Am I ahead?

[TJD]

I am trying to determine the best way to calculate the probability of me being ahead of N opponents on a flop.

e.g. I have A7o and 2 opponents and spike middle pair on K72r board.

It is easy to work out if anyone has AT LEAST 1 King but I also need to consider 77,22, 72 or a PP higher than 7's

I think against 1 opponent the chance of them having a K is 12.5% but this clearly understates the chance of me being behind.

It would be also fairly easy to work out for ONE opponent the probability that they have any hand that could be ahead of me.

However, I see no easy way to apply that to more than one opponent.

My instinct is to work out the probability for one to beat me and then do 1- P(they are ahead) to get their "fail to beat me" probability and raise this to the power of the number of opponents; take that from 1 and voila!

However, I have no idea whether this is correct, since the P's would clearly change for player 2 if we KNEW what player 1 had.

Any ideas?

Cheers

T

[TJD]

Oh dear! Is the question too hard; too easy; have I made it impossible to understand or has it been covered a squillion times before? [img]/images/graemlins/smile.gif[/img]

I searched the archives and couldn't find anything.

If it has been covered, anyone got a link?

TYIA

trevor

[aloiz]

The problem with your line of thinking is that it has very little use in a real world situation. Players don't value every hand equally; therefore, the odds that you are ahead in the givin scenario would be misleading.

[ QUOTE ]

My instinct is to work out the probability for one to beat me and then do 1- P(they are ahead) to get their "fail to beat me" probability and raise this to the power of the number of opponents; take that from 1 and voila!


[/ QUOTE ]

Once you figure out the odds of being ahead vs. one person you cannot just treat each opponents hand as being indpendent, and thus just raise the probability to a certain power. If one opponent has or does not have you beat, that "changes" the odds that another opponent does or doesn't have you beat.

If you really want to figure out the odds of you being ahead assuming completely random holdings, the easiest way would be to use some of the simulation software that's availible.

aloiz

[TJD]

Thx Aloiz

[ QUOTE ]
The problem with your line of thinking is that it has very little use in a real world situation. Players don't value every hand equally; therefore, the odds that you are ahead in the givin scenario would be misleading.

[/ QUOTE ]

I think most of the probabilty discussions could give "misleading" results. What I am looking for is the method of making the determination. Once I know it, then I can make adjustments based on likley hand values. In some cases against some very loose and passive players or where there are blinds posted, their hands are very close to random and in those cases results like this may indicate something.

[ QUOTE ]
If you really want to figure out the odds of you being ahead assuming completely random holdings, the easiest way would be to use some of the simulation software that's availible.


[/ QUOTE ]

So it is your belief that this problem is not solvable by normal mathematical means?

Thx again

trevor