How to calculate winrate and probability of winning in a day

[linuxrocks]

I guess this has been discussed on these forums, but I couldn't find it through search. I am not talking about variance. I know how to use my ROI and standard deviation to figure out the size of possible downswings etc.

What I want to know is: How do I calculate the chance of having a winning day in a week? What is the chance of having a winning session? I have the stats for number of winning sessions, ROI etc. Just want to know how to use them to calculate this stuff. Any thread pointers on winrate are also useful.

I read from Aba's blog that he has 58% winrate. What does that mean and how does one get that? Is it the chance of winning in a day?

[jay_shark]

It depends on the size of your sessions/win rate/s.d .

If each session is 100 hands and your win rate is 6bb/100 hands with a standard deviation of 60bb/100 hands , then there is a 68.27% chance that you will win anywhere from -54bb/100 to 66b/100 . We wish to find the probability that you win some positive number .

z=(0-6)/60 =-1/10 .

So there is about a 0.0398 + 0.5 = 53.98%(look up a z chart) chance that your session will be a profitable one . If you wish to extend the session , then your success should be even higher .

[Gigglegirl]

In PT click on 'Session notes' and then on 'more detail'.
This breaks down your game into indiviual days of the week and every 2 hour slot within each day.

[jay_shark]

If you're a pro and you play 800 hands a day with a win rate of 6bb/100 and s.d of 60bb/100 , then the probability that your session is a profitable one is :

Mean = 48bb/800 hands .

z=(0-48)/60 = -0.8 . There is a 28.81% chance that you will win anywhere from 0bb's to 48 bb's . Plus 50% of the time you will win more than 48 bb's / 800 hands .

P=0.5+0.2881 = 0.7881 or about a 79% chance that your session is a success .

In limit hold em , there aren't many player who will do better than to win two thirds of their sessions .

[linuxrocks]

[ QUOTE ]
It depends on the size of your sessions/win rate/s.d .

If each session is 100 hands and your win rate is 6bb/100 hands with a standard deviation of 60bb/100 hands , then there is a 68.27% chance that you will win anywhere from -54bb/100 to 66b/100 . We wish to find the probability that you win some positive number .

z=(0-6)/60 =-1/10 .

So there is about a 0.0398 + 0.5 = 53.98%(look up a z chart) chance that your session will be a profitable one . If you wish to extend the session , then your success should be even higher .

[/ QUOTE ]

Sorry for the noob questions. How did you get the 68.27% number? What does

z = (0-6)/60 mean?

[jay_shark]

68% is about one standard deviation above and below the mean . The mean is the number of BB's/100 hands . That is , if we regard one session as 100 hands .

So if our standard deviation is 60 bb/100 hands , then in 100 hands , there is a 68% chance that we will be anywhere from 6-60BB to 6+60 BB in that session .

z is a term used in statistics which may also be called the z-score .

[jay_shark]

-6/60 is the number of standard deviations to the left of the mean and all the way to 0 bb's . The area of that region is approximately 3.98% .

[linuxrocks]

[ QUOTE ]
68% is about one standard deviation above and below the mean . The mean is the number of BB's/100 hands . That is , if we regard one session as 100 hands .

So if our standard deviation is 60 bb/100 hands , then in 100 hands , there is a 68% chance that we will be anywhere from 6-60BB to 6+60 BB in that session .

z is a term used in statistics which may also be called the z-score .

[/ QUOTE ]

Thanks, I just now read it. So, you used a zero value to see if we can win 0 or more dollars in a session right? Then, we look up z-charts to find the probability of this.

In the second calculation, why did you use (0-48)/60 ? Shouldn't the mean and s.d be of same units? Do we just assume that s.d. would be 60/800 hands as well?

[jay_shark]

To answer your first question , yes that's correct .

[jay_shark]

Sorry I got the second part wrong .

Your sample standard deviation is 60/sqrt(8)

after 800 hands we have mean =48 .
standard deviation of 169.704bb/800 hands

(0-48)/169.704 = -0.2828

50% + 11.03% =61.03% .

Hope that's clear now .