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Peter-23
08-23-2005, 04:58 AM
What are the odds of running in to four of a kind when holding a house three times within a 1000 hands range.

(one paired board and holding a pocket pair)

AaronBrown
08-23-2005, 02:38 PM
Do you mean what is the chance of a full house on the flop turning to quads by the river?

If so, and you hold XX while the board is XYY, you could get four of a kind if either of the last two cards are X, or if both the last two cards are Y.

There are 47 unseen cards, 1 of which is X and 2 of which are Y. These 47 cards can be combined 47*46/2 = 1,081 ways. 46 of these ways include an X (X with any other card, including Y), so that's 46/1,081 = 2/47 = 4.3%. There is only 1 way to get both Y's, so that's 1/1,081 = 0.1%.

KJL
08-23-2005, 03:13 PM
I think he meant what are the chances of someone having quads when he has a boat. For instance, he has 55 and the board is 577K4, he wants to know the probability someone has 77. I'm not sure exactly how to do this, but I will try it later.

Peter-23
08-23-2005, 03:31 PM
No, I mean if we have a full board of five cards with one pair and the rest of different rank.

I have a pocket pair that make a set with one of the three diffrent ranks and of cause a house since there is a pair on board.

If I play 1000 hands and get at least three houses. What are the odds that someone else will get four of a kind three of those times.

This is perhaps nonsens knowlage but I just like to know how rare it is to have these kind of events occur this close together.

Actually I try to figure out the probability for extreme events to happen within a playing session.

Lets say I play 1000 hand each day, then how many days a week or moth or year will extreme events like this occur.

If you could tell me how to calculate it for my self I would be very gratefull.

(assume ten handed tables).

LetYouDown
08-23-2005, 03:31 PM
[ QUOTE ]
I think he meant what are the chances of someone having quads when he has a boat. For instance, he has 55 and the board is 577K4, he wants to know the probability someone has 77. I'm not sure exactly how to do this, but I will try it later.

[/ QUOTE ]
If you're playing heads up, the probability someone has 7-7 in that scenario is 1/C(45,2) = .10101...%

And if that's the way you want to approach this, the probability of that happening exactly 4 times in 1000 is: ~1.576%. This obviously isn't a reasonable approach but I believe it's the answer he's looking for based on his second post. This assumes you already have the full house, including a third of your card and the board is dealt.

And if you're looking for 4 or more times in 1000 hands, I believe it's ~1.955%.

AaronBrown
08-23-2005, 07:05 PM
Okay, let me try again.

If you play 1,000 hands of Hold'em, with the full board dealt, 423 of those hands will have one pair but no more on the board.

Given the board, there are 47*46/2 = 1,081 possible pocket cards. 9 of them make a full house. So you expect 9*423/1,081 = 3.5 full houses. Only one pocket combination makes four of a kind, but with 9 opponents you expect it to happen the same 9*423/1,081 = 3.5 times.

It is slightly more likely that one of your opponents has four of a kind when you have a full house, because the two cards in your hand that don't match the pair mean there are only 990 pocket combinations. So the chance of both things happening on the same deal, your full house losing to someone's quads, is (1,098,240/2,598,960)*(9/1,081)*(9/990) = 0.000032.

The chance of that happening 3 or more times in 1,000 hands is 1/188,394, less likely than being dealt a Jack high straight flush or better in five cards.

Peter-23
08-24-2005, 04:42 AM
This is great stuff, thanks.

But how do you reach the number 1/188,394.

LetYouDown
08-24-2005, 08:42 AM
Through a binomial distribution:

BINOMDIST(997,1000,1-1098240/2598960*9/1081*9/990,TRUE)

Comes out to 188394.414