What am I doing wrong?? algerbra, percent,s and roots.. oh my.

[Shroomy]

aggghhhh... Why doesnt this work??

First a bit of explanation, because maybe an asumption is wrong. Im going to try not to use poker specific language because it just complicates things.

If your chance of winning a bet against one player is 50% your odds against multiple players are (roughly) .5 ^ n where n is the number of opponents.

so... I thought I could reverse the process and if I was given the chance of winning and the number of opponents I could figure out the equivelent for the different number of opponents.

for example .5 with one opponent is roughly equal to .25 with two opponents.

Im sure I probably messed something up by now ... but I'll continue ... [img]/images/graemlins/smile.gif[/img]

so to reverse it I dragged a bit of algerbra out of the recesses of my brain and

figured I could take the Nth root (N= # Opponents) of the chance of winning against multiple opponents, and have a rough estimate of the chance you would have against one opponent.

so if I had a .40 chance or winning againts 5 opponents it would be the same as .40 ^ (1/4)

since getting the nth root of a number is the same as raising it to the power of 1/n.


but .... obviously Im wrong, or I wouldnt be posting asking for help again [img]/images/graemlins/smile.gif[/img]

but where? (is there a plural to the word where?)

[Tom1975]

I think you have the right idea. The only mistake I see is that if you're .4 to win against 5 opponents, you would want the fifth root, not the fourth to find the odds of beating a single opponent, i.e. .4^(1/5), not .4^(1/4). Of course this assumes your chance of beating each opponent is not dependent on you beating or losing to other opponents.