Daily Poker Quiz: Archive

AA odds?
2008-08-11

You have KK at a full tenhanded table. What's the risk of at least one opponent receiving pocket aces?





0.2 %
Two players could have aces, at most. There are three ways of giving AA to two people, and six ways of dealing AA to one person.

In the example with two players, there are C(46,14)*13!!*3 (13!! = 13 * 11 * 9 * 7 * 5 * 3) ways of dealing all other hands. This is a very large number. There are C(48,16)*15!!*6 ways of dealing the other hands in the second case, minus the ways in the first one.

Those are big numbers and it gets complicated. But C(46,16) means that there are 16 cards left to deal out of 46 in the deck. C(46,16) then means the number of combinations you could possibly deal. 15!! is the number of ways you can deal 16 cards to eight hands each having two cards. The product is the number of way you could possibly deal two cards to that number of players. So...

If you subtract the first from the other, you get 27 326 563 159 385 514 025 ways that at least one player has pocket aces. Divide this by the number of ways you can deal two cards to nine players (C[50,18]*17!!) and you get about 4.4 %, or roughly 22:1. Now wasn't that easy? :)

1.8 %
Two players could have aces, at most. There are three ways of giving AA to two people, and six ways of dealing AA to one person.

In the example with two players, there are C(46,14)*13!!*3 (13!! = 13 * 11 * 9 * 7 * 5 * 3) ways of dealing all other hands. This is a very large number. There are C(48,16)*15!!*6 ways of dealing the other hands in the second case, minus the ways in the first one.

Those are big numbers and it gets complicated. But C(46,16) means that there are 16 cards left to deal out of 46 in the deck. C(46,16) then means the number of combinations you could possibly deal. 15!! is the number of ways you can deal 16 cards to eight hands each having two cards. The product is the number of way you could possibly deal two cards to that number of players. So...

If you subtract the first from the other, you get 27 326 563 159 385 514 025 ways that at least one player has pocket aces. Divide this by the number of ways you can deal two cards to nine players (C[50,18]*17!!) and you get about 4.4 %, or roughly 22:1. Now wasn't that easy? :)

4.4 % (correct)
Two players could have aces, at most. There are three ways of giving AA to two people, and six ways of dealing AA to one person.

In the example with two players, there are C(46,14)*13!!*3 (13!! = 13 * 11 * 9 * 7 * 5 * 3) ways of dealing all other hands. This is a very large number. There are C(48,16)*15!!*6 ways of dealing the other hands in the second case, minus the ways in the first one.

Those are big numbers and it gets complicated. But C(46,16) means that there are 16 cards left to deal out of 46 in the deck. C(46,16) then means the number of combinations you could possibly deal. 15!! is the number of ways you can deal 16 cards to eight hands each having two cards. The product is the number of way you could possibly deal two cards to that number of players. So...

If you subtract the first from the other, you get 27 326 563 159 385 514 025 ways that at least one player has pocket aces. Divide this by the number of ways you can deal two cards to nine players (C[50,18]*17!!) and you get about 4.4 %, or roughly 22:1. Now wasn't that easy? :)

7.6 %
Two players could have aces, at most. There are three ways of giving AA to two people, and six ways of dealing AA to one person.

In the example with two players, there are C(46,14)*13!!*3 (13!! = 13 * 11 * 9 * 7 * 5 * 3) ways of dealing all other hands. This is a very large number. There are C(48,16)*15!!*6 ways of dealing the other hands in the second case, minus the ways in the first one.

Those are big numbers and it gets complicated. But C(46,16) means that there are 16 cards left to deal out of 46 in the deck. C(46,16) then means the number of combinations you could possibly deal. 15!! is the number of ways you can deal 16 cards to eight hands each having two cards. The product is the number of way you could possibly deal two cards to that number of players. So...

If you subtract the first from the other, you get 27 326 563 159 385 514 025 ways that at least one player has pocket aces. Divide this by the number of ways you can deal two cards to nine players (C[50,18]*17!!) and you get about 4.4 %, or roughly 22:1. Now wasn't that easy? :)



Result:

0.2 % 24%
1.8 % 46%
4.4 % 25%
7.6 % 6%

Totally 980 vote(s)