A neat formula for confidence intervals

[jay_shark]

If you want to know with a 95% confidence that your win rate lies between x- m.o.e and x + m.o.e , where x is your sample mean and m.o.e is your margin of error ,then use the following approximation formula :

Total number of games = (1/M.O.E)^2

So if your sample mean is 55 % and you're interested in the total number of games to determine if you're a winning player , then we would use a m.o.e of 2.5% since 52.5% is the break even point . This means that you would need to play (1/0.025)^2 = 1600 games to be 95 % certain .

Here is another example :

Suppose you've played 500 games and won x% of games but you want to be 95% confident that your win rate lies between x-m.o.e, x+m.o.e , then we would solve the equation .

Total number of games = (1/m.o.e)^2
500=(1/m.o.e)^2
m.o.e=sqrt(1/500)
m.o.e = 4.47 %

So with 95% confidence , we know our confidence interval to be x - 4.47 , x + 4.47 . So if x =60% , then our interval is 55.53,64.47 .

Pretty cool stuff .

[Schpacko]

u da man!

So, the bigger winner you are, the less games you need to know your winrate, right?

EDIT: No, wrong.

[jay_shark]

Yes , this is certainly true but by a negligible amount .

You may want to check out this link and specifically the numbers given by roundhouse .

http://forumserver.twoplustwo.com/showfl...e=2#Post7912125

[jay_shark]

To be exact , you should use +-1.96*sqrt [p*(1-p)/n]

So for 500 games and a win % of 60%, then your exact margin of error for 95% confidence is +-1.96*sqrt(0.6*0.4/500)=+-4.294%

Using the margin of error formula you get :

Total number of games = (1/m.o.e)^2
500=(1/m.o.e)^2
m.o.e=sqrt(1/500)
m.o.e = 4.47 %

The derivation of the approximation formula is easy . Just replace 1.96 with 2 and notice that p*(1-p) =~1/2*1/2 .

So we have M.O.E = +-1.96*sqrt(p*(1-p)/n) =~2*sqrt0.5*0.5/n)=~sqrt(1/n)
solve for n = (1/M.O.E)^2

[Nichomacheo]

I like. Thx.