Odds of opponent flopping trips?

13% if they have a pp

@ Targetme: Thanks. Does pp stand for pocket pair? If so, that's unfortunately not what I meant.

I meant the likelihood that one of the two hole cards of some other player matches the pair that's already on the table (among the three post-flop community cards). I want to know this likelihood given that my hole cards do not match the community pair and given that I know nothing else about anyone's cards. Could someone please help?

Well he has two unknown cards. There are 52 cards in a pack and you know what five of them are (2 in your hand, 3 on the flop). so 47 cards are unknown. 2 of those unknown cards could match the pair on the board. So odds are 2:47 or approx 4.3% that any one of his cards match the par. But he has two cards so chances that he matches with one of those cards is around 8.6%.

But I just woke up and havnt had coffee yet so I could very well be wrong.

1,35%

@ evertonroar: Thanks! Your thoughts are in line with my best guess at what the odds are: For every additional player in the game, the likelihood that anyone has a card that matches the community pair rises by app. 8 %. I think that's it. Thanks, everyone!

Hey Irving3,

Unfortunately, 8% is not correct.

You want to look for common flop odd percentages, try Google, I'm certain you'll find something very comprehensive.

To answer your question:

Flopping precisely trips relating to one of your cards is 1.35%!

Congrats to ragney for nailing it, and a big thanks to everyone for chiming in here!

All the best,

Flopping precisely trips relating to one of your cards is 1.35%!

Agreed. But I don't think thats the question OP asked. He asked if the board is paired, what are the chances that someone else has flopped trips (so after the flop). Surely this has to be higher than the preflop odds of flopping trips.

ie the chance that someone will flop a full house is quiet low. But if the flop comes QQQ, then the chance that someone has flopped a full house must be higher and related to the chance that someone has a pocket pair.

Ah yes, I see where I may have misunderstood the scenario.

The exact percentage in such scenarios can't be consistently useful, as the scenario and actions taken therein would give you more accurate information.

For example, in a 4-bet pot, the chances of your opponent having trips on 223r are far lower than AAK or AKQ boards, as a general rule.

Whereas if you min-steal from the SB you can expect your opponent to have a defending range which includes more lower card combinations as opposed to a scenario following aggressive action pre-flop, such as the 4-bet scenario.

Because we're talking about post-flop, we need to take into consideration the texture of the board, as we have this information in front of us, as such there is no general percentage to describe such a scenario accurately I believe.

Some food for thought while I give the problem more consideration

I was looking for these odds and couldn't find them, so I decided to try and work them out myself. I know it's an old thread but maybe there are other people looking for them too.

I like to look at these problems the other way around. Let's say it's a heads up scenario. If your opponent DOESN'T have those trips (or quads), then (s)he must have a hand made from the other cards in the deck. So, they have 45 choices out of the 47 remaining cards for their first card, and 44 out of 46 for the second. These odds work out to (45/47)*(44/46) = 91.6%. So, the odds that your opponent DOES have at least trips (using the board pair) is 1 - 91.6% = 8.4%.

For more than one opponent, you can't just double the odds. For two opponents, the odds that neither one of them has either card that matches the board pair is (45/47)*(44/46)*(43/45)*(42/44) = 83.5%, so the odds that at least one of them has at least trips is 1 - 83.5% = 16.5%.

I tabulated these odds for a number of opponents below.

1 opponent - 8.4%
2 opponents - 16.5%
3 opponents - 24.1%
4 opponents - 31.5%
5 opponents - 38.4%
6 opponents - 45.0%
7 opponents - 51.2%
8 opponents - 57.0%
9 opponents - 62.4%
10 opponents - 67.5%

Of course, someone may have folded one of those matching cards pre-flop, so you should only count the people who are remaining in the hand.

Good stuff. You have the same info for PLO?

Originally posted by daddynohands
I was looking for these odds and couldn't find them, so I decided to try and work them out myself. I know it's an old thread but maybe there are other people looking for them too.

I like to look at these problems the other way around. Let's say it's a heads up scenario. If your opponent DOESN'T have those trips (or quads), then (s)he must have a hand made from the other cards in the deck. So, they have 45 choices out of the 47 remaining cards for their first card, and 44 out of 46 for the second. These odds work out to (45/47)*(44/46) = 91.6%. So, the odds that your opponent DOES have at least trips (using the board pair) is 1 - 91.6% = 8.4%.

For more than one opponent, you can't just double the odds. For two opponents, the odds that neither one of them has either card that matches the board pair is (45/47)*(44/46)*(43/45)*(42/44) = 83.5%, so the odds that at least one of them has at least trips is 1 - 83.5% = 16.5%.

I tabulated these odds for a number of opponents below.

1 opponent - 8.4%
2 opponents - 16.5%
3 opponents - 24.1%
4 opponents - 31.5%
5 opponents - 38.4%
6 opponents - 45.0%
7 opponents - 51.2%
8 opponents - 57.0%
9 opponents - 62.4%
10 opponents - 67.5%

Of course, someone may have folded one of those matching cards pre-flop, so you should only count the people who are remaining in the hand.

Originally posted by Wildncrazyguy
Good stuff. You have the same info for PLO?

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