Nier raid 2b outfit rant.

[Duskane]

[Lilseph]

Please, never leave us.

[Lacan]

Please, never leave us.

I never plan on it my child.

Also, looks like someone else won their roll:

[Kobalos]

There was a post buried a few pages back where the guy said he wanted one for each retainer. Too lazy to find it though. lol.

Despite the fact you can just glam over the retainer gear

[XenociderOmega]

I won with a 93 roll
/mmambo

[TheMightyMollusk]

Lost on a roll of 99. Wow.

[Lacan]

Lost on a roll of 99. Wow.

You know who won it instead??

[KitingGenbu]

You know who won it instead??

Emel Sexy. This isn't even his final form....

As for continuously obtaining the chest, I would say that would be a more realistic and sooner seen issue to solve vs overhauling the system. All they have to do is do it like how you can't loot mount items once its already in your character's system. Probably a bit more elaborate to code since mounts just go to a tab and the clothes can be literally anywhere on your character's accessible inventories (retainer/dresser/etc). I suppose if they had that much sorted out, you'd only ever be able to have one map across all inventories.

[HyoMinPark]

As for continuously obtaining the chest, I would say that would be a more realistic and sooner seen issue to solve vs overhauling the system. All they have to do is do it like how you can't loot mount items once its already in your character's system. Probably a bit more elaborate to code since mounts just go to a tab and the clothes can be literally anywhere on your character's accessible inventories (retainer/dresser/etc). I suppose if they had that much sorted out, you'd only ever be able to have one map across all inventories.

They’d have to do some type of recoding. Even now, you can obtain numerous copies of Unique gear pieces so long as they’re all stored in separate inventories (e.g., Saddlebag, Armory chest, Retainer inventories, Glamour Dresser). I’ve ended up with copies of some of the dungeon gear because I’ve forgotten I had them shoved away in a retainer for later leveling and just never fished them back out.

Its 12.5% a run with 3 chances to try. The boxes are unique marked so if you get one you cannot get another while its in the inventory.

Your chances will go down with fewer coffers, so it’s not always 12.5%—I think that’s likely applicable only to the initial lotting where you have all 24 players rolling on all 3 coffers (3 rolls, 24 players: 3/24 = 1/8 = 12.5% chance of winning 1 of the 3 available coffers before any are awarded). The second roll, after 1 person has already won, is an odds of 2 rolls per player, with 23 players rolling (again, assuming all are rolling). So, 2/23 = 8.69% chance to win 1 of the 2 coffers now available. The final coffer will be 1 lot among the remaining 22 players: 1/22 = 4.54% chance for your roll to win that 1 coffer.
This is the math I’ve done. It’s been a while since I’ve taken any sort of probability, so this could be incorrect in terms of its figuring. But I think it’s relatively correct. If it’s wrong, someone can feel free to provide me with counter math.

Having 1 available roll against 21 other players is lower odds than 3 available rolls against 23 others assuming everyone is still rolling; which, right now, is likely to be a majority of parties since the content is still new.

[PyurBlue]

Your chances will go down with fewer coffers, so it’s not always 12.5%—I think that’s likely applicable only to the initial lotting where you have all 24 players rolling on all 3 coffers (3 rolls, 24 players: 3/24 = 1/8 = 12.5% chance of winning 1 of the 3 available coffers before any are awarded). The second roll, after 1 person has already won, is an odds of 2 rolls per player, with 23 players rolling (again, assuming all are rolling). So, 2/23 = 8.69% chance to win 1 of the 2 coffers now available. The final coffer will be 1 lot among the remaining 22 players: 1/22 = 4.54% chance for your roll to win that 1 coffer.
This is the math I’ve done. It’s been a while since I’ve taken any sort of probability, so this could be incorrect in terms of its figuring. But I think it’s relatively correct. If it’s wrong, someone can feel free to provide me with counter math.

Having 1 available roll against 21 other players is lower odds than 3 available rolls against 23 others assuming everyone is still rolling; which, right now, is likely to be a majority of parties since the content is still new.

I decided to try my hand at the math. Usually you approach a problem like this by finding out the probability of failure and then subtracting that from 1 to get the probability of success. Probability of failure in this case is losing all rolls:

((P-1)/P)*((P-2)/P-1))*((P-3)/P-2))*...

P is the number of party members. There is a lot of cancellation which leads to:

(P-R/P)

With R being the number of roll chances. In the case of Alliance Raids, P = 24 and R = 3. Our chance of failure is 21/24 = 87.5%. So our chance of success is 1 - failure, which is 12.5%.