Determining if you're a winner

[jay_shark]

I'm sure many people have asked the following question with regards in determining if they're a profitable player or not .

Lets say you've played 100 heads up games and you've won 50 of them . What is the probability that your long term success rate is 60% ? Of course there is a chance that you were unfortunate and that you're a better player than the results indicate . On the flip side , you may even be a long term loser but you were lucky enough to win 50 games . It turns out , there is an easy way to determine yourself within reason if you're a profitable player .

First lets calculate your standard deviation if your long terms success rate is 60% .

sigma^2=n*p*(1-p)

where n is the number of trials =100
p is the probability of success
1-p is the probability of failure .
sigma = standard deviation and sigma^2 is your variance .

Plug in your numbers and you get
sigma^2 = 100*0.6*0.4 =24
sigma =4.89

It turns out that there is a 68.26% chance that any sample mean will fall between 60+-4.89 .

There is a 95.45% chance that your sample mean will fall between 60 +- 2*4.89 and a 99.73% chance that your sample mean will fall between 60 +- 3*4.89 . This means that if you hypothetically assume that your true win rate is 60% , then there is a 68.26% chance that you should win between 55.11 and 64.89 games and you may round up or down since you cannot win a fractional number of games . In general , if you've won x amount of games out of 100 , then you should convert your results to a z score which is easily done .

z=(x-mu)/sigma

x is the number of games won out of 100
mu is 60
sigma =4.89

z=(50-60)/4.89 = -2.044 . This means that the probability you will win 50 or fewer games as a 60% player is 0.0207 or 2.07% .

You may find it worthwhile to find a z score table if you don't already have won . As long as you know what x, mu and sigma are , you can easily convert to a z score and then look up the probability yourself . There are other inefficient methods such as using the binomial distribution formula but this will take you forever to work out and it isn't necessary .

[jay_shark]

Here is an alternative method which uses the binomial distribution formula .

If your long term success rate is 60% , the probability you win between 55 games and 65 games is the following :


100c55*0.6^55*0.4^45 =0.0478
100c56*0.6^56*0.4^44 =0.0576
100c57*0.6^57*0.4^43=0.0667
100c58*0.6^58*0.4^42=0.0742
100c59*0.6^59*0.4^41=0.0792
100c60*0.6^60*0.4^40=0.0812 (highest point on the curve)
100c61*0.6^61*0.4^39=0.07988
100c62*0.6^62*0.4^38=0.0753
100c63*0.6^63*0.4^37=0.0681
100c64*0.6^64*0.4^36=0.0591
100c65*0.6^65*0.4^35=0.0491

Add all the numbers on the rhs and you arrive at the desired probability of 73.81% .

By the way, "won" should be spelled "one" from my original post .

[jay_shark]

Here is an alternate and much more practical solution to the above problem . We're interested in the probability that you'll win between 55 and 65 games as a 60% long term winner .

so instead of plugging 55 and 65 for your sample means , you should plug 54.5 and 65.5 .

(65.5-60)/4.89 = 1.12 and if you look at z=1.12 you get that the probability you win between 60 and 65 games should be 36.86% . Likewise the probability you win between 55 games and 60 games is around 36.86%.

(54.5-60)/4.89 =-1.12 and the result is the same .

36.86*2 =73.72% chance that you'll win between 55 and 65 games which is very close to the exact answer of 73.81% .

This solution is clearly less tedious to compute and can answer a whole variety of questions .

Feel free to reply if you're unsure about something .

[teteatot]

nice, thanks for this info