Please could someone tell me how often i will flop 2 pair (i pair my 2 hole cards) and my opponent will get trips with the remaining card. Or how about this if i flop 2 pair 1000 times out of that 1000 how many times is opponent gonna hit trips with the remaining card?
I am so sick and tired of leaving live tourneys because of this, at least half a dozen times in the last year with an average of 1 live tourney a week. I am really getting scared to bet or call a bet when i flop 2 pair! My thinking is that this should be hard to do as i really cant remember seeing it happen to anyone else and i look for it.

The other guy is gonna win 8% of the time. According to your rough estimate he is winning about 11% of the time. That seems to be within acceptable parameters.

How did you calculate this? My analysis shows a much lower number assuming we're calculating the same event probability.


For heads up, given you hit two pair, the chance your opponent has a pair and the third flop card gives him trips is

11*C(4,2)/C(48,2)*(2/46)+2*C(2,2)/C(48,2)*44/46 = 0.424%.

Here I included in the calc the chance villain has a pair and that its rank may match one of your cards. So if you flop two pair 1000 times, he will have trips between 4 and 5 times.

Of course, the above assumes heads up but since the probability is so small, you can simply multiply it by the number of players in a multi-way hand for a very good approximation. Also, I took no account of villain's decision making in seeing a flop with or without a pair.

Just a quick comment.

You will appear to see this more than the above calculation indicates since the hand is virtually guaranteed to go to showdown if one player flops two pair and another player flops a set.

But if you flop two pair and nobody flops a set (or hits a set later), the hand may very well not get to showdown.

So you remember the times that it does happen far greater than the times that it does not happen.

Maybe i misunderstood question. I thought the other guy flopped a pair and the next card or river gave him three of a kind.
Flop a k 4 villian has 4 turn OR river is 4.
Is 8 % right for that situation?

There are 2 remaining 4’s in remaining deck of 47 cards. Probability of getting exactly one more 4 on turn or river is 2* 2/47*45/46 = 8.3%, so on rounding you’re right. Don't think that was the question though.

So, the question is: How often do you flop two pair and the villain flops a set?

Seems like the percentage would be higher than set over set which I think was over 1%
according to another thread on here. I guess the calculation being heads up makes a difference.

I get a slightly different answer. I explicitly required the third card on the flop to have different rank from the other two flop cards. My approach gives the same numerators but slightly different denominators.

Maybe I am misunderstanding something or did something wrong.

I got the same result as statmanhal by counting partitions between flop card and villain's hand
(11*4*C(3,2) + 2*1*11*4*1) / (48! / (2!*1!*45!))
It looks he is using conditional probabilities which I actually like more as it reduces errors

Sorry for not explaining very well and thanks for all the answers.
this is the hand
I had QJs villain has 55, the flop is Q 5 J. He was first in on the btn with a 4x raise i was the Big Blind called. Just me and him sb folded

Here’s my breakdown:

11*C(4,2)/C(48,2)*(2/46) = probability villain trips on flop given he pairs both his hole cards and the trips is of different rank than hero’s ranks.

11-number of ranks not matching hero’s holding
C(4,2)- number of pairs in a rank
C(48,2)- 4 cards accounted for – hero’s holding and 2 pairing flop cards leaving 48. So, C(48,2) is number of possible holdings for villain.
2/46 – there are 2 remaining cards which will give villain trips on the flop given he has a pair and there are 46 unknown cards

2*C(2,2)/C(48,2)*44/46 = same as above but the trips is equal to one of hero’s ranks

2 – number of ranks hero has
C(2,2)- number of ways villain can pair one of hero’s ranks
C(48,2) - same as described above
44/46 – to avoid hero having a full house, third flop card cannot match one of hero’s ranks and there are 44 such cards

Thanks statmanhal
much appreciated soooo were talking like 1 out of 100 or more
and i'm just getting clobbered by the deck not playing it wrong

On a full table (ten handed) would the answer be 10 × .42%

For edification purposes only ...

Clearly I must be wrong but here was my train of thought for the denominator (we agree on the numerators):

Hero has two unpaired cards. (Given)

Flop has two cards each of which match one of hero's hole cards. (Given)

Third card on flop must be of a different rank. (Choose one of the 44 cards that "qualify".)

Villain has any two cards (Choose two from the remaining 47 cards of the deck.)

I guess this is not kosher. But why not?

Thanks.

It's kosher and halal.

For sharing one of hero's ranks, he did 2/C(48,2) * 44/46
You did the 3rd flop card first and then villain's cards, which should be 44/48 * 2/C(47,2)
which is the same*, so idk what you did. Nor does the order of conditional probabilities (in the product) matter for the unshared rank case. Either way it combines to the same single fraction you'd get when doing it as one probability.

*That they're the same can be seen without crunching the numbers.
C(48,2)*46 = 48*C(47,2) because they both equal C(48,3)*3

I am not being clear I guess. I get a different prob since I do not think that (44/48) is relevant for the problem we are addressing.

The prob of the third flop card not matching one of the other ranks is NOT 44/48. It is 1. It is a GIVEN in the problem:

"Given that I have two unpaired cards and the flop contains three unpaired cards two of which match my hole cards, what is the prob that villain has flopped a set?"

I get 220/47,564 as this prob.

Others get 220/51,888.

I would be happy to be wrong here. But I am not yet convinced that I am.

That was my interpretation too. I thought you were saying you tried to answer the same thing as statman but still got a different result. So you don't actually disagree with his math, then, just his interpretation of OP.
Gotta run out the door now. I'll double-check later, but I don't doubt it. I won't post unless I get a different answer.

@whosnext:
So you fix the flop and simply calculate (C(3,2) + 2*1) / C(47,2)

That is very nice and simple and correct. I personally jumped to statman's interpretation to see if I can get the same result and didn't read carefully OP's question

Nowhere did I see OP explicitly say there were three unpaired flop cards but it is not unreasonable to assume that from his statement. I specifically included the probability of not having the third card pairing either of the other two and whosnext had that as a given. I tend to think whosnext is more right, but in either case the probability is very small.

No big deal, but I don't see any other way to interpret OP other than:

"Given that I have two unpaired cards and the flop contains three unpaired cards two of which match my hole cards, what is the prob that villain has flopped a set?"

Clearly, if Hero holds J8 and flop comes J99, then Hero would flop two pair but not by pairing his two hole cards.

If Hero holds JJ and flop comes 992, then Hero would flop two pair but, again, not by pairing his two hole cards.

Finally, if Hero holds J8 and flop comes JJ8 or J88, then OP would not say that he flopped two pair!

Anyway, leaving all that aside, this silly little thread is an exemplar of the best aspects of the Probability Forum. Great people like statmanhal, kapw7, heehaww, and others helping out 2+2 members to better understand the likelihoods of various poker phenomenon.

In many cases, it turns out that coming up with a concise description of the phenomenon is half the battle!

Thanks again to everyone who contributed to this thread.

Well Said
Yes getting the description right is sometimes very difficult
Thank you gentlemen for your very able and concise answers

I don't get how any other way should even be contemplated.

I thought you were against using combinations?

I see you calculated this for heads up play. Does the solution change very much for a full table. My thinking is that most tournament play is on full table and that is when OP sees most of these coolers.

The number of opponents doesn't change the probability (or the actual frequency it occurs).