Let's say I have 578 wins and 489 losses, how do I get one

Do you need it very urgently? Because I have some material to do it but I have to give it a reading before

>>285374
I figured it out. At least I think so.

For standard deviation
>Win/loss ratio from 578 wins and 489 losses = 0.85
>total trials = 1067
>trials * win ratio * lose ratio
>1067 * 0.85 * 0.15 = 136.04
>square root of 136.04 = 11.66
>standard deviation = 11.66
right?

But how do I do binomial distribution with a 95% confidence interval based on these numbers?

95% confidence interval is defined by standard deviations around a bell curve. 95% of scores are within 2 standard deviations of the mean by definition in a normal distribution.

the way you found standard deviation is ridiculous though, what the fuck were you thinking
>Win/loss ratio from 578 wins and 489 losses = 0.85
how does this math make sense at all? looks more like 0.54 to me
so recalculate your standard deviation. then 2 std devs is 95% interval

>>285759
I should say, within 2 standard deviations of the mean. also, technically, it's just under 2 standard deviations, since 2 standard deviations is ~95.4%. you can find the actual formula somewhere else

File: Screen Shot 2017-03-25 at 3.15.16 AM.png (20KB, 349x150px) Image search: [Google]

20KB, 349x150px

>>285759
Sorry, I'm retarded. I was dividing 489/578 instead of 578/1067.
I'm just trying to replicate the results of pic related with an example, but nothing I do replicates his results.

But let me see if I got it right this time
>1067 * 0.54 * 0.46 = 265.04
>square root of 265.04 = 16.28
>standard deviation = 16.28

I'm using the formula described in this video https://www.youtube.com/watch?v=obRx4vocKFE
It's the Wald Method according to wikipedia.

>p = 0.54 (winrate)
>q = 0.39 (loserate)
>z = 1.96 (95% confidence interval)
>n = 1067 (total trials)

So
>0.54 - 1.96 * sqrt 0.54 * 0.46 / 1067
>0.54 + 1.96 * sqrt 0.54 * 0.46 / 1067

Answer is (0.51, 0.56) right?

>>285774
>0.54 - 1.96 * sqrt 0.54 * 0.46 / 1067
what a dumb equation
you calculated the standard deviation, correctly this time (16.28)
your mean is .54*1067=576.18
your 95% interval is mean +/- 1.96 std dev
so [544.27,608.088] would be your 95% interval. 95% of samples with 1067 trials should fall within this interval.
calculating a "winrate" interval is kinda goofy because your numbers still only directly apply when you're looking at a sample with 1067 trials.

also your math is a bit wrong the numbers are closer to .51,.57, they should be equally spaced around the mean. you can calculate the equivalent based on the interval above by just dividing by 1067.

>>285780
Thank you mate, I got it! As for the image, what was used? Because I applied everything you said and got different results. I got 2.87, 7.48 and 5.09. The results in the image are 3, 2 and 2.

Also, I calculated the 95% confidence interval and subtracted the difference and still got a different result from 5,5,3.

>>285792
let me just work through them. This is a polynomial distribution, a bit different from the binomial distribution that the last problem had
n=214+74+28=316

for "Win"
variance = n*p*(1-p) = 316*(214/316)*(102/316) = 69.0759
stdev = sqrt variance = 8.311
so in %, the stdev = 8.311/316 = .026

for "Loss"
variance = n*p*(1-p) = 316*(74/316)*(242/316) = 56.6708
stdev = sqrt variance = 7.528
so in %, the stdev = 7.528/316 = .0238

for "Overtime"
variance = n*p*(1-p) = 316*(28/316)*(288/316) = 25.519
stdev = sqrt variance = 5.05
so in %, the stdev = 5.05/316 = .01599

The table is correct, with rounding.
I suspect you were using the wrong formula for standard deviation. For a polynomial the stdev = n * p * (p-1). For the binomial, above, it was the same

>>285799
Thank you so much!