Hi.

I'm trying to solve this puzzle.

My win rate in games is fixed (for arguments sake) at 53%.

I play 2,000 games per month.

So the mean average of games won per month is 1060:940. +120 buy ins.

I pay 2% rake. So I lose 40 buy ins per month to rake. I average, after rake a total profit of 80 buy ins.

My problem is: what is the probability of me having a losing month where I am -1 or more buy ins down?

Thanks in advance, Zak

It seems each win is 1 buy-in, presumably these are tournaments. For losing at least one buy-in in 2000 games given you have a rake of 40 buy-ins, you cannot win more than 1019 games. Then you will have won no more than 1019-981 = 38 games net for an overall loss of at least 2 games. You cannot have an overall loss of 1 buy-in.

The probability of this with a win probability of 53% using Excel's binomial distribution function is,

binomdist(1019, 2000, 0.53,1) = 3.49%.

In general, if you play a game n times (so n=2000 here), and your probability of winning is p (so p =0.53 here), you would expect to win np = 1060 times, and your standard deviation of the number of times you win will be √[n*p*(1-p)]. So your standard deviation would be √[2000*0.53*0.47] = √498.2 ~ 22.3 or so.

Our results should be approximately normally distributed, so about 68% of the time, your number of wins will be within 22.3 of 1060 (so between 1037.7 and 1082.3), and 95% of the time will be within 44.6 of 1060 (so between 1015.4 and 1104.6). We want to know the probability that our number of wins is at least 41 below expectation, or in other words at least 41/22.3 = 1.84 standard deviations below expectation. Looking at a standard z-score table, the answer seems to be about 3.3%.