I knew you were going to say that, but somehow I thought you wouldn't at the same time
I don't know that I think computing the variance of your sample makes much sense, because it's clearly not normally distributed. But fwiw, the variance would be
(5*.5^2 + 5*.5^2 * 995*0^2)/1000
because the all-tails coins (5/1000) deviate from the mean by .5 and so do the all heads, the rest don't deviate from the mean at all.
This is variance=.0025 (or .25%)
What's the expected variance of a coin drawn from the population? I *think* this variance will be the same for any coin, whether it's all heads, all tails, or any combo in between. Why? Because if you say that heads is 0 and tails is 1, then the mean is M where 0 <= M <= 1
If we did N flips, where N is large, we expect to get MN tails and (1-M)N heads
V = MN * (1-M)^2 + (1-M)N * M^2
uh, I think that's right. That is, the tails will deviate from the mean by 1-M and the heads will deviate from the mean by M. Normally M=0.5 and 1-M = M but in your case it isn't necessarily so
V = [ MN*(1 - 2M + M^2) + (1-M)N*M^2 ] / N
V = [ MN - 2NM^2 + NM^3 + NM^2 - NM^3 ] / N
V = [ MN - NM^2 ] / N
V = [ MN(1-M) ] / N
V = M(1-M)
And this looks right to me, because, duh, it's the variance of a binomial. Which if I was thinking about it, would be sort of the obvious way to figure the variance.
So I guess the expected variance of any distribution of rigged coins is the variance of a binomial variable the same as the average of your population?
I've had a few beers so I'm not sure if the above is totally obvious, dead wrong, or somewhere in between to be frank, and I am not sure how to approach the rest of your questions.