How to calculate standard deviation of event without knowing probability. So, I've been bowling recently, and I realized I had an odd conundrum. I knew how to do binomial distributions with coin flips and whatnot, because we knew the probability.
However, when I tried to do strike chance analysis, I realized I was a bit lost.
For example, I bowled 5 games. I had 52 shots at all 10 pins, and 13 of those ended up as strikes. I know the total number of chances, and the number of successes, but since I don't know the actual probability of me getting a strike, I don't know how to find the standard deviation of my true strike rate.
If I had $A$ chances to strike, and striked $B$ times, I have a strike rate of $A/B$. Do I, for statistical purposes, presume that $A/B$ is the true probability? Or is there another formula for how to make estimates like that when the true probability is unknown, and I want to find the standard deviation of the true probability, and find the confidence intervals of my strike rate.
 A: Your model is that the number of strikes (successes) is a random
variable $X \sim Binom(n, p),$ with known $n = 52$ and unknown $p.$
The usual estimate of $p$ is $\hat p = X/n,$ where $X = 13$ is
the observed number of successes. As you point out, if you use the normal approximation to the binomial, you need
to know the SD of $\hat p$ in order to get a confidence interval 
for $p$. Of course, $Var(X) = np(1-p),$ so $Var(\hat p) = p(1-p)/n$
and $SD(\hat p) = \sqrt(p(1-p)/n.$ But we do not know $p$ so
we can't get an exact value for $SD(\hat p).$ 
The traditional 'Wald' 95% CI is of the form
$$\hat p \pm 1.96\sqrt{\hat p(1- \hat p)/n},$$
where we use an estimated value of $SD(\hat p).$
Recent studies have shown that this interval does not work well, particularly
for small $n$ because there are two approximations involved: normal to
binomial and $\hat p$ to $p.$ Specifically, the Wald CI does not have
the 'advertised' 95% coverage probability, especially for certain values of $p.$
A better procedure, due to Agresti and Coull, uses $\tilde n = n+4$ and $\tilde p = (X+2)/(n + 4)$  to get a 95% CI of the form
$$\tilde p \pm 1.96\sqrt{\tilde p(1- \tilde p)/\tilde n},$$
which performs much better. For your data this 95% CI is $(0.152, 0.384).$
As is typical of binomial CIs based on small $n$, this interval is quite wide.
For example, it takes about $n = 2500$ to estimate $p$ within $\pm .02$.
And if you bowled that much, your $p$ might change!
My Answer to a related Question on this site goes into some detail
about the poor behavior of the Wald interval and other types of CIs beyond
the ones discussed here. Yet
another Answer mentions the use of a Bayesian interval estimate for $p.$
