I was watching a NLHE poker tournament recently and a key hand arose in which two players each flopped a set. All the players' chips inevitably made their way to the center of the table, and one player felted the other. It turned out to be a key hand in determining the winner of the tournament. The broadcasters said that set-over-set is a rare phenomenon in NLHE that every poker player dreads.
Here I want to explore exactly how rare set-over-set on the flop actually is.
There are different ways to explore this question. What I have chosen to do is to look at the following:
- Hero has a pocket pair (X,X)
- Flop comes three distinct ranks (X,Y,Z) -> Hero flops a set
- What is the probability that an opponent has flopped a higher set than Hero?
For simplicity of calculations, I will assume that all opponents' cards are random (e.g., nobody ever folds pre-flop). It doesn't matter for what follows, but we can suppose that Hero had the X of spades and the X of hearts, and the flop comes X of diamonds, Y of spades, and Z of clubs.
Case 1: One Opponent
Now that we have "fixed" 5 of the cards (Hero's two hole cards and the three flop cards), there remain 47 other cards that the one opponent can have. Clearly, for Villain to have also flopped a set, he must have a pocket pair and a third card of that rank must appear on the flop. So the probability that Villain has flopped a set of Rank R is given by (for all ranks other than X):
= [C(1,1)*C(11,1)/C(12,2)] * [C(3,2)/C(47,2)]
where the first term in brackets is the prob that Rank R appears on the flop (remember that, other than X, the other two flop cards are distinct ranks not equal to X). The second term in brackets is the prob that Villain has a pocket pair of Rank R.
= (11/66) * (3/1081)
= 33/71346
= 1/2162
This is the probability that Villain has flopped a set of Rank R, for all R other than X.
The probability of Villain flopping an over-set (higher than Hero's set) is now easy to see. Clearly, an over-set is a set of higher rank than X. Using the standard rank values (Ace=14, King=13, Queen=12, Jack=11, Ten=10, ..., Deuce=2), we clearly have:
Prob(overset on flop when Hero flops a set of X's when facing one opponent):
= (14-X)/2162
Of course, when X=14 (Hero flops a set of aces), she cannot be oversetted. When X=13 (Hero flops a set of Kings), she can only be oversetted by a set of Aces, etc. The highest overset probability, of course, occurs if Hero flops a set of deuces.
At the bottom of this post we will present a table of all the overset probabilities for any number of opponents and for any X (Hero's set rank).
Case 2: Two Opponents
In this case there are now two opponents who can each flop a set of Rank R (as above the probability of any set is the same for all ranks other than X).
There are three observations that are relevant here. First, note that it is impossible for any opponent to flop the same set as Hero (there are only four cards in the deck of each rank). Second, by the same reasoning, it is impossible for both opponents to flop the same set. Third, it is possible for both opponents to flop different sets (clearly, this would occur if one opponent had hole cards YY and the other opponent had hole cards ZZ).
The Principle of Inclusion-Exclusion (PIE) tells us when there are two opponents:
Prob(at least one opponent flops an overset) = 2*Prob(either opponent flops an overset) - Prob(both opponents flop an overset)
Using the logic of the "independence" of multiple players flopping sets, as in the first case above:
Prob(either opponent flops an overset):
= (14-X)/2162
Using the logic that the only way two opponents can both flop sets (oversets) is for them to each flop a different set (overset), we have:
Prob(both opponents flop an overset):
= [C(14-X),2)/C(12,2)] * [C(3,2)*C(3,2)/C(47,4)/3]
The first term in brackets reflects that if both opponents flop sets, both must be oversets (have rank higher than X). For example, if Hero has a set of Queens, the opponents must flop a set of Aces and a set of Kings, respectively (C(2,2)=1). If Hero flops a set of Tens, say, then the opponents must both flop a set with ranks "chosen" from Aces, Kings, Queens, Jacks (C(4,2)=6). Etc.
The second term in brackets reflects both opponents each flopping a different set. Of course, the C(3,2) reflect that there are 3 cards of each rank available (one must be on the flop) for each opponent's hole cards. The C(47,4) in the denominator reflects the number of ways to choose 4 cards of the remaining 47 cards, and the last 3 in the denominator reflects the number of ways that the 4 cards can be divided into two two-card hands where the order of the cards and the order of the players does not matter.
Putting it all together, we see:
Prob(at least one opponent flops an overset):
= 2*(14-X)/2162 - [C(14-X),2)/C(12,2)] * [C(3,2)*C(3,2)/C(47,4)/3]
Case 3: Three Opponents
In this case there are now three opponents who can each flop a set of Rank R (as above the probability of any set is the same for all ranks other than X).
Let's revisit our three observations from Case 2. First, note that it is impossible for any opponent to flop the same set as Hero (there are only four cards in the deck of each rank). Second, by the same reasoning, it is impossible for any two opponents to flop the same set. Third, it is possible for two opponents to flop different sets (clearly, this would occur if one opponent had hole cards YY and another opponent had hole cards ZZ). A fourth observation is also relevant (though obvious): it is impossible for more than two opponents to flop a set (there are only two non-X cards on the flop).
The Principle of Inclusion-Exclusion (PIE) tells us when there are three opponents (and using the logic reflected in the observations):
Prob(at least one opponent flops an overset): = 3*Prob(any opponent flops an overset) - Prob(two opponents flop an overset)
We saw in the first case above, and via the logic of the "independence" of multiple players flopping sets:
Prob(any opponent flops an overset):
= (14-X)/2162
Using the logic that the only way two
specific opponents can both flop sets (oversets) is for them to each flop a different set (overset), we have:
Prob(two specific opponents flop an overset):
= [C(14-X),2)/C(12,2)] * [C(3,2)*C(3,2)/C(47,4)/3]
We must capture that there are more than one pair of "specific" opponents that can each flop an overset. Of course, given three opponents who can each flop an overset, there are C(3,2)=3 pairs of opponents who can both flop an overset.
So we have:
Prob(two opponents flop an overset):
= C(3,2) * [C(14-X),2)/C(12,2)] * [C(3,2)*C(3,2)/C(47,4)/3]
Putting it all together, we have:
Prob(at least one opponent flops an overset):
= 3*(14-X)/2162 - C(3,2) * [C(14-X),2)/C(12,2)] * [C(3,2)*C(3,2)/C(47,4)/3]
Case N: N Opponents
The same logic can easily be extended to any number of opponents. The generalized formula becomes:
Prob(at least one opponent flops an overset out of N opponents):
= N*(14-X)/2162 - C(N,2) * [C(14-X),2)/C(12,2)] * [C(3,2)*C(3,2)/C(47,4)/3]
Upon inspection it can be seen that the three earlier cases are special instances of this formula (N=1, N=2, N=3).
Table of Results
The table below presents the probability of being over-setted on a flop of three distinct ranks based upon two factors: (1) the rank of Hero's set (what we called X above); and (2) the number of opponents also seeing the flop.
Hero's Set | 1 Opponent | 2 Opponents | 3 Opponents | 4 Opponents | 5 Opponents | 6 Opponents | 7 Opponents | 8 Opponents | 9 Opponents |
---|
A | 0.000% | 0.000% | 0.000% | 0.000% | 0.000% | 0.000% | 0.000% | 0.000% | 0.000% |
K | 0.046% | 0.093% | 0.139% | 0.185% | 0.231% | 0.278% | 0.324% | 0.370% | 0.416% |
Q | 0.093% | 0.185% | 0.277% | 0.370% | 0.462% | 0.555% | 0.647% | 0.739% | 0.832% |
J | 0.139% | 0.277% | 0.416% | 0.555% | 0.693% | 0.831% | 0.970% | 1.108% | 1.246% |
T | 0.185% | 0.370% | 0.555% | 0.739% | 0.924% | 1.108% | 1.292% | 1.476% | 1.660% |
9 | 0.231% | 0.462% | 0.693% | 0.924% | 1.154% | 1.384% | 1.614% | 1.843% | 2.072% |
8 | 0.278% | 0.555% | 0.831% | 1.108% | 1.384% | 1.659% | 1.935% | 2.209% | 2.484% |
7 | 0.324% | 0.647% | 0.970% | 1.292% | 1.614% | 1.935% | 2.255% | 2.575% | 2.895% |
6 | 0.370% | 0.739% | 1.108% | 1.476% | 1.843% | 2.209% | 2.575% | 2.940% | 3.305% |
5 | 0.416% | 0.832% | 1.246% | 1.660% | 2.072% | 2.484% | 2.895% | 3.305% | 3.714% |
4 | 0.463% | 0.924% | 1.384% | 1.843% | 2.301% | 2.758% | 3.214% | 3.668% | 4.122% |
3 | 0.509% | 1.016% | 1.522% | 2.027% | 2.530% | 3.032% | 3.532% | 4.031% | 4.529% |
2 | 0.555% | 1.108% | 1.660% | 2.210% | 2.758% | 3.305% | 3.850% | 4.393% | 4.935% |
I believe the above derivations and results are correct, but confirmation/refutation is welcome.
Last edited by whosnext; 10-08-2017 at 03:20 PM.