What are the odds of pocket aces beating pocket aces?

[sfetaz]

This is a two part question. The first part is what are the odds if you have pocket aces, that one opponent will also have pocket aces. I am pretty sure I know the answer but want to confirm. Second, what are the odds of that occuring, and in that same hand a 4 flush appearing?

[jay_shark]

It's really simple actually . If you hold pocket aces , then the probability one of n opponents holds aces is n/50c2 . At most , one other player may hold aces other than yourself and so the events are mutually exclusive .

However , if you didn't hold aces , then the probability two players get dealt aces is 6/52c2*1/50c2*nc2 ~0.00016622

Given that you do hold aces , the probability you will lose to another player with aces is :

2*[n/50c2*12c4*36/48c5]+2*[n/50c2*12c5/48c5] ~ 2n*0.000008873114

So if n=9 , then the answer is ~ 0.000159716

[jay_shark]

Also , we must multiply the above by the probability that you do get dealt aces which is 6/52c2 .

Our answer becomes 0.00000072269

[sfetaz]

Ok I totally lost what you are saying here. I am used to seeing things in terms of fractions but my fractions are not working out because they are either too often or not often enough, and thats just for two people being dealt pocket aces. Can you possibly explain your formula? I have no idea what C2 means especially and where you get your numbers from.

[TrvChBoy]

If all you want to know is the answer, then CardPlayer.com has an easy-to-use odds calculator

http://www.cardplayer.com/poker_odds/texas_holdem

If you are wanting to know the math behind the answer, then I applaud you and suggest Sklansky's book "Getting the Best of It".

[radradrad]

Jay, I don't think he is asking what you are answering. He is not asking to have the odds of one or two people getting pocket Aces. I THINK the question is what are the odds of either person winning (instead of tieing or being beat) WHEN two people have pocket Aces.
IF that is the question, the answer is 2.17% for each person to win flat out. That would include flushs or straight flush possibilities.

[DavidG1966]

You are all missing the 2 questions...

1) 1 in x hands 2 players will be dealt pocket aces during the same hand at a 10 player table... 1 in 10,000... 1 in 100,000... etc...

and

2) Given the same sinerio 1 in x hand will that play out with someone losing poket aces to pocket aces do to a flush.. 1 in 1,000,000 hands??????

Thats what he is asking... Does anyone know... I am curious also!

[DarkMagus]

My numbers agree with jay. So at a 10 handed table, approximately once in 6000 hands will two players get AA, and about 4% of the time either one will win by a flush. So in total the whole series of events occurs once in every 139,000 hands.

For a 6-handed table, two players get aces once in 18,000 hands, and one of them wins once in 416,000 hands.

[jay_shark]

n/50c2=n/1225 is the probability that if you hold aces , then one other player will hold aces, against n other opponents .

50c2 is the number of ways of selecting two objects from a collection of 50 objects .You may also find this button on most calculators so just look for n(c,r) or nCr .


For your second question , did you mean four flush on the flop ?

If that's the case, given that two players hold aces , it should be
2*12c3/48c3=2.54%

To radrad:

I already showed how one would get 2.17% .

2*[12c4*36 + 12c5]/48c5=2.173913%

[mykey1961]

Actually he asked 3 questions.

1) The title question.

It's not specific about who's AA wins or loses, so the answer is more like 4.34619%

2) How often will one opponent have AA if you have AA.
either 1/1225 or N/1225 depending on how you interpret "one opponent"

3) A combination of #2, and how often will the board show 4 of one suit given that both of you have AA.

if you assume "4 flush" means 4 on the board, and the A making a 5th.

(4*12*11*10*9*36*5)/(48*47*46*45*44*1225) = 0.003398%

or

(n*4*12*11*10*9*36*5)/(48*47*46*45*44*1225) = n*0.003398%

again depending on how you interpret "one opponent".

if you assume 4 flush to mean came up short of making a flush.

(12*11*10*36*35*20*4)/(48*47*46*45*44*1225) = 0.05286110%

or

(n*12*11*10*36*35*20*4)/(48*47*46*45*44*1225) = n*0.05286110%

again depending on how you interpret "one opponent".

#3 doesn't ask how often your AA will win or lose.