This problem came up in a game I was playing and I was curious how I would apply conditional probability here, or if conditional probability is even appropriate.
Problem: Imagine a turn-based game where you are up against four enemies. Their health points are as follows:
- Enemy A: 5
- Enemy B: 8
- Enemy C: 1
- Enemy D: 11
You have an option to play an attack that strikes randomly twice. It may strike the same enemy twice if their health pool allows for it (e.g., striking enemy D twice), or it may strike two different enemies. Each time it strikes, this attack deals 8 points of damage. What is the probability that you kill two enemies?
I tried working this out by hand, and here's what I got:
- Kill A: $\frac{3}{4}$ chance; then $\frac{2}{3}$ chance of killing another = ($\frac{3}{4}) \times (\frac{2}{3}) = \frac{1}{2}$
- Kill B: $\frac{3}{4}$ chance; then $\frac{2}{3}$ chance of killing another = ($\frac{3}{4}) \times (\frac{2}{3}) = \frac{1}{2}$
- Kill C: $\frac{3}{4}$ chance; then $\frac{2}{3}$ chance of killing another = ($\frac{3}{4}) \times (\frac{2}{3}) = \frac{1}{2}$
- Kill D: $0$ chance, then $\frac{4}{4}$ for the second strike since all enemies are within killing range = $0 \times \frac{4}{4} = 0$
We can end up in any one of these four states, so summing them, we get: $\frac{1}{2} + \frac{1}{2} + \frac{1}{2} = \frac{3}{2}$ which... makes no sense because it exceeds $1$.
I also tried applying Bayes' Theorem, but that didn't get me very far as I had trouble identifying what the two probabilities are:
$$ P(A|B) = \frac{P(A)}{P(B)}P(B|A) $$
Where:
- $P(A|B)$ is the probability of killing a second enemy given that you already killed one
- $P(A)$ is the probability of killing an enemy on your second hit
- $P(B)$ is the probability of killing an enemy on your first hit
- $P(B|A)$... doesn't seem to make sense in the context of this problem
Where am I going wrong? Should I not be treating this as a conditional probability problem?
I don't know if I'm missing something, but I think it's quite simple. To kill 2 enemies, you cannot hit enemy D. Hence, for the first strike it's $\frac{3}{4}$, and then you need to hit one of the remaining two with less than 8 points of health, i.e. $\frac{2}{3}$. Which gets you to $\frac{3}{4} \times \frac{2}{3} = \frac{1}{2}$