# How much redundancy is needed to comfortably ensure an outcome of a probabilistic event

probability

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?

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.