Odds of Losing with 4 of a Kind in Texas Hold'em

[Eezekiel]

I recently got beat in playing poker in the worst way that I have ever seen before. I was playing on a poker site online (pokerstars) and got beat with four of a kind to a higher four of a kind. The game was texas hold'em. For those of you who don't know texas hold'em, that means you get dealt 2 hole cards and combine the 2 hole cards with 5 community cards to make your best 5 card hand. As you all probably know there are 52 cards in the deck with 4 different suits and 13 different rankings for each suit. Could someone please tell me the odds of losing with 4 of a kind, four 3's to be exact, in the game of texas hold'em.

 

[Feldoh]

What beats 4-3's in Texas Hold Em'?

4-anything higher
flush
royal flush
full house

Does a high enough two pair beat four threes?

 

[Eezekiel]

4 of anything higher beats 4 3's.

not that might not need to know but the hands go as follows

royal flush-10JQKA of one suit
striaght flush-5 cards in sequential order of same suit
4 of a kind-2222,3333,...AAAA- AAAA being highest
full house-55522 three of a kind with a pair
three of a kind-QQQ
2 pair-33JJ
1 pair-AA
high card


card rankings-A K Q J 10 9 8 7 6 5 4 3 2
suits-clubs diamonds hearts spades

 

[Feldoh]

Hmm that's some crazy math.. I don't think I could do it, but http://en.wikipedia.org/wiki/Poker_probability_%28Texas_hold_%27em%29 might be able to give you some more info

 

[nicksauce]

Let's assume you hold 33 and on the board is 33XYZ... your opponent holds AB. So we need to find the probability that 4 out of {A,B,X,Y,Z} are equal. This would be
(47/47) * (3/46) * (2/45) * (1/44) * 5C4 ~= 1/3036. There will need to be a slight correction since four 2's lose to four 3's.

Now, let's assume you hold 33 and the board is 33XXY and X > 3. We need to find the probability your opponent holds XX. This would be (2/47) * (1/46) ~= 1/1081.

Now let's assume you hold 33 and the board is 33XXX and X > 3. We need to find the probability your opponent holds XY. This would be ~= 1 / 23.

So it all depends on what you define as your given variables.Maybe some mistakes here, my probability is a little rusty.

 

[nicksauce]

Using Monte Carlo simulations with the program "pokerstove" http://www.pokerstove.com/download/

I found the following probabilities

Probability of winning given that you hold 33 and the board is 33XYZ -> 99.955%
Probability of winning given that you hold 33 and the board is 33XXY -> 99.710%
Probability of winning given that you hold 33 and the board is 33XXX -> 95.556%
Probability of winning given that you hold 3X and the board is 333YY -> 99.899%

 

[quasar987]

What was the board when you played?

 

[nicksauce]

Note also that the calculated probabilities are valid for being against one opponent and you should (roughly) multiply the probability by itself N times where N is the number of opponents in the hand.

 

[Eezekiel]

the board was AA333. I, of course, was holding Q3. So the odds of him flopping a quads are incredible, let alone me catching runner runner 33 for my quads. I have literally played millions of hands and have never seen that before in texas hold'em. Of course in other games like 5 card draw with wilds and so forth I have seen quads up against quads. I play poker maybe 5 days a week sometimes putting 5-12 hrs in a single session and I get quads maybe about once or twice a month. So you could imagine my disbelief when the guy flipped over quads to my quads. The worst part about it was that I was a big blind,ie. forced bet,
so i had no choice but to play my hand. The guy didn't raise before the board cards were put out; I was doomed no matter what lol. (quads= four of a kind)

 

[quasar987]

You had 1 chance in 1 thousand to lose that hand.

I hope that wasn't no limit hold'em! :P