Fighting stormtroopers

Let’s call $q$ the probability that a stormtrooper blasts us. In the problem, $q=\tfrac{1}{1000}$. Let’s call $P(N)$ the probability that we will blast all of the $N$ stormtroopers and survive to tell the tale. Each round, several things can happen. We can blast a stormtrooper (or not) and we can get blasted ourselves (or not). One thing for sure is that we can only advance if we don’t get blasted.

• The probability that we blast a stormtrooper is given as $q K\sqrt{N}$.
• The probability that one stormtrooper doesn’t blast us is $1-q$ so the probability that we don’t get blasted by any of the $N$ stormtroopers is $\left(1-q\right)^N$.

Assembling these probabilities, we can write the following recursion:
\begin{align}
P(0) &= 1 \\
P(N) &= \left(1-q\right)^N\left[ \left(1-q K\sqrt{N}\right)P(N) + q K\sqrt{N}\, P(N-1) \right]
\end{align}The initial condition is that when there are no stormtroopers left, we have a 100% chance of blasting them all so $P(0)=1$. This recursion can be rearranged so that it’s in a bit easier to work with:
\[
P(N) = \left( \frac{\left(1-q\right)^N\left(q K\sqrt{N}\right)}{1-\left(1-q\right)^N\left(1-q K\sqrt{N}\right)}\right) \, P(N-1)
\]Now it’s clear that we can substitute $P(0)$ and solve for $P(1)$, and then solve for $P(2)$, and so on. So the probability is given by the formula
\[
P(N) = \prod_{n=1}^N \frac{\left(1-q\right)^n q K\sqrt{n}}{1-\left(1-q\right)^n\left(1-q K\sqrt{n}\right)}
\]

Exact formula

There doesn’t appear to be any way to simplify the product above, so we resort to a numerical solution. Here is a plot that shows the probability of blasting all the stormtroopers as a function of the number of stormtroopers $N$ and the blasting efficiency ratio $K$.

The plot above makes sense; our chances of blasting all the stormtroopers increases as $K$ increases (we get better at shooting them) or as $N$ decreases (there are fewer stormtroopers). We can also estimate that to blast 9 stormtroopers 50% of the time (without getting blasted ourselves!) we should have $K\approx 25$. In order to get a more precise estimate of $K$, we must do a bit more work.

Expanding the recursion as a function of $K$, the solution for $P(9)$ turns out to be the following rational function:
\[
P(9) \approx
\tfrac{K^9}{K^9+19.37 K^8+165 K^7+810 K^6+2524 K^5+5176 K^4+6972 K^3+5943 K^2+2904 K+619}
\]The only approximation I used here was in rounding the coefficients in the denominator. This function starts at zero when $K=0$ and increases monotonically and asymptotically to 1 as $K\to\infty$. This makes sense; the better we shoot, the likelier we are to blast all the stormtroopers! Here is a plot of $P(9)$ as a function of $K$:

Solving $P(9)=0.5$ numerically for $K$, we obtain $K\approx 26.797$. So in order to have a 50% chance of surviving an encounter with 9 stormtroopers, we should be roughly 27 times more accurate.

Note: It’s also possible that we blast all the stormtroopers but on our last shot, we also get blasted. I did not account for this possibility because I assumed that “blasting all the stormtroopers” meant that we also survived the encounter!

Approximate solution

In this scenario, $q$ is very close to zero and $N$ is small as well. So we can approximate $q^2 \approx 0$. This is effectively a first order Taylor expansion. This allows us to write $(1-q)^n \approx 1-nq$, for example. By applying $q^2 \approx 0$ in the original formula, we obtain the simpler product:
\[
P(N) = \prod_{n=1}^N \frac{K}{K + \sqrt{n}}
\]Note that $q$ actually cancels! The formula above represents the limit as $q\to 0$, which we’re assuming is a good enough approximation given that $q = \tfrac{1}{1000}$ is so small. Solving it numerically, we obtain $K \approx 26.707$, which is pretty close to the true solution of $K=26.797$. Unfortunately, this is still a problematic expression and there doesn’t appear to be a closed-form solution, but we can approximate further… Taking logs of both sides, we obtain:
\[
-\log P(N) = \sum_{n=1}^N \log\left( 1 + \tfrac{\sqrt{n}}{K} \right) \approx \frac{1}{K}\left(\sum_{n=1}^N \sqrt{n}\right)
\]The approximation we made stems from the fact that the expression on the right is of the form $\log(1+x)$ and $x$ is relatively small (we expect $K\approx 25$ and $n < 10$, therefore $x = \tfrac{\sqrt{n}}{K} < 0.12$). The approximation we used is a first order Taylor approximation $\log(1+x) \approx x$. If we want to find the $K$ such that $P(9) = 0.5$, we can solve this easily and obtain: \[ K = \frac{1}{\log 2} \sum_{n=1}^9 \sqrt{n} \approx 27.85 \]This is reasonably close to the $K\approx 26.707$ we were looking for and a lot easier to work with! The reason our approximation isn't better is because $x$ isn't actually that close to zero. To get a better approximation, we can use the second order Taylor approximation $\log(1+x)\approx x-\tfrac{1}{2}x^2$ and we obtain the equation:
\[
-\log P(N) \approx \frac{1}{K} \left( \sum_{n=1}^N \sqrt{n} \right)-\frac{1}{2K^2} \left( \sum_{n=1}^N n \right)
\]This is a simple quadratic equation in $K$ and solving it with $N=9$ and $P(9)=0.5$ yields the solution $K\approx 26.63$, which is much closer to the $K\approx 26.707$ we were approximating and still pretty close to the true solution $K=26.797$.