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 .