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AK diamonds vs AK clubs vs AA AK diamonds vs AK clubs vs AA

09-16-2014 , 01:25 AM
Tonight in a tournament i was holding AK diamonds, i raised double big blind, guy calls, lady doubles my bet everyone else folds, i shove AI, guy calls, lady calls, we turn them over lady has AK clubs guy has pocket rockets, if i remember correctly there were 7 of us left at the table, what are the odds of this scenario happening? (Don't think it matters but none of us were in the blinds)
AK diamonds vs AK clubs vs AA Quote
09-16-2014 , 02:33 AM
To anticipate one of the standard responses, are you asking what are the odds that two opponents on a 7-handed table have AKs and AA when you have AKs or are you asking what are the odds that AKs, AKs, and AA all are dealt on a 7-handed table?
AK diamonds vs AK clubs vs AA Quote
09-16-2014 , 12:25 PM
Quote:
Originally Posted by whosnext
To anticipate one of the standard responses, are you asking what are the odds that two opponents on a 7-handed table have AKs and AA when you have AKs or are you asking what are the odds that AKs, AKs, and AA all are dealt on a 7-handed table?
Indeed that would have been my response

I'll just answer both.

#1: C(6,2) * 3 / C(50,4) / 3!! = 1 in 15353.333...

#2: C(7,3) * C(4,4) * C(4,2) / C(52,6) / 5!! = 1 in 1,454,180 exact

There's only one valid way to group the cards since the AK hands have to be suited.
AK diamonds vs AK clubs vs AA Quote
09-26-2014 , 03:01 PM
Here's another way to get to the answer.

For this to happen, all four Aces and two Kings must dealt to three hands. There are C(52,6) = 20,358,520 ways to select 6 cards out of the deck, 6 of them consist of four Aces and two Kings.

Given that this happens, 1 time in 5 both Kings will be in the same hand (the guy who gets the first King gets 1 of the 5 other cards, 1 of which is a King). Given that two different players get the Kings, the guy with the first King has 1 chance in 4 of getting the suited Ace, if he does, the gal with the second King has 1 chance in 3. So the chance of this happening given that the required 6 cards are dealt to three players is 4/5 x 1/4 x 1/3 = 1/15. Another way to think about this is to say the guy who gets the first King needs to get 1 card out of 5 (the suited Ace) and the gal who gets the second King needs to get 1 card out of the 3 remaining (the suited Ace) and 1/5 x 1/3 = 1/15.

So with three players, the chances are 1/20,358,520 x 6 x 1/15 = 1/50,896,300.

At a 7-handed table, there are C(7,3) = 35 ways the three players can be chosen. 35/50,896,300 = 1/1,454,180, as heehaww computed.
AK diamonds vs AK clubs vs AA Quote

      
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