Hello, I am trying to figure out how to calculate the probability of being a certain amount from EV. I feel my math is wrong somewhere, that is why I am curious what others have to say.

Here is how I went about the calculation. In my example I am using 9230 hands. The amount of money lost in this sample was $292.81. The amount that was lost in EV was $65.29. A gap of $227.52. Since I am only interested in the probability of the gap between the actual winnings and the EV I will compute the mean and the standard deviation of the difference between the amount won/lost in each hand and the EV won/lost in each hand. The results are a mean of -0.007074 and a standard deviation of 1.2297097. Finally I will compute the average actual amount lost per hand in the same. This number is -0.031724. An average amount lost per hand of roughly 4.5x more than expected.

With all this I will calculate the Z score. Z = x - μ / σ

[-0.031724 - (-0.007074)] / 1.2297097 = -0.02

Z = -0.02

A Z score that close to zero implies that this gap between actual winnings and EV is very likely. P ~ 50%. It feels like a much more improbable run than this, but the math says otherwise.

Not trying to complain, just trying to get a handle on how probable this actually is.

You’re looking at the difference in means of two distributions, but you only used the standard deviation of the larger varying distribution, that of the actual win rate. Look up the Behrens-Fisher problem. Not sure if this is the best approach because there is correlation between the two variables.

Thank you for the response. After reading about the Behrens–Fisher problem all i can really say is WHOOSH that is way over my head LOL. Is it really that difficult to calculate the probability of the gap between the actual winnings and EV? There are all kinds of threads where people say they are running X amount below EV, but there doesn't seem to be anything about how likely it is to be X amount below EV.

Perhaps a better way to look at this would be to filter for hands that were all in on the turn or earlier. Look at the equity percentage of each hand. compute mean and std deviation, and compare it with showdown win %?

Is $1.23 the standard deviation of your profit per hand minus expectation? Or the standard deviation of your average profit per hand minus expectation over a sample of 9,230 hands (in that case, the standard deviation per hand would be $1.23 times the square root of 9,230, or $118)? Asking this question another way, are you playing $0.05/$0.10 or $5/$10 poker?

In the former case, you should multiply your z-value by the square root of 9,230, giving you -1.93, which is on the border of conventional statistical significance.

But your test is not the most powerful one. You really should measure deviation from expectation hand-by-hand, and divide by the standard deviation hand-by-hand, before aggregating the numbers. This will tell you much more precisely whether or not your model of EV per hand is correct.

To do this, you have to make some strong assumptions about the variances.