#1
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AA vs. AA 2 out of 3 hands?
I was playing at a B&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? |
#2
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Re: AA vs. AA 2 out of 3 hands?
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. |
#3
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Re: AA vs. AA 2 out of 3 hands?
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 |
#4
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Re: AA vs. AA 2 out of 3 hands?
[ QUOTE ]
For there to be AA vs AA, the odds are 3/51 * 2/50 * 1/49 = .00004801% [/ QUOTE ] Where did this come from? |
#5
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Re: AA vs. AA 2 out of 3 hands?
[ 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. |
#6
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Re: AA vs. AA 2 out of 3 hands?
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.
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#7
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Re: AA vs. AA 2 out of 3 hands?
[ 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. |
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