At the Planet Hollywood game last night there seemed to be an unusual number of double-paired board (e.g., something like K-K-2-7-7). It got me to wondering how often one should expect that to happen, on average.
There are 2,598,960 possible five-card boards possible in hold'em, because C(52,5) = 2,598,960.
The tricky bit, then, is counting up how many possible double-paired boards there are--a task I undertake here with more trepidation than usual, having just minutes ago put up a post illustrated by a book that appears to be titled Counting for Dummies. I feel motivated to get it right to avoid getting the taunts turned on me! So I'm figuring this out as I go. Follow along, if you care to.
Consider the highest possible double-paired board: A-A-K-K-x. How many ways can such a board occur? (We don't care about the order the cards come in for this question.) There are six different combinations of A-A that can be part of it (Ac-Ad, Ac-Ah, Ac-As, Ad-Ah, Ad-As, and Ah-As). Similarly, there are six different combinations of kings that can constitute the other pair. That's a total of 36 different combinations of those four cards. Then we have left in the deck 44 other cards, any of which can serve as our fifth card. 36 x 44 = 1584. So there are 1584 A-A-K-K-x boards.
Given 13 ranks of cards, there are 78 pairings of different ranks that can be made (A-K, A-Q, etc., down to 3-2). We accounted for only one of them in the example just completed. So the final total should be 1584 x 78 = 123,552, for the number of different possible double-paired boards.
The frequency of this occurrence, then, is 123,552 / 2,598,960, which equals 0.0475, or about 4.8%. Double-paired boards should happen just a smidgen under 5% of the time. That roughly corresponds to my general experience, as I think back on things.
This surely can't be the first time anybody has ever run this calculation, but a quick search of my poker math books and these interweb thingies doesn't locate any previous pioneers to give me confirmation. So I think it's right, but if I screwed something up, let me know, and I'll fix it in an addendum, and call myself a dummy in the process.
The image above is "Two Pears," by Georgia O'Keeffe, 1921.
Wednesday, December 03, 2008
Double-paired boards
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4 comments:
It appears you are including the boards that would be a complete full house (e.g. AAAKK). While these are technically also double paired boards, it seems in your calculations you normally exclude the occurrences of better made hands, or at least discuss the rationale for inclusion or exclusion of them.
No, I excluded the full-house boards by counting only 44 possible cards for the fifth slot. That is, after accounting for two pairs on the board, I excluded all 8 cards from those two ranks as the fifth card. E.g., the "x" in AAKKx cannot be either A or K; there are 44 cards left that are not A or K.
The good news is that you are correct. The bad news is that this is not an original discovery. On balance though, I think you escaped the dummy label.
To find existing examples of this, you can reframe the question. Instead of viewing it as the board in Hold 'Em, view it as the initial deal for a hand in 5 card draw. Now if you Google 5 card odds, you will get many examples showing that there are 123,552 ways to be dealt a hand ranked as 2 pair. Links:
http://www.poker1.com/mcu/tables/Table12.asp
http://www.pokerhands.com/poker_odds.html
Thanks, Zot. It didn't occur to me that it was equivalent to a 5-card draw scenario, but that seems obvious once it's pointed out. I didn't really think I could be the first one to have done the calculation, because it seems an obvious thing to wonder about. Nice to know that once in a while I can get the math right, even in the middle of the night.
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