How many recursive calls does it take to compute binomial probability weights?
If we let $T(n,k)$ represent the number of times that binomial1 is called when computing binomial1(n, k, p) we'll have $T(n,k) = T(n-1, k) + T(n-1, k-1)+1$ and $T(-1, k) = T(k, -1) = T(0,0)=1$. The closed form expression for $T(n,k)$ is a bit messy, but fortunately we don't need it.

Note that the binomial coefficients $\binom{n}{k}$ are defined by the nearly similar recurrence $\binom{n-1}{k} + \binom{n-1}{k-1}$ so we'll have $T(n,k)\ge \binom{n}{k}$ and in particular $T(100, 50)\ge \binom{100}{50}$ and by Stirling's approximation we'll have $T(100,50)\approx 2^{100}/\sqrt{50\pi}$, which is pretty much what you guessed.

[In fact, it turns out that $T(n,k)$ is quite a bit larger than $\binom{n}{k}$.]