Please do something about it already.

[Iscah]

If I'm doing maths right on my phone calculator... (0.875) (x^y) (32) = 0.014 = only a 1.4% chance of doing 32 runs and not getting it once.

0.875 being your 7-in-8 chance of not getting it on a single run.

Bad luck just statistically happens sometimes.

[ItMe]

Theoretically I could run the Drowned City of Skalla forever and never get the fending coat.

Theoretically I could send my retainer out forever and never have them come back with the special untraceable onion helm (the one you only get from them, not the other one).

Theoretically you could never get the special ring for getting 1st prize at the jumbo cactpot.

Man ... Thinking about it there are quite a few things we do in FF14 that do not guarantee the reward you're shooting for. In some ways that's scary, but in others it helps some activities (like fishing) feel kinda "at home" for me.
But I got the fending coat.
My retainer brought back that helmet.
AND YOU CAN GET THIS CHEST!

[Mhaeric]

If I'm doing maths right on my phone calculator... (0.875) (x^y) (32) = 0.014 = only a 1.4% chance of doing 32 runs and not getting it once.

0.875 being your 7-in-8 chance of not getting it on a single run.

Bad luck just statistically happens sometimes.

The math isn't quite right since there's 3 drops per run so they had 96 chances rather than 32. They're also likely rolling against more the 7 other people. Going by what they just said in the last post they were usually rolling against 10-15 people.

It's also muddied by not being a 1 in n players roll so the base chance of losing isn't exactly (n-1)/n. E.g. rolling a 2 has a chance of winning and 98 has a chance of losing, but 2 has a much lower chance of winning than 98 does of losing. The same can be said for 3 and 97, 4 and 96, etc, so the overall weighting is unevenly skewed towards losing on average. Over time and looking at all players rolls it would be a (n-1)/n probability, but from the perspective of a single player, their probability would depend heavily on their individual rolls and only rolls above 90 would have a reasonable chance of actually winning.

All that said, yeah that's still unlucky to not have won it in 32 runs.

[Iscah]

The math isn't quite right since there's 3 drops per run so they had 96 chances rather than 32. They're also likely rolling against more the 7 other people.

I incorporated that. Three chests per run means that three out of 24 people are leaving that raid with a chest, so you have a 1-in-8 chance of being one of those three people in any single run (assuming everyone rolls - your odds can only improve if some are skipping). Therefore 32 runs = 32 1-in-8 chances.

[Mhaeric]

I incorporated that. Three chests per run means that three out of 24 people are leaving that raid with a chest, so you have a 1-in-8 chance of being one of those three people in any single run (assuming everyone rolls - your odds can only improve if some are skipping). Therefore 32 runs = 32 1-in-8 chances.

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.

This way of looking at it is still not the whole picture, however, because it doesn't take into account the competing roll situation where you have to include the chance of your own roll being good enough to win compared to other rolls. Yes, this averages out over time to the above calculation when accounting for everyone's rolls and 1000's of attempts, but we only have a sample size of 32 here so the probability of you rolling high enough to have a reasonable chance to win becomes a significant factor from an individual player perspective. Here's how quickly the chance of winning on a given roll against 23 other players drops off (stolen from a reddit source since I didn't want to actually create a spreadsheet just for this):



Even if all 3 of your rolls were 90 for example, you'd still have a roughly 78% chance of not winning any of them. (I say roughly because only the first roll would follow that table above since it's against 23 players and the probability would be slightly higher for the second and third rolls against 22 and 21 players.)

[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.