How much redundancy is needed to comfortably ensure an outcome of a probabilistic event? I'm playing a game where meteor showers periodically strike the surface of the planet. For protection, you can install defense systems that can fire one projectile per meteor shower which have an 80% chance to hit and destroy a meteor. In case of a miss, there is ample time for any redundant systems to shoot at any missed meteors.
However, the number of meteors in each shower is random. There is a 50% chance of one, 25% chance of two, 12.5% chance of 3, 6.25% chance of 4, and so on. What equation represents the probability that no meteors strike the planet in a meteor shower given $n$ defense systems?
 A: The number of succesful defence systems can be seen as a $\operatorname{Binomial}(n,0.8)$ variable, although some of the defence systems might be shooting at thin air if there are no meteors left. The distribution of meteors follows a $\operatorname{Geometric}(\frac12)$ distribution, so the probability of having no meteor impacts would be
$$\mathbb{P}(X \geq M) , \quad  X \sim \operatorname{Binomial}(n,0.8)  , \quad M \sim \operatorname{Geometric}(\frac12),$$
where $X$ and $M$ are assumed to be independent. Note that we can calculate this using the law of total probability
\begin{align*}
\mathbb{P}(X\geq M) &= 1-\mathbb{P}(M>X) \\
                    &= 1- \sum_{k=0}^n \mathbb{P}(M>k\: | \: X=k)\mathbb{P}(X=k) \\
                    &= 1- \sum_{k=0}^n \mathbb{P}(M>k)\mathbb{P}(X=k) &\text{(by independence)} \\
                    &=1 - \sum_{k=0}^n (\frac{1}{2})^{k} \cdot {n \choose k} 
(0.8)^k(0.2)^{n-k} &\text{(inserting the CDF and PMF)} \\
                    &= 1 - \sum_{k=0}^n  {n \choose k} 
(0.4)^k(0.2)^{n-k} \\
                    &= 1 - (0.4 + 0.2)^n &\text{(using the binomial formula)}
\end{align*}
We see thus, that the probability of not getting hit by a meteor would be $1-(0.6)^n$. Note in particular, that the probability goes to $1$ as the number $n$ of defence systems goes to infinity.
A: I'll make a few additional assumptions to fully specify the scenario:

*

*The meteorites come in some "order" (one after the other)

*The defense systems employ a "greedy" approach: they always try to shoot down the meteorite that's closest to the surface. If they succeed, they move on to the second closest, and so on.

*To shoot down a single meteorite, the defense systems take turns in shooting until one succeeds or the projectiles run out (meaning that the meteorite strikes the surface).

Let $p$ be the probability of one trial: a single projectile taking down a single meteorite.
Then $1 - p$ is the probability of missing.
We want to compute the probability $\Pr(S)$ of a meteorite striking the surface given that there are $n$ defense systems ($n$ projectiles).
This could happen in a few different ways: let $A_k$ denote the event in which the meteorite shower has $k$ meteorites and one hits the surface, and let $B_{k,i}$ where $1 \le i \le k$ denote the event in which the $i$-th meteorite strikes the surface.
Then $S = \bigcup_{k=1}^\infty A_k$ and $A_k = \bigcup_{i = 1}^k B_{k,i}$, moreover these are disjoint unions.
For the first meteorite to hit the surface, all $n$ systems have to fail:
$$\Pr(B_{k, 1}) = (p - 1)^n $$
If it's the second one that strikes the surface, the first one must have been shot down after $j$ trials, and the rest must have failed on the second meteorite:
$$\Pr(B_{k, 2}) = \sum_{j = 1}^{n} (p-1)^{j-1}p(p-1)^{n - j} = \sum_{j = 1}^{n} (p-1)^{n-1}p = np(p -1)^{n-1} $$
We can generalise this.
If the $i$-th meteorite hits the surface, the first $i - 1$ must have been shot down.
There are ${n \choose i - 1}$ ways to choose the $i - 1$ projectiles in the sequence that were successful.
All the other projectiles missed.
Therefore:
$$\Pr(B_{k, i}) = {n \choose i - 1} p^{i - 1}(1 - p)^{n - (i - 1)} $$
Now we just have to sum up all these disjoint cases:
$$\Pr(A_k) = \sum_{i = 1}^k \Pr(B_{k, i}) = \sum_{i = 1}^k {n \choose i - 1} p^{i - 1}(1 - p)^{n - (i - 1)} = \sum_{i = 0}^{k-1} {n \choose i} p^{i}(1 - p)^{n - i}$$
As a sanity check, we can see that if $k > n$, i.e. if we have at least one more meteorite than there are projectiles ($k \ge n + 1$), then $\Pr(A_k)$ becomes $1$ (using the convention that ${n \choose i} =0$ if $i > n$):
$$\sum_{i = 0}^{k-1} {n \choose i} p^{i}(1 - p)^{n - i} = \sum_{i = 0}^{n} {n \choose i} p^{i}(1 - p)^{n - i} = (p + (1 - p))^n = 1^n = 1$$
I'm not sure how similar this is to the other answers posted here.
