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Old 10-08-2017, 02:53 PM   #1
whosnext
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Probability of being Oversetted on the Flop

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 Set1 Opponent2 Opponents3 Opponents4 Opponents5 Opponents6 Opponents7 Opponents8 Opponents9 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.
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Old 10-08-2017, 07:48 PM   #2
statmanhal
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Re: Probability of being Oversetted on the Flop

Quote:
Originally Posted by whosnext View Post

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?


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)]
[C(1,1)*C(11,1)/C(12,2)]

Are you sure of this term? If you look at cards instead of ranks and assuming hero has KK you would have C(1,0)C(4,1)C(44,1)/C(49,2), which is a lower number.

Am I using a different conditioning?
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Old 10-08-2017, 09:10 PM   #3
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Re: Probability of being Oversetted on the Flop

If Hero has KK and we are given that the flop has a King and two other cards of distinct ranks (and neither another king), then let's calculate the prob that there is an Ace on the flop.

By using cards, I get that this prob is:

[C(4,1)*C(44,1)] / [C(48,1)*C(44,1)/2]

The C(4,1) in the numerator is the number of ways to draw exactly one Ace from the deck.

The C(44,1) in the numerator is the number of ways to draw exactly one card from the remaining 11 ranks (third card on flop cannot be an Ace or a King).

The C(48,1) in the denominator is the number of ways to draw the second card on the flop; it cannot be a King so 52-4 is 48.

The C(44,1) in the denominator is the number of ways to draw the third card on the flop, given that it cannot be a King or the same rank as the second flop card.

The 2 in the denominator is needed since we do not care about the order of the flop cards. That is, 7s 4d is the same as 4d 7s.

Returning to where we were at the top:

[C(4,1)*C(44,1)] / [C(48,1)*C(44,1)/2] = 176/1056 = 1/6 as we found in the original post.

I think this is correct, but of course I am a "biased" observer.

If anybody thinks this is wrong or has other questions/comments/corrections, do not hesitate to post them.
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Old 10-08-2017, 10:42 PM   #4
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Re: Probability of being Oversetted on the Flop

I think we are using different conditions. You used 48 cards for the 2nd flop card then 44 for the 3rd, predicated on the condition that a 2nd king (or whatever hero’s rank is) hasn’t flopped.

I didn’t make that assumption but included the probability of no quads or full houses for either player. I suppose if you are given villain has a set and not better, then your condition is correct.
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Old 10-08-2017, 11:05 PM   #5
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Re: Probability of being Oversetted on the Flop

Yes, I am restricting my analysis to being exactly overset on the flop, so full houses or quads are not allowed (on the flop).

Thanks much for your comments.
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Old 10-09-2017, 01:30 PM   #6
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Re: Probability of being Oversetted on the Flop

Quote:
Originally Posted by whosnext View Post

For simplicity of calculations, I will assume that all opponents' cards are random (e.g., nobody ever folds pre-flop).
This is an important assumption. If an opponent sees the flop, he is more likely to have a pair than what random hands provide. Therefore, you should consider the table values to be lower bounds for an overset.
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Old 10-09-2017, 02:55 PM   #7
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Re: Probability of being Oversetted on the Flop

That's a very good point that should not be overlooked.

A similar assumption is typically made in many threads in this forum. If someone asks for the probability of making a flush in 7-card stud, the answer will most likely consider all possible 7-card hands, not attempting to take into account that some terrible starting hands would be folded much the time.

Or in answering how often does a straight flush occur in Omaha (the subject of several threads in this forum over the years), we typically assume all hands are equally possible.

First, assuming randomness in this fashion makes the math easier. Second, that is typically the type of answer the questioner is looking for when coming to the Probability Forum.

Only when "non-randomness" is key to the question (such as when specific hand ranges are vitally important) do we typically abandon the assumption of pure randomness.
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Old 10-10-2017, 01:33 PM   #8
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Re: Probability of being Oversetted on the Flop

Good point.

Upon further reflection, oversetting does indeed seem like a poker phenomenon for which hand ranges and card removal are vitally important.

I will try to get to a more realistic treatment later this week.
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Old 10-11-2017, 09:26 PM   #9
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Re: Probability of being Oversetted on the Flop

I’m a little late to this thread, but I just discovered this forum, saw this question, and decided to challenge myself in my spare time. I wanted to piggyback off the math whosnext did, so I took some time (or 3 hours...) to create a Monte Carlo simulation of the question, “what are the odds of getting overset on the flop, given that I have a pocket pair?”.

Model Assumptions:

1) Hero is dealt a random pocket pair in every hand.

2) It’s a 9-handed table where the other 8 opponents were all dealt 2 cards totally at random (i.e. hero is guaranteed to have a pair, opponents are not). I have the functionality to change the table size in my script, but for the sake of computation time, I only ran the numbers on a 9-handed table.

3) All iterations of the simulation saw a flop (in other words, there are no scenarios where the hand ended with a raise or re-raise pre-flop).

4) All iterations ONLY saw a flop (no turns or rivers, although I can probably go re-run that later).

5) There’s 4 possible outcomes: nobody flopped a set, hero flopped the only set, hero flopped an overset, or hero flopped an underset.

6) I ran the simulation 803,526 times. If that sounds like a really random number, then it is. I was going for a cool million, but it sounded like my computer was sucking wind, so that was the point where I gave it a rest.

Model Explanation:

The model randomly picks out a pocket pair for the hero, deals two random cards to each of the 8 opponents at the table, checks if any of the opponents have a pocket pair, and then if there is at least one other pair it will evaluate if a set over set situation occurred. If hero was dealt the only pocket pair, then it will test to see if his pair flopped a set before moving on to the next hand. This process was repeated 803k times.

Validation:

As a way to validate the results, I checked that the hero was flopping a set an appropriate amount of the time. It's been fairly well established that a pocket pair flops a set about 11.8% of the time. After 803k simulations, the hero flopped a set 94,442 times, or 11.753% of the time. I think that should at least somewhat prove that the random number generation was working fine.

Results:



We can see that the hero was oversetted ~.0025% of the time (1987 times).This comes out to about 1 in every 405 hands. If we change the question to ask how often any set over set situation happens (including when hero is the oversetter), then it's 1 in every 206 hands.

I believe whosnext's calculations involved the hero having already flopped a set, whereas my simulation only guaranteed the hero being dealt a pair with a random flop to come. Just using some back-of-the-napkin math in conjunction with an eyeball test, it looks like the table he posted in OP seems to center around the 2% area of an overset when hero has already flopped a set. If you then take 11.8% of that (how often a pair flops a set), then you're probably at least somewhere in the neighborhood of the .0025 my model came up with.

Note that there is a small difference in the number of trials where hero was overset and underset, but I'm guessing that's just a product of small sample (only ~3900 trials where a set over set situation happened). One would assume those numbers would eventually converge to even probability. The derived probability of .0025% is likely very close to correct though, as we saw the percentage normalize around that area as more trials went on (this graph is limited to only the first 50k trials, but I think it proves its point).



Any questions or observations, then please let me know. Thanks to whosnext for ruining my night with this challenge
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Old 10-11-2017, 10:19 PM   #10
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Re: Probability of being Oversetted on the Flop

Here is my approach to derive realistic estimates of the probability of a player being over-setted on the flop, taking into account the rank of Hero's set and the number of opponents seeing the flop. (As in my original post, I take as GIVEN that Hero has flopped a set and the flop comes with three distinct ranks.) Herein I posit a realistic hand range for all players, rather than assuming all hands are equally likely as I did above.

I have assumed that all players restrict themselves to a Top-25% Hand Range pre-flop. Players see a flop with the following hand range: 22+, A2s+, A7o+, A5o, K9s+, KTo+, Q9s+, QJo, J8s+, JTo, T8s+, 98s. There are 334 combos in all in this range which represents 25.2% of the C(52,2)=1326 total possible hole card combos.

The following table presents the over-set probabilities I have derived thus far based upon players using these hand ranges. (For ease of computation, I did not use the complement hand-range for any other players at the table not seeing the flop. I do not envision that the results would be significantly different if folded hands were taken into account but that is an open issue for now.)


Hero's Set 1 Opponent 2 Opponents 3 Opponents
A
0.00%
0.00%
0.00%
K
0.20%
0.29%
0.34%
Q
0.38%
0.67%
0.88%
J
0.60%
1.10%
1.53%
T
0.77%
1.43%
2.03%
9
0.90%
1.71%
2.44%
8
1.06%
2.05%
2.99%
7
1.21%
2.37%
3.49%
6
1.34%
2.72%
4.07%
5
1.55%
3.14%
4.80%
4
1.68%
3.46%
5.35%
3
1.84%
3.81%
5.98%
2
2.01%
4.24%
6.68%


I should note that the results of Hero being over-setted vs. only one opponent are exact (subject to the proviso mentioned above). There are few enough cases that a computer can run through all cases in brute force manner. Each of the results from the other columns are based upon simulations of 1,000,000 deals with Hero flopping a set of that specific rank, the flop having three distinct ranks, and all opponents having a hand in the Top-25% Hand Range. These numerical results are likely to be accurate within .02%.

Roughly speaking you will see that the probabilities in the first column (vs. one opponent) of this table are approximately four times larger than the corresponding probabilities in the original table posted in the thread. Of course, the original table used a 100% hand range for all players, and 100/25 = 4. Mathematically, in the 100% range case pocket pairs represent 78/1326 = 5.9% of the range, whereas in the 25% range case pocket pairs represent 78/334 = 23.4% of the range.

Turning our attention to the second column of the table above, a few things stand out. First, the "card removal" effect is fairly strong at the top of the column. Whereas with a set of Kings Hero can be expected to be over-set when facing exactly one opponent (clearly only a set of Aces oversets a set of Kings) 0.20% of the time, the top-heavy overlapping hand ranges of multiple opponents makes the probability that Hero holding a set of Kings is likely to be over-set when facing exactly two opponents 0.29% of the time. The reason that it is not closer to 2*0.20% = 0.40% is that there just aren't enough Aces to go around. Since now there are two opponents with Top 25% hand ranges (top-heavy with Aces), the flop is dealt from a deck stub "poor" in Aces. This is a reflection, of course, of the well-known "bias" in flop ranks that Spadebidder and others have discussed over the years.

Run your eye down the second column and compare it to twice the entry in the first column. You'll see, as we just mentioned above, that the second column entries start far below twice the first column entries. But around the middle of the column the entries "catch up". And by the time the bottom of the second column is reached, the entries noticeably exceed twice the corresponding entry in the first column. Why is that happening?

For the same reason that the card-removal effects make the deck stub from which the flop emanates "poor" in high cards (such as Aces and Kings), the stub must be "rich" in low cards (such as deuces and threes). Thus, there is a slightly elevated chance that a low card appears on the flop, and is thereby "available" to make a set with low pocket pairs held by any opponent.

The same set of phenomenon occur in the third column of results for when Hero and exactly three opponents see the flop. The chance of Hero being over-setted when holding a set of Kings barely rises (going from .29% to .34%) when the number of opponents seeing the flop goes from two to three. (It will be interesting to see how this probability changes when a fourth opponent using an Ace-heavy hand range is added to the mix.)

Similarly, the "card removal" over-set phenomenon see-saw is balanced (there are roughly the same number of over-sets that are positively and negatively affected by the card removal phenomenon) around the point where Hero flops a set of 7's. Again, at the very bottom, if Hero flops a set of Deuces, the probability of being over-set by three opponents is noticeably greater than three times than if Hero is only facing one opponent.

Keen observers will notice that the table only reports three columns of results. This is due to the fact that the one million trials takes a lot longer for the cases in which there are more than three opponents seeing the flop. If and when I ever obtain additional results, I will append the results to the table at a later date.

If anybody is interested in the simulation methodology, the reason that the number of opponents has such a large effect on the runtime is that I am being scrupulously "fair" (or stupid, depending upon your perspective) in drawing hands with the correct multi-distributions. For example, rather than simply starting with four opponents each with hands in the designated hand range, I distribute cards completely at random (other than Hero's hand and the first card of the flop) and then only inspect whether any opponent has over-setted Hero. Of course, with a limited hand range for each opponent, it takes an inordinate number of random deals to finally obtain a deal in which all four opponents each have hands in the designated hand range. I didn't anticipate that this would be much of an issue, but the very lengthy runtimes have taught me otherwise.

Last edited by whosnext; 10-12-2017 at 01:36 AM. Reason: fixed another typo (thanks go to ngFTW for alerting me to it)
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