Probability question (math nerds welcome)

[Renton]

why are you using the standard deviation for one trial with a sample of 100 trials? Don't you need to use the standard deviation for 100 trials?

In the case of 100 wins the mean is 2 so it should be clear that your bounds were meaninglessly large.

The mean for 100 trials is the expected value for 100 trials. Since the EV for one trial is zero, so should the EV for 100 be. And the EV outcome of winning all 100 would be 200, not 2. I think you're considering the wrong sort of mean, otherwise I am.

Can you address what the results of my long form calculation at the end of the post would be? Would it be the same as the 95% confidence interval or something different entirely. If the latter, then confidence interval is definitely not the value that interests me.

[jackvance]

The arithmetic mean (or simply "mean") of a sample , usually denoted by , is the sum of the sampled values divided by the number of items in the sample:

If the latter, then confidence interval is definitely not the value that interests me.

This is probably the conclusion to be drawn here. You seem to be getting a few concepts confused.

[MadMojoMonkey]

Can you address what the results of my long form calculation at the end of the post would be? Would it be the same as the 95% confidence interval or something different entirely. If the latter, then confidence interval is definitely not the value that interests me.

I think the result from this would be exactly 0, because it's a long-form EV calc.

Excell could answer this one in a couple of minutes.

EDIT: It would not be exactly 0 because it is not weighted by the binomial distribution. If the combinatorial weights of each outcome are taken into account (which is your intent, I believe), then it is a long-form EV calc.

Without applying the binomial distribution, it is NOT related to the variance or any other meaningful statistic, AFAIK.