only be counted exactly one time. Which is why we need to subtract off 2 times the number of 3-way BBJs from simply taking 3 times the number of 2-way BBJs.

The same thing applies as more and more players are at the table. You wind up needing to alternate adding and subtracting terms since you alternate including too much, then subtracting off too much, adding back too much, subtracting off too much, etc. The ultimate formula involves combinations and factorials to properly account for each deal once and only once.

If you are interested in learning more about it, I recommend looking up "Principle of Inclusion Exclusion" on the internet (the Wiki page is pretty good).