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View Full Version : AA chop, becoming a habit

dcarlc
10-05-2005, 11:04 PM
My last two sessions playing I have been dealt AA once in each session. Both times I ended up chopping the pot(and not much of one)with another pair of Aces. Just wondering what the odds are of another player being dealt the other 2 aces if you have 2, and what are the odds of it happening twice in a row?

Thank You.
BruceZ
10-06-2005, 05:54 AM
[ QUOTE ]
Just wondering what the odds are of another player being dealt the other 2 aces if you have 2

[/ QUOTE ]

(1225/N - 1) to 1, where N is the number of opponents.


[ QUOTE ]
and what are the odds of it happening twice in a row?

[/ QUOTE ]

(1225/N)^2 - 1 to 1.
dcarlc
10-06-2005, 10:05 AM
[ QUOTE ]
[ QUOTE ]
Just wondering what the odds are of another player being dealt the other 2 aces if you have 2

[/ QUOTE ]

(1225/N - 1) to 1, where N is the number of opponents.


[ QUOTE ]
and what are the odds of it happening twice in a row?

[/ QUOTE ]

(1225/N)^2 - 1 to 1.

[/ QUOTE ]

I'm not very bright. What does N stand for here? There are 9 players in game if that matters.
RiverTheNuts
10-06-2005, 10:22 AM
[ QUOTE ]
(1225/N - 1) to 1, where N is the number of opponents.


[/ QUOTE ]

Hope that helped
Abbaddabba
10-06-2005, 01:35 PM
A closer look...

Given that you have AA, there are 2 aces in the deck. Order does not matter.

2/50*1/49 of the other player holding AA, given that you have AA as well. Or 1/1225 for an individual opponent. That's what the other guy already wrote, but in terms of a ratio instead of fractions.

If you want to add another condition... that the pot is not split, you have to consider the probability of there being 4 to a flush on the board, since that's the only way that it wont be a chop. The probability of this happening is a bit more difficult to calculate. I'm not sure the exact value of how likely it is for 4 to a flush being dealt on a board.

Once you have that, the probability of it happening two times in a row is simply a matter of squaring it.
LetYouDown
10-06-2005, 02:03 PM
Out of 1712304 boards, 74420 will result in a pot that goes in one direction only.

So 2.17% of the time you will lose, 2.17% of the time you will win and the rest you will split.

Note: you have to consider board flushes as well, with the exception of straight flushes.

4 Flush:
[4 * C(12,4) * C(36,1)] = 71280

5 Flush:
4 * C(12,5) = 3168

28 Straight Flushes that split the pot.

[4 * C(12,4) * C(36,1)] + [4 * C(12,5)] - 28 = 74420