AA vs AA heads up

osirisdean

osirisdean

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last night i was playing a live tournament at a local watering hole and it got down to me and another guy heads up. at some point we were both dealt pocket aces. on the same hand. hilarity ensued. :)

i was just wondering, what are the odds of getting AA vs AA (or any other pair vs same pair) preflop, heads up?

(and boy it wouldve sucked for him had i hit my flush. i had 4 to it but the last one didnt fall. still won it though. ;) )

d
 
diabloblanco

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How exactly did the hand history look from the best of your recollection? I'm just curious.

How exactly did the hand history look from the best of your recollection? I'm just curious. Also, I noticed you're from Atlanta, me and Irish Dave are playing a tournament tomorrow in Marietta. Its at the Elk's lodge and the buyin is 150. 21K paid to the final table with first prize being a seat in the 2006 wsop Main Event and accomodations or the cash equivelant. Maybe you ought to come check it out if you have the cash to play.
 
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Dorkus Malorkus

Dorkus Malorkus

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4/52*3/51 = 0.077*0.0588 = 0.00453 = 0.45% = P(Plyr1 has AA)

2/50*1/49 = 0.04*0.0204 = 0.000816 = 0.082% = P(Plyr2 has AA | Plyr1 has AA)

0.00453*0.000816 = 0.0000037 = 0.00037% = P(Plyr1 & Plyr2 have AA)

0.0000037*12 = 0.000044 = 0.0044% = P(Plyr1 and Plyr2 have same pocket pair)

Been a while since I did stats so I might be off, dunno.
 
t1riel

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Dorkus Malorkus said:
4/52*3/51 = 0.077*0.0588 = 0.00453 = 0.45% = P(Plyr1 has AA)

2/50*1/49 = 0.04*0.0204 = 0.000816 = 0.082% = P(Plyr2 has AA | Plyr1 has AA)

0.00453*0.000816 = 0.0000037 = 0.00037% = P(Plyr1 & Plyr2 have AA)

0.0000037*12 = 0.000044 = 0.0044% = P(Plyr1 and Plyr2 have same pocket pair)

Been a while since I did stats so I might be off, dunno.
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How in the hell did you come up with this? Wait, don't tell me. It might confuse things even more.
 
Dorkus Malorkus

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4 Aces in deck, 52 cards, so probability that Plyr A's first card is an A = 4/52
3 Aces left, 51 cards, so probability that Plyr A's second cards is an A given that his first card is = 3/51
Multiply together to get the probability that both Plyr A's cards are Aces.

2 Aces left, 50 cards, so probability that Plyr B's first card is an A given that Plyr A has AA = 2/50
1 Ace left, so probability that Plyr B's second card is an A given that his first is an A and Player A has AA = 1/49
Multiply together to get the probability that both Plyr B's cards are Aces given that both Plyr A's cards are Aces.

Then the probability both Plyr A's cards are Aces multiplied by the probability that both Player B's cards are Aces given that both Player A's cards are Aces gives the probability that both players are dealt AA in any given hand.

Multiply by 13 (not 12 like I originally said, too late to edit post now) for any pocket pair because there are 13 possible pocket pairs.
 
osirisdean

osirisdean

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Dorkus Malorkus said:
4 Aces in deck, 52 cards, so probability that Plyr A's first card is an A = 4/52
3 Aces left, 51 cards, so probability that Plyr A's second cards is an A given that his first card is = 3/51
Multiply together to get the probability that both Plyr A's cards are Aces.

2 Aces left, 50 cards, so probability that Plyr B's first card is an A given that Plyr A has AA = 2/50
1 Ace left, so probability that Plyr B's second card is an A given that his first is an A and Player A has AA = 1/49
Multiply together to get the probability that both Plyr B's cards are Aces given that both Plyr A's cards are Aces.

Then the probability both Plyr A's cards are Aces multiplied by the probability that both Player B's cards are Aces given that both Player A's cards are Aces gives the probability that both players are dealt AA in any given hand.

Multiply by 13 (not 12 like I originally said, too late to edit post now) for any pocket pair because there are 13 possible pocket pairs.
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yep, dorkus is dead on. i suppose i couldve done the math, but enh, im lazy. :) and it was more a "holy crap, AA vs AA, how often do you see THAT?" kinda rhetorical question. but i was curious about the statistics, so grazie! :)
 
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