Easy question. Answer with formula or spreadsheet preferred:

[Baulucky]

A player plays LHE 300 hands per hour, a maximum of 6 hours per day. He makes 3 BB/100 with an SD of 16 BB/100.

If he makes 100 BBs at any point during the day, he takes the rest of the day off, otherwise he plays on for the 6 hours.

How many losing days per 100 can he expect and what is the average loss of these days?. How many days can he expect to finish early and how many hours on average does he play on those 100+ BB days?.

Thanks.

[AaronBrown]

Let's start by letting him play the entire six hours. That's 1,800 hands, with a mean of 54 BB, 3*1,800/100, and a standard deviation of 68, 16*SQRT(1,800/100).

The probability of being below zero is Normdist(0,54,68,TRUE) or 21%. The chance of being over 100 is 1-Normdist(100,54,68,TRUE) or 25%. Of course, if he's over 100 at the end of six hours, this means he hit 100 at some point and stopped playing, so he has exactly 100.

Now we have to figure out the probability that he hit 100 and went back down. There is no simple formula for this. If his expected win rate was zero, the probability is the same as the probability of being over 100. But in this case its lower, about 21%. 1% of the time he will go over 100 at some point but end up below zero.

So he has a 46% chance of winning 100 BB, 20% of the time he will end up below zero and 34% of the time he will play the full six hours and end up a winner, but of less that 100 BB.

His expected winnings are 46BB, so we know his average playing time is 46/54 times six hours, or just over five hours.

[Baulucky]

Thanks for taking a shot.