4 of a kind

[kingstalker]

Anybody know the odds of a player getting 4 of a kind two hands in a row in Texas hold them? Got busted out in a tournament today by a player who halved my stack when I had K-K vs Q-Q and we got it all in before the flop, and then got kncoked out on the very next hand when I had A-Q vs his 8-8 and he floped 4 of a kind again.

[anduril]

I'm not sure of the odds but I don't think that could possibly have sucked anymore than it did. I feel you man, right here

[kingstalker]

what really sucked about the first one was that the board came down 10-Q-8-Q and then to add insult to injury a king came down on the river giveing me a useless boat, on the very next hand when the same guy got 4 of a kind again I let out a weird maniac laughter when I saw him get the 4 again, time to go to the doctor for a check up!

[hukilai]

The odds of having four of a kind is 0.00024. So, to have four of a kind in two hands in a row is about one chance from 16,000,000 (sixteen million) [img]/images/graemlins/smile.gif[/img]

[bigpooch]

In holdem, if someone has a pocket pair, his chances of
making four of a kind (using his pocket pair) are

C(48,3)/C(50,5) = 17296/2118760 = 0.0081633

Squaring the above yields 0.0000066639. Of course, there
are less likely scenarios such as having quads on board or
making quads with just one card. For the case of making
quads with two different ranks in the holdem hand, the
number of combinations is:

2xC(47,2) = 2162

So for someone to have quads two hand in a row, the chances
(with this added possibility) are about 0.000084340. Of
course, your opponent did you in by the usual way!

[kingstalker]

Im not very good at math, is your number the same as the other guys 1 in 16 million? Basically I want to know so I can tell my friends about my bad beats and the odds of a certain occurance happening.

[bigpooch]

Actually, the first number is exactly 2/245 so it is
15,005.25 to 1 against, not as rare as the other poster had
suggested (I think he had in mind a game other than holdem).

[bigpooch]

Sorry, my post did not go as I intended. It's actually
15005.25 to 1 against someone with a pocket pair to make
quads twice in a row.

[BruceZ]

C(48,3)/C(50,5) = 17296/2118760 = 0.0081633

Squaring the above yields 0.0000066639. Of course, there
are less likely scenarios such as having quads on board or
making quads with just one card.


It is actually MORE likely that you will make quads by using just 1 card from your hand than with a pair in the hole when you factor in the probability of getting dealt a non-pair which is 16 times the probability of getting dealt a pair. You can also get a pair on one round and a non-pair on the other.

[ (1/17)*C(48,3) + (16/17)*2*C(47,2) ]^2 / C(50,5)^2
= 481,866 to 1

This includes everything except the split pot with quads on board.

If we only consider pairs in the hole, then it is (1/17)^2 to get dealt a pair twice in a row, times 1/1,5006.25 to make quads after we get the pair, or 4,336,805 to 1 all together to get dealt a pair and make quads twice in a row. So it is much less likely to do it with a pairs alone than with non-pairs by about a factor of 10.

[BruceZ]

It's actually 15005.25 to 1 against someone with a pocket pair to make quads twice in a row.

This assumes we already have the pairs twice in a row. The probability of being dealt a pair is 1/17, and it is (1/17)^2 to be dealt a pair twice in a row. So all together it is (1/17)^2*(1/15006.25) or 4,336,805 to 1.

If we consider the cases where we make quads with just 1 hole card, it becomes about 10 times more likely. See my other post in this thread.