SD/100

[DBowling]

if by winrate is 2bb/100 and my sd/100 is 16bb/100 is it possible to find the probability of me hacing an upswing/downswing of x amount?
also, is SD of 16/100 normal?

[BruceZ]

[ QUOTE ]
if by winrate is 2bb/100 and my sd/100 is 16bb/100 is it possible to find the probability of me hacing an upswing/downswing of x amount?
also, is SD of 16/100 normal?

[/ QUOTE ]

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) shows you how to compute your swings for different numbers of hours.

Of course if you play long enough, any size swing will become likely to occur, but it doesn't matter if you are a winning player, as the size of the swing will be small compared to your bankroll at the time that it occurs.

Conventional wisdom (Mason, et. al.) say that your SD should be 10 times your win rate per hour for hand dealt games, and for a good hand reader it can be even less. So a 1 bb/hr winner would have a SD of 10 bb for 1 hr. Now 100 hands is roughly 3 times the number of hands that would be hand dealt in 1 hr, so a win rate of 3 bb/100 would correspond to a SD of sqrt(3)*10 = 17 bb/100. Your win rate of 2 bb/100 would correpond to an SD of 11.5 bb/100. However, it is possible that this rule of thumb should be modified to some extent to account for the large number of beginners now playing online. You will have a higher SD if you are playing in shorthanded or very aggressive games. A higher than normal SD could also mean that you need to improve your hand reading skills.

[DBowling]

thanks for the posts, but i admit i havent taken a math course in a while and im sort of confused. if you wouldnt mind helping, i would appreciate it.

SD= 16bb/100
win= 2bb/100
avg win for 2000 hands = 44bbs
i think my SD for 2000 hands is 72bbs = 16*sqrt(20)

If everything is right up to know, im having a hard time figuring out how to look at this information in relation to a table of the standard normal distribution (http://www.isixsigma.com/library/content/zdistribution.asp).

Basically, i would like to find out the probability of myself having a 230bb upswing in 2000 hands

[BruceZ]

[ QUOTE ]
thanks for the posts, but i admit i havent taken a math course in a while and im sort of confused. if you wouldnt mind helping, i would appreciate it.

SD= 16bb/100
win= 2bb/100
avg win for 2000 hands = 44bbs
i think my SD for 2000 hands is 72bbs = 16*sqrt(20)

[/ QUOTE ]

avg win for 2000 hands should be 2*20 = 40 bbs.
SD for 2000 hands is 16*sqrt(20) =~ 71.55 bb.


[ QUOTE ]
If everything is right up to know, im having a hard time figuring out how to look at this information in relation to a table of the standard normal distribution (http://www.isixsigma.com/library/content/zdistribution.asp).

Basically, i would like to find out the probability of myself having a 230bb upswing in 2000 hands

[/ QUOTE ]

That table is laid out differently from the one that I described in my post. Yours gives the probability from 0 to z, rather than minus infinity to z. You would have to add 0.5 to each entry to get the table that I described.

230 bb would be 190 bb above average for 2000 hours, and this is 190/71.55 =~ 2.66 standard deviations above the average. The value in your table for 2.66 standard deviations is 0.4961. For this table, this means that the probability is 49.61% that your winnings will be between your average and 2.66 standard deviations above the average. You want to know the probability that your results will be this far above the average or greater, so to convert the number in the table, add 0.5 to it to get 0.9961. This means that the probability is 99.61% that your result will be between 2.66 standard deviations above the average and minus infinity. So the probability that it is greater than 2.66 standard deviations above the average is 1 minus this or 0.39%.