[TNixon]

Ok, I have a variance result that I have calculated over ten hands, and what I want to do is attempt to figure out what sort of edge and average pot size it would take for a $100 stack playing 0.5/1 NL to have the same amount of variance.

Train of thought follows (I cribbed this from somebody else, but it turns out he made an algebra mistake that, when accounted for, gave the problem zero non-imaginary solutions, so I took a different approach, trying to figure out what's going on. It turns out my approach was simpler than all the algebra he had to do, but we both end up with an unsolvable equation)

Over ten hands, we're going to look for an average pot size (A) and a player edge (p, which represents the percentage of the time our 100BB player wins) that will give the same variance as our pre-calculated result (which was also over ten hands)

I'm laying out every step, so that hopefully somebody can point out my error, because the result is confusing me.

p is the percentage of times player is going to win the pot (expressed from 0 - 1). A is the average pot size. Therefore our mean should be pA. We're making the assumption that every pot is average-sized, because I have no clue how to go about approaching this if we don't make that assumption. [img]/images/graemlins/smile.gif[/img]

Anyway, the variance formula is:

10p((A - pA)^2) + 10(1-p)((0-pA)^2)
...
10p(A^2 - 2pA^2 + p^2A^2) + (10 - 10p)(-p^2A^2)
...
10pA^2 - 20p^2A^2 + 10p^3A^2 + 10p^3A^2 - 10p^2A^2
...
10pA^2 - 30p^2A^2 + 20p^3A^2
...
Since the actual value of the variance we're trying to solve for doesn't seem to matter (it's unsolvable no matter what the exact value for variance is, I'm going to call it V.
...
A^2(10p - 30p^2 + 20p^3) = V
...
A^2(10p - 30p^2 + 20p^3) - V = 0
...
Substituting P for the expression 10p - 30p^2 + 20p^3, to make things more clear for a minute
...
PA^2 - V = 0
...
Solving for A, using the quadratic equation, where a = P, b = 0, and c = -V
...
A = +/- sqrt( -4 * P * -V ) / 2P
...
We can pull 4V out, leaving P by itself in the sqrt, again to simplify further figuring, and sqrt(4) cancels the 2 in the denomiator.
...
A = +/- sqrt( V ) * sqrt( P ) / P
...
We know V is a positive real number, but P is an equation dependent on the player's edge. We will not be able to determine an average pot anytime P is less than zero, so solving for p = 0
...
20p^3 - 30p^2 + 10p = 0
...
The roots of this equation are 0, 0.5, and 1. We only care about the ranges 0 to 0.5 and 0.5 to 1, but when plugging a few numbers in to see where P comes out positive, and where P comes out negative, this starts to get *very* strange:
...
for p = 0.2, P = 0.96, indicating that P is positive in the range from 0 to 0.5.
But for p = 0.7, P = -0.84.

Since this is not dependent on the actual variance value we're comparing to, I have to believe that I've done something horribly, horribly wrong, because this equation cannot find an average pot size that would match *any* variance figure, unless the player's edge is less than 50%.

Help?