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therockofgibraltar
12-31-2004, 04:25 AM
I raise preflop. Everybody folds to a player who reraises me. I cap with my AK.

Me and one opponent see the flop.
A-6-4 rainbow
I bet-he raises-I reraise-he caps.

turn is 8.
i bet - he raises - i went into calling mode and finally won the pot, he had A9s.

BUT the question ain't related to my "bad" play. It is a probability question. If I have AK and the flop contains one A, what is the chances that my opponent has AA??

Stallion714
12-31-2004, 05:22 AM
1 in 220 hands. Regardless of what you hold.

niin
12-31-2004, 05:42 AM
I know that's the chance he has AA before the flop, but can't you go back and recalc, given the fact that you know 2 aces are gone? He'd have to have 2 specific cards (and not just 2 out of 4 specific cards). Since we have more information, you should be able to get a better chance.

Just like people say "well, there's 2 jacks on the board, so the chances of him having a jack is reduced". Since 2 aces are accounted for, the chances of them having an ace is reduced?

So, given that 5 cards are known, would the chance be 2/(47*46)?

12-31-2004, 06:21 AM
Looks correct to me.

The 1/220 comes from having 4 aces available from 52 cards:

4/52 * 3/51 = 1/221

But now there are 2 aces available from 47 cards:

2/47 * 1/46 = 1/1081

I believe this is correct.

BruceZ
12-31-2004, 07:38 AM
Correct, assuming he holds any 2 random cards with equal probability.

gaming_mouse
12-31-2004, 07:42 AM
[ QUOTE ]
1 in 220 hands. Regardless of what you hold.

[/ QUOTE ]

This is not true. You cannot ignore known information. The easiest way to see this is to imagine that you hold AA yourself. Do you still think your opponent's odds have not changed?

However, to calculate the probability we need to assume that your opponent's hand is a random hand from the unseen cards in the deck -- and this assumption is false given his PF three bet, which limits his range of hands unless he's a total maniac. So the calculation is more of an exercise. Nonetheless....

There are 47 unseen cards in the deck. There are two aces left. So the chance that your oppo holds them is:

1/(47 choose 2) = 1 in 1081.

By the way, I don't think you played this poorly given an unknown opponent. An aggressive oppo might have played pocket eights this way, and a true LAG might even play pocket fours or sixes this way. Make a note, though, and next time keep playing back.

HTH,
gm