Please do something about it already.

[Iscah]

The rolls are independent so they technically can't be lumped together like that. I will definitely grant you that it's reliant on a person who wins deleting their winning coffer before rolling on the next one so that's an unlikely scenario and can be discounted from the probability calculation but it would technically be a (23/24)^3 chance to fail for one run. 7/8 is a good enough approximation assuming all players rolled and didn't delete their coffers.

Thinking this all over... the thing is, you can do all this complex maths about your chances and how three separate rolls is different to one roll with three winners and your chances once you've rolled your number and you're waiting to see what everyone else got... but at the end of the day, 24 people walk into the raid and three come out with a coffer. Therefore you have a 1-in-8 chance that you will be one of them.

Also, on the maths side of things - you said "(23/24)^3" (or about 88%) are your chances of not-winning if everyone can roll on each chest.

But because they can't, the competition for the second one is 22/23 and the third is 21/22... and (23/24)x(22/23)x(21/22)=0.875 exactly.

So. Logic first and the maths backs it up.

[Mhaeric]

"(23/24)^3" (or about 88%) are your chances of not-winning if everyone can roll on each chest. But because they can't

You must have missed the part where I explained how it's possible to win all 3 coffers and talked about how that actually plays out in reality. It was in the post with the table in it if you want to take a gander. If you don't, it boils down to me conceding the 7/8 value as being a more realistic one.

Thinking this all over... the thing is, you can do all this complex maths about your chances and how three separate rolls is different to one roll with three winners and your chances once you've rolled your number and you're waiting to see what everyone else got... but at the end of the day, 24 people walk into the raid and three come out with a coffer. Therefore you have a 1-in-8 chance that you will be one of them.

Yes, that's the probability of being a winner for a single run if you ignore what the rolls were. That's not being questioned. It does nothing, however, to describe what the probability of rolling poorly over the course of x runs has on the probability of not winning over those x runs. If they only had a handful of rolls in the 80's as their highest rolls, for example, it becomes much less surprising that they didn't win.

[Iscah]

You must have missed the part where I explained how it's possible to win all 3 coffers and talked about how that actually plays out in reality. It was in the post with the table in it if you want to take a gander. If you don't, it boils down to me conceding the 7/8 value as being a more realistic one.

You said (and I quoted) that 7/8 is "a good enough approximation". It's not an approximation, it's the exact output of the calculation of the more realistic scenario and it leads back to the logic: three people in that raid are going to win a chest.

It doesn't matter what the exact rolled numbers were. It doesn't matter what your chances are of winning chest #2 once you have rolled a 93 and you're waiting to find out if anyone got higher. Those momentary chances are going to fluctuate but they don't matter in the end.

And the fact that someone technically can be eligible for all three rolls by deliberately deleting each chest as they win it is irrelevant on an average run. There would have to be someone out to do it and they'd have to be the winner of every roll for it to make even a minor difference.

[Mhaeric]

You said (and I quoted) that 7/8 is "a good enough approximation". It's not an approximation, it's the exact output of the calculation of the more realistic scenario and it leads back to the logic: three people in that raid are going to win a chest.

I was referring to 7/8 being a good enough approximation in the context of the (23/24)^3 value being a technically more accurate even though unrealistic value. I can see how I phrased it was ambiguous.

Those momentary chances are going to fluctuate but they don't matter in the end.

I'm not sure why you seem to think I'm saying otherwise since I've brought up sample size rendering the roll meaningless in the long term several times.

And the fact that someone technically can be eligible for all three rolls by deliberately deleting each chest as they win it is irrelevant on an average run. There would have to be someone out to do it and they'd have to be the winner of every roll for it to make even a minor difference.

Yes, that's what I said.

[Iscah]

Yes, that's what I said.

So why are you basing your maths on that scenario instead of the realistic one where the first two winners are locked out of subsequent rolls, for which the equation leads to exactly 0.875 instead of a rough approximation?