# What's the chance?

In my game, I wear a set of armour which gives me 20% chance of not losing ammo when I shoot from my gun. But my gun also gives me the 33% chance of not losing ammo. So, what's the overall chance of not losing ammo?

As far as I understand, the chance is 20% $$or$$ 33%, which is $$0,2 + 0,33 = 0,53$$ (53%)? Am I right? If so, what if my armour gives me 50% chance and the gun gives me 50% chance? The chance of not losing ammo will be 100% ?

• This is not determined (mathematically) from the given data. It could, of course, be determined by game rules or internal settings.
– lulu
Commented Jun 2, 2022 at 20:05
• This is entirely up to the programmers who made the game. They have chosen (consciously or not) exactly how two percentages should interact through how the game is coded. It does remind me of back in the day when my theorycrafting friend tried to optimise his World of Warcraft build and wondered whether a miss could crit, for basically this reason. Fun times. (Or maybe I'm misunderstanding and this is not a computer game you're playing but a game you are making. In that case, it's entirely up to you.) Commented Jun 2, 2022 at 20:05
• Need more information, for instance whether they are independent or dependent. Also may be more relevant to ask here: gaming.stackexchange.com so you can include what game it is and such. Commented Jun 2, 2022 at 20:05
• I have seen games where these percentages would add; others where the corresponding decimals multiply instead; yet others where we just take the max or min. There's too much left unknown Commented Jun 2, 2022 at 20:10

I agree with the comments posted so far. However, there is another issue.

Let $$E_1$$ denote the event that the armour prevents ammo loss.

Let $$E_2$$ denote the event that the gun prevents ammo loss.

For any two events, $$R,S$$, let $$(R,S)$$ denote the event that both events $$R$$ and $$S$$ simultaneously occur.

For any event $$R$$, let $$R'$$ denote the complementary event that event $$R$$ did not occur.

Given events $$E_1, E_2$$, there are $$4$$ mutually exclusive ways that the two events might interact with each other:

• $$(E_1, E_2)$$
• $$(E_1, E_2')$$
• $$(E_1', E_2)$$
• $$(E_1', E_2')$$.

Of the $$4$$ possible events, the only one that has the ammo loss not being prevented, is the last one.

So, the probability that ammo was not lost is

$$1 - p(E_1',E_2').\tag1$$

which equals

$$p(E_1,E_2) + p(E_1,E_2') + p(E_1',E_2) ~=~ ~\text{[for example]}~~ p(E_1) + p(E_1',E_2). \tag2$$

You assumed that the expression in (2) above was equal to

$$p(E_1) + p(E_2).$$

You can see, from the RHS of (2), that this is wrong.

In order to determine the probability that the ammo is not lost, you have to evaluate either (1) above or (2) above.

As others have pointed out, there is insufficient information to determine this.

Suppose that it was given that events $$E_1$$ and $$E_2$$ were independent. This would imply that events $$E_1',E_2'$$ were independent.

Then, you could calculate
$$p(E_1',E_2') = p(E_1') \times p(E_2') = \dfrac{4}{5} \times \dfrac{2}{3} = \dfrac{8}{15}.$$

This would imply that the expression in (1) above evaluates to $$\dfrac{7}{15}.$$

More specifically, while you are given $$p(E_1)$$ and $$p(E_2)$$, you are not given either $$p(E_2|E_1)$$ or $$p(E_2|E_1')$$.

In other words, you know the $$p(E_2)$$, but you do not know the probability of event $$E_2$$ occurring under (for example) the assumption that event $$E_1$$ does not occur.