Poker Odds Question

Last night in our SDMB Poker game it was down to three handed, myself and two other players. We get dealt pockets and I see pocket Queens. Suh-weet! And even sweeter is that the betting has escalated to the point where one of my opponetns is all in. Flip the cards and…wha? We both have pocket Queens.

What are the odds that in a three handed game two players would have identical pocket pairs?

I don’t know, but odds are really good that when it happens to me they other guy hits a flush on the river. I’d even make a side bet on it. :stuck_out_tongue:

Hmm… if there were only two players, it’s pretty easy to calculate: 1 in 20,825
You’re only dealing four cards then, and they have to all be the same rank, though that can be any rank. So, the first card can be anything. Second card has to be one of three remaining in that rank: 3/51. Then 2/50 and 1/49 for the next two. Cross-multiply and you get 6/124950, which comes out to 1/20825

with three people, six cards, I’m not quite sure how to organize it yet. Hmm. Lemme think on it.

The odds of 1 person getting Overcard Pocket pairs is about 1.8% or 1 in 56, with a 55:1 odds of winning that hand.

The odds od 2 people getting that same hand is .09% or 1 in 200, with 100:1 split the pot or some one catches a flush.


Well, I’ve done some calculations, and they seem to indicate that the odds would be triple what I calculated before for two people: 3/20825 or one in 6,941.67
And, for the hold-em impaired, exactly what is overcard pocket pairs??

“Overcard” usually means higher in rank than any card on the board, but I’m not familiar with the expression “overcard pocket pair.”

The highest pocket pair on the table. If I have QQ and you have 77 and Otto has JJ, we all have pocket pairs, and I have the overcard pocket pair.

Not a common expression.

C(3,1)*C(13,1)*C(4,2)*C(2,1)*C(1,1)C(2,2)/C(52,6) = 2.310[sup]-5[/sup]

This represents choosing one player out of three, choosing one value of card, choosing 2 of the four suits, choosing one of the remaining players, choosing that suit, then choosing the two remaining cards, divided by the total number of cards that could be dealt.

Sure, not only do you give yourself the best hand, you stick me with the fuckin’ Jacks. I hope I at least have position on you.

Oh, hush. My luck, I’ll bet my queens like crazy, and then you’ll pick up the third J on the flop.

Bill The Cat, I think there’s a problem with your answer (I know counting things can be tricky, somebody ought to check my work, too).

For example, there aren’t just C(52,6) ways to deal the cards, since those 6 cards still have to be dealt to the three (indisctinct) people. There are C(6,2)C(4,2) / 3! = 15 ways to do this, so the total number of ways to deal the cards is 15C(52,6) = 305,377,800.

From here, first pick the rank of the shared pair; there are 13 ways to do this. Next, pick the other 2 cards that will be dealt–C(48,2). Finally, count the number of ways the shared pairs can be split: C(4,2).

The product of all that is 87,984. Taken out of 305,377,800 different deals, this gives a probability of .028811524609843937575030012004802 %.

I think I just caught a mistake in my own work:

That just isn’t true. Once we’ve chosen our 6 cards to be dealt, four of the same rank, the other two can be whatever else (and assuming the three people are indistinct), there are only 3 ways to split the pairs: Somebody has to have a pair with either (Spades Diamonds) (Spades Clubs) or (Spades Hearts); once that’s been picked, the rest of the deal is completely determined.

So my answer should have been half my original result: .014405762304921968787515006002401 %

Somebody still probably ought to check that.

Done some more scribbling, looking at all the cases (well, all the relevant cases,) dealing one card to each player and then their second card, and I’m sticking by my answer of 1 in 6,941.67, which works out to agree nicely with cabbage’s second figure when expressed as a probability, even though we’ve been attacking the problem in different ways, so it’s looking pretty good.

Chris' worksheet

checking all cards in order as they are dealt: A, B, C, A, B, C

1st card to a: Can be anything, 52/52. For simplicty, call it QH

1st card to b: might be a Q, same rank as 1a, QS. 3/51 Case 1 or might be something else, call it 4C, 48/51 Case 2


1st card to c: might be one of the remaining Q's, 2/50. In which case we're dead.
Or might be any other card, 48/50, call it 9D

2nd card to a: if it is a 9 or any other card but a Q, we are dead
2/49 to continue

2nd card to b: if it is a 9 or any other card but a Q, we are dead
1/48 to deal the right card to B, and then we are gold. 2c does not matter

1st card to c: might be a remaining Q, 3/50 CASE 3 might be a remaining 4, 3/50 CASE 4 42/50 we're dead.


2nd card to a: needs to be one of the Q's or else we're sunk: 2/49

2nd card to b: can be anything but the last Q: 47/48

2nd card to c: must be the last Q: 1/47


2nd card to a: anything but a remaining 4: 47/49

2nd card to b: one of the remaining 4's: 2/48

2nd card to c: the last 4: 1/47

3/51 * 48/50 * 2/49 * 1/48

48/51 * 3/50 * 2/49 * 47/48 * 1/47

48/51 * 3/50 * 47/49 * 2/48 * 1/47
3/51 * 1/50 * 2/49

1/51 * 3/50 * 2/49 

1/51 * 3/50 * 2/49