View Full Version : Holdem: You have wired Kings, Opp. has wired Aces, What are the odds?

01-10-2005, 11:52 AM
I'm just talking about the odds of this occuring, not the the outcome of the "race". Having said that.......
My question to the geniuses of this forum: Is this a problem solved by the principle of parlays? Such as 220 to 1 multiplied by 220 to 1? Or is there some other force at work here?

01-10-2005, 12:50 PM
For one, are you talking about the odds you have Kings AND your opponent has Aces, or the odds that your opponent has Aces GIVEN you have Kings. Much different situations.

Also, how many opponents are we talking here? Is it 10-handed and you're looking for the odds any of them has Aces? Or are you heads-up?

01-10-2005, 01:25 PM
kyro, maybe it would help if I described the situation as it happened, as it was not a hypothetical question.
10 handed SnG Holdem game, everyone folds to the small blind, he puts in a raise. I'm in the big blind (with KK) and this time I decided to try to let him bet off his chips. He bet the flop and went all in on the turn. I call and to my surprise he turns over the pocket aces! Interesting hand indeed!
What are the odds of us having KK against AA preflop as in the situation I just described, and how do you figure it out?

01-10-2005, 01:41 PM
There was just a recent thread on this, but a formula I devised and use at the table is this:

7n - xn/2

Where N is the number of players left to act/participate in a hand. X is the relative strength of your Pair (A=14, K=13, etc...)

This gives you a % chance that at least one other player holds an overpair to your pair.

Yes this is just an apporximation with some fudging in the algebra to simplify the equation, but another poster showed that the accuracy is good enough to be used at the tables.

So in your case, there is 1 other player in the hand, you have KK so,

7(1) - 13(1)/2 = 7-6.5 = 0.5% chance of there being AA in the SB. Not exact, but ballpark. In other words, not much of chance.

01-10-2005, 01:48 PM
The odds that someone else at a 10 person table has AA given that you have KK = 9*C(4,2)/C(50,2) - C(9,2)/C(50,4) =~ .0439

The odds that you are dealt KK and someone else has AA = 6/C(52,2) * (9*C(4,2)/C(50,2) - C(9,2)/C(50,4)) =~ .0002


01-10-2005, 02:06 PM

I know you have answered my question exactly, and I understand there is a substantial difference between the two answers.
But I do not understand the difference in the two questions.
"AA given that you have KK" and,
"dealt KK and someone else has AA"

01-10-2005, 02:40 PM
You are only dealt KK 1/220 hands. Therefore
1) is much more likely than 2).

01-10-2005, 02:43 PM
The first answer "AA given that you have Kings" is the probability that someone at your table has a pair of Aces irrespective of your hand. You could have anything as long as it doesn't have an ace in it. Your hand does not matter.

The Second answer "You are dealt Kings and someone has Aces" is the probability exactly that you are dealt Kings, and one of your opponents are dealt Aces.


01-12-2005, 06:44 PM
Since it is given that KK has been dealt to one person, the next step is to calculate the odds that at least one of the other players is dealt AA from 50 unseen cards...