# Distribution for Fraction of Success in a Binomial Setting

So the actual original question I am trying to solve is a little bit different than the title:
In a binomial setting with probability of success $$p$$, I keep examining observations until a fraction of success $$a is reached. What is the probability of stopping after exactly $$N$$ observations. This is equivalent of asking $$P(G_N|\bar{G_0},...,\bar{G_{N-1}})$$ where $$G_N\equiv a.

As a first step to solve this, I want to compute $$P(G_N)$$, i.e. the probability of getting a fraction of success $$a after $$N$$ observations. One way I can think of is to use a continuous estimation of the binomial distribution, and then integrate for $$a. But I have this feeling that there should be an easier way. Is there?

Also, I realize that after computing $$P(G_N)$$, I still have to compute the original conditional probability, which would be complicated. Is there a more clever way to think about the original question?

Computing $$P(a < f_N < b)$$ is the same as computing $$CDF_B(bN) - CDF_B(aN)$$, where $$CDF_B$$ is the cumulative distribution function of the binomial distribution with $$N$$ trials. So really all you need is an efficient way to compute the CDF of the binomial distribution.

If you're looking for a way to compute this numerically, e.g., because you want to plot $$P(a < f_N < b)$$ exactly for different values of $$a$$, $$b$$, and $$N$$, there probably exists a library for computing distribution CDF's, including the binomial CDF, in whatever programming language you're using. If you're looking for an analytic approximation, you might try something like this, where $$CDF_N$$ is the CDF of the normal distribution:

\begin{align}CDF_B(bN) - CDF_B(aN) &\approx CDF_N(bN) - CDF_N(aN) \\ &\approx PDF_N \left( \frac{a+b}{2}N \right) (b-a)N \\ &= \frac{(b-a) \sqrt{N}}{\sqrt{2 \pi p q}} e^{-\frac{1}{2\sqrt{Npq}} ((\frac{a+b}{2} - p) N)^2 }\end{align}

E.g., the probability that the observed fraction of successes is between 0.49 and 0.51 after flipping a fair coin 100 times is approximately $$0.2 / \sqrt{\pi/2} \approx 0.16$$.

The first approximation is the standard approximation of a binomial distribution by a normal distribution:

$$Binomial(n,p) \approx Normal(np, \sqrt{npq})$$

The second approximation is a midpoint approximation: for any differentiable function $$f$$,

$$f(b) - f(a) \approx f'(\frac{a+b}{2}) (b-a)$$

In the last line, we substitute the definition of $$PDF_N$$ and simplify.

All together, this approximation should be accurate so long as $$N$$, $$pN$$, and $$qN$$ are large (so that $$CDF_B \approx CDF_N$$), and $$(b-a)N$$ is small (so the midpoint approximation is accurate).

EDIT: $$p$$ is the probability of success, and $$q = 1-p$$ is the probability of failure.