hold em odds questions

[rheaume]

got a couple math question to help me get a better grasp on "on the fly" hold em odds.

* heads up in an SNG. you're dealt 2 rags. what are the odds your opponent has at least one ace? how do you calculate that?

* there's a flop of AAA. now how do you do that math? what if you're not heads up, what if there are 5 people in the hand?

* same question above, but with 2 aces instead of 3 ex. AA3

thanks for your help

[AaronBrown]

Since you have two non-Aces, there are 50 unknown cards and 4 Aces. Those 50 cards can be arranged 50*49/2 = 1,225 ways. You could select a hand with one Ace by picking any of the 4 Aces, combined with any of the 46 other cards (this does not include AA). 4*46 = 184, so the chance of having an Ace is 184/1,225 = 15%. However, with 10 players at a table there is only a 15% chance that none of them has an Ace.

You have to be careful in the second question. Once you see the flop, if you still care about the answer, the other player has bet. That changes your assessment. However, if he bet blind, or for some other reason you have no information about his hand, there is now only 1 Ace out of 47 unseen cards. He could pair it with any of 46 other cards, so the answer is 46/(47*46/2) = 2/47.

[rheaume]

hey thanks a lot for replying, i really appreciate it