Not sure how that could possibly offend anyone lol I was just answering what I thought that person was asking. I pointed out the difference in our answers so that he didn't get the idea that our answers conflicted (they don't conflict since they're answers to different questions).

I'm not sure this has been directly answered or if you are still looking for one. Anyway, GIVEN you have exactly a set and a single opponent with a random hand sees the flop, the probability he also has flopped a set is 0.48%.

This is a lower bound for the fact that villain has seen the flop will likely significantly increase the chance he has a pair over the nominal unconditional 6%.

Isn't this 2*C(3,2)/C(47,2) = 6/1081 = 0.555%?

Maybe I am not doing the same calculation you are or maybe I did it wrong.

I excluded paired flops so flop has to have three different ranks, one of which matches "your" pocket pair. So there are two other ranks available for opponent to have a pocket pair of. Once flop comes down, opponent has any two of the remaining (unknown) 47 cards.

I often do these types of calculations wrong, so don't hold back if I did it wrong.

Yes.

With N villains seeing the flop, it's

2N*C(3,2) / C(47,2) -
C(N,2)*C(3,2)^2 / C(47,4) / 3

= (6N / 1081) - 3*C(N,2)/C(47,4)

With 5 villains that gives an answer of 2.7584%

I got the same result for villain to have a matching pair for a flop of type x,y,z, where z matches hero's pair and is a given. But, you have to multiply that result by the probability of an x,y,z flop, which is

C(12,2)*16/C(50,2)=0.862 to give 0.48.


I also did it another way, viz:

Given you have a set, a single opponent will have a set if

1. He has a pair (which has to be of a different rank if you have a set

and

2. One of the two remaining cards on the flop matches his pair rank and the other doesn’t match either rank

Opponent pair. There are 12 remaining ranks, each with 6 possibilities for a pair or 72 combos for villain to have a pair out of C(50,2) = 1225 combos.

One flop card has to match his pair. There are 2 such cards that give him a set, another flop card cannot match either pair (11 ‘good’ ranks) and the third flop card is given to match hero’s rank. So the number of success flops is 2*11*4 = 88 out of C(47,2) = 1081 possibilities. So

P(opp has set given you have a set) = 72/1225 * 88/1081 =0.48%.

Either I'm wrong twice or we're not solving the exact same problem.

Not really, but your response is greatly appreciated. (I'm not being sarcastic).

Thanks.

Buzz

That's what I got for the flop. (And that's how I did it).

My original answer was for the river. I used 12/1081 = 0.00111, or 1.11%. (And then I rounded down to about 1% to get what I thought was close enough).

Thanks.

Buzz

Isn't 0.00555 for one opponent small enough to simply multiply by the number of opponents to get a good approximate answer?

In other words, why not simply 0.00555*N?

Buzz

I think this is the key phrase. Given that the flop has three ranks and you have flopped a set, what is the probability that a single opponent has also flopped a set?

The "dynamics" of the situation are significantly changed if the flop has a pair on it. And, of course, poker terminology differentiates between flopping "a set" and flopping "trips".

P.S. I guess it is obvious but just in case it needs to be said, if you have 88 and the flop comes down 855, you have flopped a full house not a set. And, of course, if you have 88 and the flop comes down 885, you have flopped quads not a set.

Sorry, statmanhal, in re-reading your post it is clear that you are indeed calculating the probability under the scenario that the three flop cards are all different ranks. So I apologize for my misunderstanding. But why do we get different answers then?

I attempted to write a simple computer program to run through all the cases but I am not sure I did it right since I get a different answer than either of the previously reported results. Permit me to review the code and post when things are clearer to me.

Alright, I am back. I reran my computer program and here are the results.

Given "you" have a pocket pair and a third card of the rank appears on the flop, there are 1,141,536 ways to fill out the other two cards of the flop and two cards to your single opponent in which the flop is not paired.

Of these 1,141,536, there are 6,336 ways in which the opponent also flops a set. This fraction simplifies to 6/1081, a result found above by several posters. [6,336 = 72*88 as statmanhal derived above.]

Where does the 1,141,536 come from? Start with C(49,4) since we need to select 4 of the remaining 49 cards (excluding "your" three set cards). There are 6 ways for these 4 cards to form the remaining two flop cards (C(4,2)=6), with the other two cards, of course, forming the opponent's hand. 6*C(49,4) = 1,271,256.

But not all of these result in a three-ranked flop (unpaired). We need to exclude the fourth card of your set appearing on the flop. There are C(48,3) combos of these other 3 cards, and we need to multiply by 3 since any of the 3 cards could fill out the flop. Also, we need to exclude a different pair from appearing on the flop. There are C(12,1) ranks for the pair, C(4,2) of the rank appear on the flop, and C(47,2) appear in the opponent's hand.

So we need to exclude 3*C(48,3) + C(12,1)*C(4,2)*C(47,2) = 51,888 + 77,832 = 129,720 cases from the 1,271,256. This leaves 1,141,536 ways to fill out the flop, without having a pair on the flop, and the opponent's two cards.

Thus, if this is correct, the probability of your opponent flopping a set when you flop a set on an unpaired flop is 6,336/1,141,536 = 0.555% (which was previously derived as 6/1081).

He didn't assume the flop was xyz, he multiplied by P(xyz flop). But like you said, the only way to flop a set is for the flop to be xyz, so if we're assuming we flopped a set then we're also assuming the flop was xyz.

Yeah. For 5 opponents it would only be off by 0.168%, and for 9 opponents by 0.06%. At the table, N*.555% would be the formula to use.

But the reason it's accurate isn't because .00555 is small, it's because the chance of being against 2 sets is small (that's what's being double-counted when you take N*.555%).

Ah. Thank you. Very helpful... a much better way to think about it. Thank you.

Intuitively it seems like it should be off more when you have more opponents. (With more opponents there's a greater chance of being up against 2 sets).

Thanks.

Buzz

Sorry good catch, I meant .0168% for 5 villains.

This was a silly thing to say. At the table, both formulas are useless, as are all the stats itt. Villain is not playing randomly, and if he were, we wouldn't need to consult a formula to get it in with bottom set against a range of ATC.

I disagree. There's quite a lot to playing poker well, more than it probably seems to a newb.

Poker is a game of deception.

I think to be long term successful you want to know the poker game you're playing. There's quite a lot to knowing the game well. And knowing how likely you are to be up against something is part of knowing the game well.

At any rate, I greatly appreciate the mathematical assistance you and others give on this forum. I don't think it's silly at all to have a quick, easy way to estimate the likelihood an opponent betting as though he has a higher set actually would have been dealt cards leading to a higher set.

Agreed.

But the range of any Villain, even a maniac, is not likely "any two cards."

That's just my opinion, based on my experience. Whether you agree with me or not, I greatly appreciate your opinion.

Thank you.

Buzz

I think it's much more useful to know that the chance Villain has a set is 6 / [total combos in range]. No one's range consists of 1081 combos except this one guy who blind-called me preflop and then blind-shoved the flop (I called with Q high and it held up).

But you have a point about stats providing a feel for the game. A novice might not know how strong bottom set is until he sees the .555% stat (among others).

While that's almost always true, I'm never folding a set to a wide/maniacal range and I definitely don't need to do any math to make the call. But the math would be quick: we could stop counting his range after just a few trash combos because it only takes a few for our call to be correct.

Good point. I agree knowing an opponent's range and thus the total number of combos in that range would be more useful, however for me it's unrealistic to know any particular opponent's range.

First, there's considerable variance in the hands with which most opponents oppose me. What I mean is from the same position, sometimes they'll play a particular combination and other times they won't.

Second, I mostly play live... I could not realistically collect and correlate enough data on each particular opponent to have much assurance about the range he/she plays.

Third, Omaha-8 is my game of choice and starting hands are more complex than in Texas hold 'em.

Interesting anecdote.

Thanks for sharing.

Buzz