Re: My chance of being a winner?
Giving it some more thought, shouldn't the divisor be sqrt(214.15)?
Both win rate and sd are calculated for sets of 100, so reasonably we should count how many sets of 100 I have?
That would give a 95% confidence interval that I am within a deviation of 2.63+-1.96(16.7454/sqrt(214.15))=2.63+-2.24
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