# What is the probability of killing two enemies out of four with two random strikes if you know the damage that you deal and the enemies' hitpoints?

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?

• Not sure the rules are clear. If, say, you strike $C$ first, then $C$ is gone (right?)...does that mean you can't strike them again? If so, then to kill $2$ enemies, you just have to avoid striking $D$. Or have I misunderstood? Of course, if you are allowed to uselessly strike at a dead enemy, then you also have to avoid that.
– lulu
Commented Aug 2, 2022 at 13:59
• The probability of killing $A$ on the first hit is 1/4, not 3/4, and similarly for $B$ and $C$. Commented Aug 2, 2022 at 14:02
• @lulu Yeah, that's correct: If you hit C, then they die and you are left with only three enemies. I was under the impression that this is conditional probability because your candidate pool for the second strike depends on which enemy you hit on the first strike. Commented Aug 2, 2022 at 14:06
• @eyeballfrog Oops, you're right! Commented Aug 2, 2022 at 14:06
• It is very confusing. From where have you got chances of 3/4 each for A,B,C, and 0 for D ? These have been put in the solution, but aren't there in the question ! Commented Aug 2, 2022 at 14:40

## 1 Answer

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 $$\frac{3}{4} \times \frac{2}{3} = \frac{1}{2}$$

• That's the answer I got as well, for the same reasons (+1). Like you, however, I feel like we are missing some key part of the question.
– lulu
Commented Aug 2, 2022 at 13:59
• Oh... I may have been overthinking this. I thought I would need to use conditional probability here because I was under the impression that the first strike changes the outcome of the second strike (because your enemy pool might change). Commented Aug 2, 2022 at 14:08
• As others mentioned, they way you worked it out is fine, but the probability to strike each one of A,B,C,D is $\frac{1}{4}$ on the first strike. This gets you the same result if you add up the three cases where 2 enemies are killed, but there is no need to distinguish those three, as they are all killed by a single strike.
– Alex
Commented Aug 2, 2022 at 14:53