How do I calculate the odds of a winning session?

over x hands
with y standard dev
with z winrate

So, assume that winrate and SD are per hand. Then, the total expected winnings for a given session is x*z. Assuming that winrates can be modeled using a normal distribution through the central limit theorem, the variance of the sample is y^2/x, so the standard deviation is Sqrt(y^2/x). Then, your probability of a winning session is P(Z<(x*z/Sqrt(y^2/x)), where Z is the Z-score from a normal random variable with mean 0 and standard deviation Sqrt(y^2/x). If z (winrate) is positive, this model would say that P(winning session) is strictly greater than .50.

The variance is x*y^2, so the standard deviation is y*Sqrt(x). The standard deviation of the win RATE or standard error is Sqrt(y^2/x), which decreases with the number of hands x. Here we want the standard deviation of the WININGS which inceases with the number of hands x.


If the original normal distribution X has mean μ = x*z and standard deviation σ = y*Sqrt(x), then Z is a standard normal distribution (with mean 0 and standard deviation 1) which comes from subtracting the mean and dividing by the standard deviation (normalizing) to get

Z = (X-μ) / σ) = (X - x*z) / [y*Sqrt(x)].

P(X > 0) = P(Z > -x*z / [y*Sqrt(x)] )

By symmetry of the bell curve:

P(X > 0) = P(Z < x*z / [y*Sqrt(x)] )

as you said. This can be evaluated with a table of the standard normal distribution, or with the Excel function NORMSDIST

P(X > 0) = NORMSDIST(x*z / [y*(Sqrt(x)] )

or

P(X > 0) = NORMSDIST(z*sqrt(x) / y).

Note that if y and z are given in units per 100 hands as is usual, then x should be total hands /100.