variance formula question

[Scott_Baio]

Let's assume your natural winrate is Xbb per hour,

what is the statistical likelyhood of being +/- YBBs in a Z hour session?

Is there a straight forward formula for this?

[AaronBrown]

The other thing you need to make this computation is the standard deviation of your win rate. Let's say you win X BB per hour on average, with a standard deviation of 50 BB per hour. After Z hours, you expect to be up Z*X BB, with a standard deviation of 50*SQUARE_ROOT(Z).

If you play 100 hours, Z = 100, the square root of Z is 10. So you expect to earn 100*X, with a standard deviation of 500 BB.

About 1 time in 44 you will be up more than 2 standard deviations from your expectation, and 1 time in 44 you will be more than 2 standard deviations below it. About 1 time in 3 you will be above expectation, but within one standard deviation, another 1 time in 3 you will be below expectation but within one standard deviation. The remaining times will be split equally between 1 and 2 standard deviation above or below expectation.

[BruceZ]

[ QUOTE ]
Let's assume your natural winrate is Xbb per hour,

what is the statistical likelyhood of being +/- YBBs in a Z hour session?

Is there a straight forward formula for this?

[/ QUOTE ]

You need the standard deviation for 1 hour = sigma. Then the probability is:

NORMSDIST[(Y - Z*X)/(sqrt(Z)*sigma)] - NORMSDIST[(-Y - Z*X)/(sqrt(Z)*sigma)]

Where NORMSDIST(x) is an Excel function meaning the probability of a variable distributed by the standard normal distribution being less than x.

See this post (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&Board=probability&Number=192 2208&Forum=,All_Forums,&Words=&Searchpage=0&Limit= 25&Main=1921919&Search=true&where=&Name=197&datera nge=&newerval=&newertype=&olderval=&oldertype=&bod yprev=#Post1922208) for more info.