- Thread starter Spiderman
- Start date

The figure they're giving you may be for opening up

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

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

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

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

1 - (0.95^100) = 0.9940794707796659745170750351176. About 99.4%.

Does that help?

Uh... right?

From what I remember of the statistics I took in school, the error in thinking that plagues some students is going past that point, revealing more information, and then treating it as though it's still that same original probability. For instance, let's say that I've got 200 chests and the probability of getting the desired item in each chest is, independently, 0.5%. A quick calculation shows that the probability of getting my item in there somewhere is 1 - (0.995^200) ≈ 63.3%. So it's in the realm of not really being close to a sure thing, but still pretty likely. Now, someone who is thinking things through properly and who knows that the starting probability is 63.3% would realize that as you go through the chests one-by-one and keep coming up with failures, it becomes increasingly likely that you're not in your desired 63.3% chance of success, but rather that you happen to have landed in the undesirable 36.7% part of the map. Once you've opened up 199 chests, it's easy to see that your probability of winning is not 63.3%, but instead is exactly 0.5% (which is extremely bad, albeit still not completely impossible). And it shouldn't be hard to realize that at every point along the way, our calculated probability would have been some number between 0.5% and 63.3%. Our chances steadily dropped as the failures racked up. It's a version of the "Gambler's fallacy" to hold onto that original 63.3% from when we had less information and to think, "We had a better than even chance when we started and we've come up dry so far, and that means success is just around the corner." And of course it doesn't work that way. Like you said in your first post, these are independent events. Failing to score 199 times in a row doesn't make that 200th chest any more or less likely to give your item. The difference between opening up 200 all at the same time and failing vs. opening them up one-at-a-time and failing is that the latter is slower and more agonizing. The end result is the same and is, from our starting point, equally likely.