% Villain's AA probability %

killing_random

killing_random

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My hand is Ax, one ace lies on the desk.
What ??.??% probability of villain having AA:questionm

:hmpf: Idk how to calculate it
 
Edu1

Edu1

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per example - holding AK - a A in the flop, the probability of the villain have AA is 0.02%
I don't know the formula, I saw in another poker site
 
killing_random

killing_random

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per example - holding AK - a A in the flop, the probability of the villain have AA is 0.02% I don't know the formula, I saw in another poker site
I think it's definitely can't be more than that, cos probability of having even one of two aces that left, on a river:

52-2-5 = 45 2/45 = 4.444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444...%
then 4.4444444444%*4.4444444444% = 0,1975308641935802%

Sorry, look like I'm just been lazy :icon_geek that was actually not that hard to do...

Edit:
Oh wait. I've got 0.2%, not 0.02%
i'm confused
 
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SharkFinn

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The probability of villain's first card being an A is as you've said above 2/45 =4.4%. But now you've removed another A so only one left, so for villain's second card it's 1/44 = 2.2%. Then multiply those to get 0.1% probability that they have AA
 
killing_random

killing_random

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Got it)

2/45*1/44= 0,001010101010101 = 0.1%
 
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fundiver199

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If its on the flop, there are 47 unseen cards, which gives 47 x 46 / 2 = 1.081 combos, of which only 1 is AA. So not exactly 0,1% but pretty close. Also this is ignoring the fact, that action took place preflop, and that most opponents dont see the flop with all those 1.081 possible combos, because they would have folded combos like 74, 83, T2 etc. In the hand, you shared, SB raised OOP against a limp, and he probably does that with something like 200 combos at most rather than 1.081.
 
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