prob of beating N opponents given prob of beating 1 opponent
 

I want to make sure that I have this right for texas hold'em.
If I know my pocket cards and any common cards then I
should be able to compute the odds of beating one opponent.
[In fact I have written code to use poker-eval library to
do this.] If the probability that I beat one opponent is p,
then is the probability that I bean N opponents roughly N*p?

Re: prob of beating N opponents given prob of beating 1 opponent
 

I think you mean p^N, p*N can be greater than 1, so that can't be a probability.

Re: prob of beating N opponents given prob of beating 1 opponent
 

respondant is correct.... of course, your odds of beating the second guy won't be same as beating the first, might be ok for an assumption though in certain circumstances.....

nice thing is that if you have TT, and one opponent has AK and another AQ, then their sharing of the aces cuts down on their cumulative winning percentage. if they have QJ and AK against your TT, then that's a big problem.

Re: prob of beating N opponents given prob of beating 1 opponent
 

Right. I meant p^N. But I think it's only an approximation.
I would like to understand if this is indeed an approximation
and if so, how good is it.

Re: prob of beating N opponents given prob of beating 1 opponent
 

[ QUOTE ]
Right. I meant p^N. But I think it's only an approximation.
I would like to understand if this is indeed an approximation
and if so, how good is it.

[/ QUOTE ]
In general, it's a terrible approximation.

Multiplying probabilities is correct for independent results. The probability that the next 10 coinflips are all heads is the product of the probabilities of a head on each toss.

In a competition, how you compare against one opponent is not independent of how you compare against another opponent. This is particularly true for drawing hands. A strong draw which hits is likely to beat all opponents. If you have the nut flush draw on the flop, most of your wins against made hands come from making your flush. To beat 9 opponents, you don't need to hit 9 flushes in a row, just one.

You can't determine how well a hand will do against n random opponents from how well the hand does against 1. You have to know how the hand wins. 22 is about 50% against 1 random opponent since 22 unimproved often wins. Against 3 random opponents, 22 unimproved is highly unlikely to win, and the probability of hitting a set is too low (but greater than 0.5^3). 22 wins 21.5%. 98s also wins about 50% against a random hand, but a larger fraction of its wins come from making big hands like straights and flushes which are very likely to win against 3 opponents. 98s wins 28.1% against 3 random opponents.

Re: prob of beating N opponents given prob of beating 1 opponent
 

yeah it's bad for the reasons you stated, just figured that was his typo.