Bad-Beat Jackpots: The Basics

Brian Alspach

Poker Digest Vol. 3, No. 23, November 3 - 16, 2000

I received a letter from Mr. Dave Ruoen of Las Vegas in which he asked for the approximate probabilities for the occurrences of a variety of bad-beat jackpots. This article is the first of four that will answer his questions.

First, we shall lay the groundwork and illustrate it with a simple example, and then consider what happens when the qualifying conditions are weakened or the number of players changes. In subsequent articles we take detailed looks at hold'em, Omaha and seven-card stud.

To illustrate the approach to finding some of the probabilities involved, we consider an example that's about as simple as it gets when considering bad-beat jackpots. Let's look at the probability that two players in a 10-handed hold'em game end up with four-of-a-kind -- given that both players must have pairs in their hands. We are interested in the probability that this occurs, but don't care which particular players are involved. Thus, we are interested in semi-deals.

The first number we need is the total number of semi-deals possible in a 10-handed hold'em game. The number of possible boards is C(52,5). There are then C(47,20) ways to choose the 20 cards to be dealt to the players and there are 19!! = 654,459,425 ways to partition the 20 cards into 10 hands of two cards each. The product of these three numbers is the total number of semi-deals in 10-handed hold'em and this product is X = 16,611,851,842,086,615,628,485,972,000.

All that remains to do is to count the total number of semi-deals for which two players end up with four-of-a-kind (each has a pair in the hole), and then divide by X to get the probability. In order for this to occur, the board must have rank multiset $\{x,x,y,y,z\}$. There are 123,552 such hands (see the file entitled 5-Card Poker Hands at For a given board with two pair on it, two players will end up with four-of-a-kind provided that the remaining two cards of rank x are dealt to some player and the remaining two cards of rank yare dealt to another player. Thus, in forming semi-deals, we are choosing 16 cards from 43 and partitioning them into eight further hands. Hence, the number of semi-deals which give two players four-of-a-kind for the given board is C(43,16)15!! The number of semi-deals producing two players with four-of-a-kind is then the latter product multiplied by 123,552. Altogether this gives 66,413,011,087,292,099,630,400. Dividing the preceding number by X yields the probability of .0000039979 that two players will make four-of-a-kind with pairs in their respective hands given that no one ever folds pocket pairs. This is approximately 1 in 250,000.

We now examine how the above particular probability changes as we change conditions in the game. The reader can extrapolate how similar changes will affect the probabilities we provide in future articles as this is the only time we shall explicitly deal with these issues.

First let's see the effect of allowing only one of the player's cards to play. This introduces the possibility of three cards of rank x being on the board. All we then need is for some player to have the remaining card of rank x in her hand. However, for a second four-of-a-kind to occur, there must be a pair of some rank y on the board as well. That is, the board must be a full house. There are 3,744 full houses. In order to have two players with four-of-a-kind hands, both cards of rank y must be in the same hand and the remaining card of rank x must be among the other 18 cards dealt to the players. Thus, we are choosing 17 cards from 44 and partitioning the 18 cards containing the card of rank x in 17!! ways into nine hands. Altogether, the product of C(44,17), 17!! and 3,744 yields 88,550,681,449,722,799,507,200 semi-deals producing two players with four-of-a-kind when the board is a full house. Adding the previous number of semi-deals to the number just obtained and dividing by X gives us a probability of .0000093285 that two players will get four-of-a-kind when one or two cards from the player's hand are allowed for qualifiers. So the probability is more than doubled but is still only roughly 1 chance in 100,000.

We conclude this article by examining how the number of players in the hand affects the probability. You might think by going through all the computations again, the relationship is complicated but we shall see a simple relationship. Let's return to the case where both of the players must have a pair in their hands to qualify. In counting the total number of semi-deals for 11 players, C(47,22) replaces C(47,20) above and 21!! replaces 19!! Upon making the appropriate replacements, the total number of semi-deals becomes $(27\cdot 26 X)/22$. In counting the total number of semi-deals producing two players making four-of-a-kind, C(43,18) replaces C(43,16) and 17!! replaces 15!! Making the replacements increases the number of semi-deals by $(27\cdot 26)/18$. When we carry out the division to get the probability, there is a lot of cancellation and the probability for 11 players is simply 22/18 times the probability for 10 players.

When I tell you that the probability for 10 players is, in fact, 20/16 times the probability for nine players are you surprised? Do you see a pattern emerging here? If so, you are correct in assuming the pattern holds. I remind you that this is in the particular case of 2 players getting 4-of-a-kind, where each player must have a pair in her hand, but it still gives you an idea of the effect wrought by changing the number of players.

This has implications for the expectation of getting back some of that bad beat jackpot money you have donated over many sessions.

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