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A binomial process is a random process that represents the total number of positive outcomes of probability $p$ in a binary (yes-or-no) trial repeated $t$ times. For each $t$, $\mathcal{P}(t)$ generates a probability of:
\[ \binom{t}{x}p^x (1-p)^{t-x} \]
This is a binary trial, because we are only interested in whether we get a five-star ranked Servant, or not. Then according to the pull info, the probability of getting one of those is $1\%$, which means $p = \frac{1}{100}$. Therefore:
\[ \mathcal{P}(t) = 99^{t-x} 100^{-t} \binom{t}{x} \]
For all $ 0 \leq x \leq t$. The term $\binom{t}{x}$ is equivalent to $\frac{t!}{x!(t-x)!}$.

Since we are interested in knowing the probability of someone getting $k$ or more pulls, we have to sum up the probabilities from $k$ to $t$ (since logically, one cannot get any more pulls than the number of attempts made). Thus:
\[ \sum_{x = k}^{t}\left( \frac{99^{t-x} 100^{-t} t!}{x!(t-x)!} \right) = \frac{99^{t-k} 100^{-t} t!}{k!(t-k)!} {}_2F_1\left(1,k-t,1+k,-\frac{1}{99}\right)\]
Now, while this may look intimidating, especially with the hypergeometric function thrown in there, it turns out that this expression simplifies greatly if we take note of the fact that $k$ and $t$ are integers, and require that $ 0 < k \leq t$, then:
\[ P( x \geq k ) = \frac{\mathrm{B}\!\!\left(\frac{1}{100};k,1-k+t\right)t!}{(k-1)!(t-k)!} \]
The incomplete beta function $\mathrm{B}(w;r,s)$ is defined as:
\[ \int_0^w u^{r-1} (1-u)^{s-1} \mathrm{d}u \]
Using the tool of numerical integration, we now have everything we need to get a quantified probability. Substituting $t = 10$ and $k = 7$, the exact result is $\frac{29219479}{25000000000000000000}$, or about $1.2 \times 10^{-12}$.

For comparison, that's about half the chance of being canonized as a saint (1 in 20,000,000) after suffering an injury due to fireworks (1 in 20,000).

Better break out the Catherine wheels.

Wed, 03 Jan 2018 08:48 GMT