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math problem against 2 opponents math problem against 2 opponents

12-12-2008 , 02:21 PM
An Example
Lets say you raise first in on the button with pocket Kings SB and BB calls

SB and BB have exactly the same calling range here,

A2o-AQo A2s-AQs
K2s-KQs K8o-KQo
22-JJ
All suited connectors
97s+ T7s+ J7s Q7s+
QTo+ JTo+

flop comes Ah Ad 6c


trough stoxpoker combo I can find out that ONE of the opponent holds trips or better 27% 81 combos of 301

The question is how often does anyone of them holds trips or better?
I know I cant multiply 27% with 2 it must be something between 1 and 2
so it should be 27% * 1.x

but how do I count that?
And how do I count it against 3 opponents?

edit:
I tried to do the same thing in pokerstove against 2 opponents with these calling ranges

against one opponent his equity was 28% mine 72%
against two opponents their equity together was 48% and mine was 52%

so the equity 2 opponents has against me is multiplied by something like 1.7 (28 * 1.72 = 48)

Last edited by jeck; 12-12-2008 at 02:36 PM.
12-12-2008 , 02:26 PM
50/50. They have it or they don't.
12-12-2008 , 02:42 PM
Depends on the stakes. If it is high stakes like you are posting about right now, they never have it.
12-12-2008 , 08:33 PM
Wrong forum, you should have posted this in the probability forum.

If the probability of them having a better hand individually is 28% then to find the probability of either of them having a better hand you find the probability that they both missed and take it away from 1 (or 100%). So:

1 - (0.72)(0.72) = 1 - 0.5184 = 48.16%

The reason you can't add 0.28 to 0.28 is that it double counts the probability that they both hit. So you can also work that out and subtract it:

0.28 + 0.28 - (0.28)(0.28) = .28 + 0.28 - 0.0784 = 48.16%

For three players it's:

1 - (0.72)(0.72)(0.72) = 1 - 0.373248 = 62.68%
12-13-2008 , 12:44 AM
In Before...

<------------ THIS
12-13-2008 , 03:08 AM
Quote:
Originally Posted by novel20
Depends on the stakes. If it is high stakes like you are posting about right now, they never have it.
It's hard to make a pair in HSNL!
12-13-2008 , 06:41 AM
i dont even know what u wrote and i beat high stakes. take that math.
12-13-2008 , 08:25 AM
Quote:
Originally Posted by RedJoker
Wrong forum, you should have posted this in the probability forum.

If the probability of them having a better hand individually is 28% then to find the probability of either of them having a better hand you find the probability that they both missed and take it away from 1 (or 100%). So:

1 - (0.72)(0.72) = 1 - 0.5184 = 48.16%

The reason you can't add 0.28 to 0.28 is that it double counts the probability that they both hit. So you can also work that out and subtract it:

0.28 + 0.28 - (0.28)(0.28) = .28 + 0.28 - 0.0784 = 48.16%

For three players it's:

1 - (0.72)(0.72)(0.72) = 1 - 0.373248 = 62.68%

This is close but not 100% correct. The probablility of one of them hitting is not independent of whether the other hit. If one of them hit the ace it's less likely that the other one did as well, so in reality they hit a little bit more than 48.16%.
12-13-2008 , 10:41 PM
thread title is misleading, what you have sir is kings.

      
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