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.
Last edited by whosnext; 01-09-2018 at 02:45 AM.
Reason: filled in two missing numbers in formula for P(10)