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View Full Version : AA vs. AA 2 out of 3 hands?

Mr. Curious
09-15-2005, 11:58 AM
I was playing at a B&amp;M last night when I saw this happen. The first hand I had pocket aces and the guy to my right did as well. Two hands later two other people had them. So I thought I would take a stab at trying to figure out the probability of it happening.

For there to be AA vs AA, the odds are 3/51 * 2/50 * 1/49 = .00004801%

For this to have happened again, it is just .0004801^2, right?
I don't need to figure out the odds of it not happening in the second hand, do I?

LetYouDown
09-15-2005, 12:56 PM
Odds of A-A vs. A-A on a given hand, full ring...I believe:

C(10,2) * 1/C(52,4) = 45/270725 = .0166% or about 6015 to 1.

For this to happen exactly 2 times in 3 hands:

=BINOMDIST(2,3,45/270725,FALSE) = .00000828738% = ~12,066,538.7-to-1

Someone correct me if I'm wrong.

eOXevious
09-15-2005, 01:03 PM
I think the answer to your problem must have a more specific question... Every hand dealt, AA vs AA has the same chance of comming up as before. So its a .00004801% every hand. Just like flipping coins, Heads could come up 5 times in a row, but its still a 50/50 chance on the 6th try.
I think you are missing information to ask the correct question... I think we need to know how many people are at the table, and how many hands... you said Two hands later, so I'm thinking you meant you played 3 hands total, and twice in that time, AA was vs AA....that can be calculated.
But for AA vs AA to have just happened again its still a .00004801%
Someone tell me if I'm wrong

LetYouDown
09-15-2005, 01:55 PM
[ QUOTE ]
For there to be AA vs AA, the odds are 3/51 * 2/50 * 1/49 = .00004801%

[/ QUOTE ]
Where did this come from?

Mr. Curious
09-15-2005, 04:17 PM
[ QUOTE ]
[ QUOTE ]
For there to be AA vs AA, the odds are 3/51 * 2/50 * 1/49 = .00004801%

[/ QUOTE ]
Where did this come from?

[/ QUOTE ]

I thought that was the odds that AA would be dealt to two different people preflop, but now I see that it is wrong. That is the likelyhood that any one card will be the first four cards dealt.

LetYouDown
09-15-2005, 04:30 PM
Ok. So if you change it to C(52,4) or 4/52 * 3/51 * 2/50 * 1/49 for the specific case of AA vs. AA, I believe the math presented should hold up. Really not likely. Granted, this is only the odds that you sit down at a table and this happens on the first three hands. If you played 20,000,000 hands, this situation would be much more likely to occur at some point.

chiachu
09-15-2005, 05:54 PM
[ QUOTE ]
Odds of A-A vs. A-A on a given hand, full ring...I believe:

C(10,2) * 1/C(52,4) = 45/270725 = .0166% or about 6015 to 1.

For this to happen exactly 2 times in 3 hands:

=BINOMDIST(2,3,45/270725,FALSE) = .00000828738% = ~12,066,538.7-to-1

Someone correct me if I'm wrong.

[/ QUOTE ]

the math looks right to me.