Chance of nuts beating second nuts with same ranks, suits?
 

I played an amazing limit hold'em heads-up hand, http://www.pokerhand.org/?1684069 , (already posted to BBV http://forumserver.twoplustwo.com/sh...umber=12897887 ) in which my opponent and I had hole cards of the same ranks and the same suits, and in which the the nuts beat the second nuts.

What is the chance of this happening? That is, in a heads-up game, each pocket has the same suits and same ranks, and one player wins?

What about the chance of this happening where neither pocket is the big slick?

Also, I think this configuration is interesting enough to name it.

I propose the term "chiasmus" for a hand in which each player has the same ranks and suits, and where one player wins at showdown.

Thus, I am asking for the odds of a chiasmus in which nut beats second nut.

Re: Chance of nuts beating second nuts with same ranks, suits?
 

The chance is approximately:
680304/2781381002400

Re: Chance of nuts beating second nuts with same ranks, suits?
 

Odds of being dealt a non-pair: 16/17
Odds of them being unsuited: 3/4
Odds of opponent drawing the one exact hand to fit the requirements: 2/50 * 1/49 = 1/1225
Odds of a result other than a tie: 25,334 in 1,712,304

16/17 * 3/4 * 1/1225 * 25,334/1,712,304 = 1 in 117,295.

Odds of me calculating the odds of it being nuts vs second nuts: 0

Re: Chance of nuts beating second nuts with same ranks, suits?
 

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Odds of me calculating the odds of it at being nuts vs second nuts: 0

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It's actually pretty easy. For suit to matter it's got to involve a flush. For the double-suiting that means 4 or 5 flush. We can skip straight-flushed boards b/c there is no second nut, that leaves 11 choose 5 - 1 flush boards, and 13 choose 4 times. For any partitcular set up like that, there is exactly one first and one second nut card, so there are 6 ways (choose the second suit, and who's got the first nut) to have it happen.

Re: Chance of nuts beating second nuts with same ranks, suits?
 

Not sure of the logic behind skipping the straight flush boards. One opponent can hold the next higher card for the nuts, and any two other cards play the board for the trivial second-nuts, no?

You DO have to skip all the five-flush boards where the board's flush counterfeits both players' hands (AKJT9s on board, 8s7h and 8h7s in hand) - and there are quite a lot of those.

The basic idea -- count all 4-flushed boards, and the 5-flushed boards that don't counterfeit both player's hands -- is correct.