I remember that BigBlue was a math person. Forget if we had anyone else...
The figure they're giving you may be for opening up
all of those chests. Each one is independent, but we can still compute the probability of getting the desired item from a set number of chests.
Let's say that we have a simple scenario. There's 1 chest and it has a 50% chance of containing the type of item we want. Well, then our probability of getting the item is 0.5, or 50%. But let's say we have 2 chests and each of them has that same chance. What it means for those to be independent events is this: once we've opened up the first chest, the outcome has no effect on the probability of getting an item in the second chest. It's the same as if we were flipping a coin or rolling a die. The fact that we flipped heads first doesn't mean we're more or less likely to flip heads on the next event.
But if all we care about is getting an item of the type we're looking for, how do we find the probability of that when there are multiple independent events? Consider our simple 2-chest scenario with an independent 0.5 probability of success for each chest. There are 4 total ways this could go...
- Win, Lose
- Lose, Win
- Win, Win
- Lose, Lose
Each of those is equally likely (we know that because each opening is an independent event). But in light of our requirement, the only one of these we actually care about is #4. We specified that we want that damn item, and it doesn't matter to us when in the series of openings it pops up. It doesn't matter how many times it pops up. Just one occurrence is enough. So outcomes 1, 2, and 3 are, for our purposes, identical. So it's easy to see that our probability of failure, of not getting an item in the set of 2 chests, is 0.25 or 25%. And from that it follows that the probability of winning, of those 3 possible good outcomes combined, is 75% because 1 - 0.25 = 0.75.
It turns out we can extend this to larger numbers of chests. We don't need to map out all possible patterns of failed and succeeded openings. We just need to know the probability of getting
all failures and then we subtract that probability from 1 to find our probability of any successful outcome.
Suppose we have some smaller probability of getting the item we want per chest. Let's say it's only 5%. That's 0.05. So the probability of
not getting the item is 0.95 per chest. 0.95 × 0.95 = 0.9025. We have a 90.25% chance of both chests coming up empty, which means the chances of success are...
1 - 0.9025 = 0.0975 = 9.75%.
Now let's try it with 3 chests (using that same 5% success rate). 0.95 × 0.95 × 0.95 = 0.857375. 1 - 0.857375 = 0.142625 = 14.2625%.
With 4 chests? 0.95^4 = 0.81450625. 1 - 0.81450625 = 0.18549375 = 18.549375%.
5 chests? 1 - (0.95^5) = 0.2262190625 = 22.62190625%
And so on...
6 chests: 0.264908109375
7 chests: 0.30166270390625
8 chests: 0.3365795687109375
9 chests: 0.369750590275390625
10 chests: 0.40126306076162109375
But maybe I
really want that item. So I'm going to open up 100 chests...
1 - (0.95^100) = 0.9940794707796659745170750351176. About 99.4%.
Does that help?