Mean Positive/Negative Session

[mantasm]

Given a winrate in BB/100, a SD, and a # of hands, how would I figure out the mean positive or negative session? I know there has to be a way to calculate this, but I can't figure it out.

[BruceZ]

[ QUOTE ]
Given a winrate in BB/100, a SD, and a # of hands, how would I figure out the mean positive or negative session? I know there has to be a way to calculate this, but I can't figure it out.

[/ QUOTE ]

First compute the average win for a session, call that u, and the SD for a session, call that sigma. Then, assuming that your session results are normally distributed, your mean positive session is:

u + sigma*EXP[-u^2/(2*sigma^2)] / [SQRT(2pi)*phi(u/sigma)]

and your mean negative session is:

u - sigma*EXP[-u^2/(2*sigma^2)] / [SQRT(2pi)*(1-phi(u/sigma))]

where phi(u/sigma) is the cumulative standard normal distribution integrated from minus infinity to (u/sigma), and is the same as the Excel function NORMSDIST(u/sigma).

For example, say your win rate is 3 bb/100 hands, and your SD for 100 hands is 20 bb, and you want to compute your average winning and losing session, where a session is 400 hands. For 400 hands your average win would be u = 12 bb, and your SD would be sigma = sqrt(4)*20 = 40 bb. Then your average winning session is:

12 + 40*EXP[-(12^2)/(2*40^2)] / [SQRT(2pi)*phi(12/40)]

= 36.7 bb.

Your average negative session is:

12 - 40*EXP[-(12^2)/(2*40^2)] / [SQRT(2pi)*(1-phi(12/40))]

= -27.9 bb.

This example serves to indicate the kinds of swings that should be expected even for a winning player with a relatively small standard deviation.

I derived the above formula for the mean positive session by computing the expected value of a random variable x distributed as a normal distribution with mean u and standard deviation sigma, with x constrained to be positive. The integral of this normal distribution from 0 to infinity is the same as the integral of the standard normal distribution from minus infinity to u/sigma, or phi(u/sigma) which is the fraction of positive sessions (draw pictures). Since x is constrained to lie between 0 and infinity, we must divide by phi(u/sigma) up front to normalize the distribution so that the integral of the distribution alone is 1. The expected value of the positive sessions then is:

1/[phi(u/sigma)*sqrt(2pi)*sigma]*integral[0 to infinity]x*exp[-(x-u)^2/(2*sigma^2)]

The mean negative session can be obtained by integrating from minus infinity to 0, with a normalization of 1-phi(u/sigma), or it can be obtained from the mean positive session by noting that phi(u/sigma)*(mean positive) + [1-phi(u/sigma)]*(mean negative) = u.

If instead of the mean you wanted the median positive and negative sessions, which are the values of the win and loss that are exceeded by exactly half of the winning or losing sessions, these can be obtained without doing the above integral. The fraction of winning sessions is phi(u/sigma) = 0.618 in our example, and the percentage of losing sessions is 0.382. The median winning session occurs at 0.382 + 0.5 * 0.618 = 0.691. From the table of the standard normal distribution, this is 0.5*sigma above the mean or 12 + 0.5*40 = 32 bb. The median losing session occurs at 0.5*0.382 = 0.191. This is 0.87*sigma below the mean, or 12 - 0.87*40 = -22.8 bb.

You might also be interested to know that the mean swing about the mean is sqrt(2/pi)*sigma =~ 0.8*sigma, which we can get from an integral calculation similar to the one above. Setting u = 0 in the above integral and in the resulting formula will give this mean swing. The median swing about the mean, which is the swing which will be exceeded exactly half the time, can be obtained directly from a table of the standard normal distribution, and it is equal to 0.67*sigma.

[mantasm]

Good stuff, thanks for your help.