What are the odds of having 2 Quads in the same hand?

Question

Just watched a video that went as follows:

• Player 1 (Button): Q♥Q♠
• Player 2: 9♥9⋄

• Player 3: 2♣2♠

• Flop: Q⋄9♠Q♣

• Turn: 5♠
• River: 9♣

So player 1 slow plays everything, player 2 has 9s full of Qs post flop and a quad by the river. I was talking about it in the office and said that's possible the worst bad beat I've ever seen and someone piped up with he seeing stuff like that, and I quote, 'all the time'.

Is there anyone could calculate the odds of 3 people getting pocket pairs, then one of them flopping a quad and someone else getting a full house at the same time and him ending up being beat on the river despite hitting quad himself?

I don't think those odds are very good...

Answer 1

Here is the maths for you. Or well my maths anyway.

With 7 cards to choose from in hold'em, your hole cards and the board, the odds of making quads is about 1 in 595. (13 * (48 choose 3)) / (52 choose 7) which = 0.00168067227 or 1 in 595. This is over the entire 7 cards.

So for another person to have quads in the same hand we figure out how many possible hands are left. We know the player who has quads has 4 of the same cards, 2 in his hand 2 from the board. Leaving 2 from the other players hand and 3 more from the board giving 5 cards. This leaves 1,712,304 possible hands with 5 more cards. (48 choose 5) / 5! = 1,712,304.

So player one has effectively knocked out a card from the original calculation to get quads, so rather than 13 quads to choose from there is only 12 possible quads left. We can work out the odds of this player getting quads, which gives up 528 possible hands to give him quads in this situation, 12 types of quads left, 44 cards to choose from 12 * 44 = 528.

So now that we know the other player has 528 possible hands to give him quads over 5 cards, and we know the total number of possible hands left to be dealt, 528 / 1,712,304 = 0.031% or 1 in 3243.

This situation only factors in heads up like the above hand, and if both players have pocket pairs. It would change if the second player needed 3 cards on the board for their quads.

Answer 2

What is the probability that two players will have different four-of-a-kinds in Texas Hold’em?

ANONYMOUS

Between two players there are 9 total cards. These must consist of two four of a kinds and one singleton. The number of combinations for this is combin(13,2)*44 = 3432. The total number of ways to pick 9 cards out of 52 is combin(52,9) = 3,679,075,400. So the probability you have the right cards, but not necessarily in the right order, is 3432/3,679,075,400 = 1 in 1,071,992.

However just because the cards are AAAABBBBC doesn’t mean both players will have different four of a kinds. The number of ways to arrange them into a 5-card hand and two 2-card hands is 9!/(5!*2!*2!) = 756. Following are the ways those 9 cards can fall.

Four of a Kind Bad Beat Combinations 36

Player 1

Player 2

Flop

Mirror Patterns

Combinations per Pattern

Total Combinations

AA

BB

AABBC

2

72

AA

AB

ABBBC

4

48

192

AA

AA

BBBBC

2

6

12

AA

AC

ABBBB

4

12

48

AA

BC

AABBB

4

24

96

AB

AB

AABBC

1

144

144

AB

AC

AABBB

4

48

192

Of these only the first and the fifth group result in both players having a different four of a kind. So the probability that an AAAABBBBC set of cards results in two different four of a kinds is 168/756 = 22.22%.

So the answer to your question is (3432/3,679,075,400)*(168/756) = 1 in 4,823,963. On a more practical note Party Poker has a bad beat jackpot for a losing hand of four eights. Given that there are two four of a kinds the probability that both are eights or greater is combin(7,2)/combin(13,2) = 21/78 = 26.92%. So the probability that any one hand of two players will result in this bad beat jackpot is 1 in 17,917,577.

Answer 3

This is the odds of quads vs. quads (any pairs) happening in a hand of HE

The probability of this can be done using combinatorics. There are 2 ways that quads over quads can happen

Scenario 1 Player 1 (x x) Player 2 (y y) Board (x x y y z)

Scenario 2 Player 1 (x x) Player 2 (y z) Board (x x y y y)

So odds of this happening are the addition of these two scenarios

Scenario 1 Player 1 Combos = C(13,1)*C(4,2) = 78 Player 1 % = 78 / C(52,2) = 5.88235% Player 2 Combos = C(12,1)*C(4,2) = 72 Player 2 % = 72 / C(50,2) = 5.87755% Board Combos = C(4,4)*C(11,1)*C(4,1) = 44 Board % = 44 / C (48,5) = .00026% Player 1 % * Player 2 % * Board % = .0000089%

Scenario 2 Player 1 Combos = C(13,1)*C(4,2) = 78 Player 1 % = 78 / C(52,2) = 5.88235% Player 2 Combos = C(12,2)*C(4,1)*C(4,1) = 1056 Player 2 % = 1056 / C(50,2) = 86.20408% Board Combos = C(2,2) * C(2,1) = 2 Board % = 2 / C (48,5) = .00012% Player 1 % * Player 2 % * Board % = .0000059%

Scenario 1 + Scenario 2 % = .0000089% = .0000059% = .0000148% This is equivalent to 1 in 6,753,548

Please correct me if my math is flawed

Answer 4

Here are the odds of this exact hand happening exactly the way it did (any 3 pocket pairs)

QQ = C(13,1)*C(4,2) / C(52,2) = .05882 99 = C(12,1)*C(4,2) / C(50,2) = .05878 22 = C(11,1)*C(4,2) / C(48,2) = .05851 Flop QQ9 = C(1,1)*C(1,1)*C(2,1) / C(46,3) = .00013 Turn Blank = C(42,1) / C(43,1) = .97674 River 9 = C(1,1) / C (42,1) = .02381

Multiply these percentages together and you get 1 in 1,613,347,458

If you don't care the order the board runs out then its 1 in 161,334,746

If you are talking the exact pairs and the order the board runs out in doesn't matter then its 1 in 276,850,423,850

Answer 5

11 million to 1 (ignoring the 3rd pair)

Two players holding a pair
(1/1)(3/51)(48/50)*(3/49) = 0.003457 = 288.2 : 1

Now 48 cards left and need to fill in by the river
how many ways to take 5 cards from 48 = combination(48/5) = 1712304
one way to get the the exact 4 cards needed
44 other to fill in the 5th
1 * 44 / 1712304 = 0.0000257 = 38915 : 1

Net 0.003457 * 0.0000257 = 0.00000008884 = 11,255,912 : 1

That is like 100 hands a day for 300 years

As the cards come off one has to have a full house at some point so that means nothing

actually you could have quads without a pair
hand 1 xx hand 2 yz
board xxyyy
If you allow for that the possibility of two quads is
535,995 : 1
13 * 12 * 44 / (52 / 9)
no that is wrong as it could include
hand 1 xy hand 2 yz
board
xxxyy
hey, it is a big number

I think the answer from Grinch91 is correct 1,929,584 : 1
One hand must have pocket pair and the other is not required to have a pockets pair

Answer 6

My quad kings on the flop of (KKA) were beaten by pocket aces for AAAA over KKKK. I calculated the odds of that happening were @1:56,000,000. It's one hand I'll never forget just because of the outlandish probabilities involved. However, I did also, in a live cash game, play 57 suited to straight flush a rival A hi flush AND a full house just like James Bond years before the movie came out. So it's not impossible. I've played over 2 million hands of poker and the game has never lost intrigue.

Answer 7

Please someone correct me if I am wrong this is my simpel solution to your question probability to hit quads IF YOU play against yourself is 3/51*(2/50+2/49+2/48+2/47)*1/46 or about one hand in 5000 so to two players hit quads is aproximatley something like 5000^2 it is something less because to be able to do this you need two starting hands to be pocket pairs but anyway it is to little to ever fold pocket quads.

Answer 8

Yes, the probability of that is very low: <0.01%.

And, things like that usually happen in online poker. You won't see this stuff much in live poker.

Answer 9

the odds of getting quads are 1 in 421 not 1 in 595 as the internet tells you not true at all worked it out myself and i wprked out all the odds in 5 card stud before the internet knew it. I have also worked out the odds of 2 handed heads up 2 hands of quads the odds are 1 in 2.25 million any more players than 2 of course it becomes more likely to occur. I actually had it happen to me 3 handed I had four tens on the flop and they had four nines with a full house on the flop so anything is possible.

Answer 10

p(dealer getting two pairs in 5 cards)= 1/17 x 1/17 x 44 app. 1/6

P(player getting one of the same pairs (2x 46C2 )/ 52C4 app. 1/78

p (2players getting Quads ) 1/6 x 1/78 x 1/78 =app 1/36504

Answer 11

Just watched the video and player 2 with the quad 9 before the showing cards. If he calculated it in his head the only way he is beaten with the cards out are quad Queens. Cuz no flush, and no straight, available. So what is the likely of player 1 having 2 queens in hand. 2 queens available out of 52 card deck minus what he already sees of his hand plus out on the table. [2/(52-7)] and the chance of having other queen 1 available out of 52 card deck minus 7 seen plus drew the first one. [1/(52-8)] so (2*1)/(45*44)= 2/1980= 0.00101 or 0.101% chance of losing. So with the 99.899% winning he goes all in and loses. That's sad. Feels bad man.