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Old 10-08-2002, 09:15 AM
SittingBull SittingBull is offline
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Default Hello,Analysts! Probability question in hold\'em...

Let's assume the turn is 10 6 2 A
U have AJ
What's the probability that at least one of your other two opponents have U beat on the turn??

Just wondering,

Sitting Bull
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  #2  
Old 10-08-2002, 12:20 PM
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Default Re: Hello,Analysts! Probability question in hold\'em...

Heya SB,

I'll take a stab at this.

We have 46 unknown cards (A - 2; J,T,6,2 - 3 each; K,Q,9,8,7,5,4,3 - 4 each).

Assuming that the board is rainbow, the following 71 hands are beating you:

AA (1), AK (8), AQ (8), AT (6), A6 (6), A2 (6), TT (3), T6 (9), T2 (9), 66 (3), 62 (9), 22 (3)

(the numbers in parathesis are how many of those hands are possible)

If your opponents are holding random cards, then the pool of possible hands that they can have contains C(46,2) = 46*45/2 = 1035 hands. The chance that you are ahead of a single opponent is 1 - (71/1035) = 93%. Roughly speaking, the chance that you are beating two opponents is 93%*93% = 87%, or there's a 13% chance that at least one of them has you beat.

But how many of our opponents play random cards? I'd say less than 60%. So, let's check out some more likely scenarios...

Lets say our opponent will continue to the turn with any pair or better, two overcards, or a three straight involving the top card on the flop. This means the pool of possible hands now consists of 511 hands:

J9 (12), QJ (12), KJ (12), KQ (16), AJ (6), 89 (16), Tx (105), 6x (105), 2x (105), any pocket pair that doesn't make a set (51), plus all of the 71 hands that beat you
(x = any card but an A, T, 6, or 2)

The percentages are now that you beat a single opponent 86% of the time, and two opponents 74%. So, at least one of them beats you roughly 26% of the time.

This is still a pretty loose pool of starting hands though (29o through the flop, etc.). Let's assume that the opponets won't play any of the 2x hands, half of the Tx, 6x hands, a pair under the second pair on the board, or a three-straight. This drops the pool of possible hands to 254 hands. Now your chance against a single opponent is 72% and two opponents is 52%. So, you're beat 48% of the time.

Finally, if they will only play top pair (52), an overpair (15), 2 overcards with an ace (6), or any of the hands that have you beat (71), then there are only 144 possible hands for your opponents. You are 51/49 against one and 26% against two. So, there's a 74% chance that at least one of the has you beat.

And so, my considered answer is that you are beat anywhere from 13% to 74% of the time. Or, as someone may have said before, "it depends".

Cheers,
PP
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  #3  
Old 10-08-2002, 12:55 PM
SittingBull SittingBull is offline
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Default Thanks,PP,for your...

thorough analysis. Since this hand occurred in an upper middle limit game,I assume the players were rated at least "good" in their division.
Hence,your conservative figure is probably more correct than the liberal one's.
One of the players was Mason.
He held AJo against this rainbow board and decided to call the turn instead of playing it aggressively("Cardplayer" article). Based PURELY upon your figures,I can now understand Y--he was about a 3 to 1 dog in a 3-way action game(from your conservative figures)where the risk [img]/forums/images/icons/confused.gif[/img] of being re-raised by a better hand was too great.

Thanks again for your analysis!
Sitting Bull
[img]/forums/images/icons/ooo.gif[/img]
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  #4  
Old 10-08-2002, 01:22 PM
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Default Re: Thanks,PP,for your...

First of all, it's nice to know that with 3 or 4 hours of spare time, a high-end SPARC workstation, two buddies correcting my mistakes, and no money on the line, I can make the same decision as Mason at least some of the time.

Keeping in mind that I know nothing about upper middle limit games, isn't a better question to ask: "What better hands might be pushed off by a raise?" (AK, AQ, perhaps?) Or even: "What worse hands than mine would call a raise?" (assuming that he's 2nd to act)

PP
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