[NY] Gang member issue

[[UVA]Acrope]

Hey guys! I have no admin ability so I have to post here.

[$dollar$]Combboy~ contacted me today about an issue with NY gang saying that [Deadly]Jonas^ joined without consent and they want him out.

I refused to remove anybody and told them that they needed to either come to a conclusion on who belongs, or I can make [$dollar$]Combboy~ and *BLESST* a new gang without jonas.

I was just wondering if there was a way to validate their claims on who was the original gang owner?

if needed, this conversation happened around 9PM CDT

here is *BLESST* voice on the matter. keep in mind, I dont speak spanish so this is translated..

*BLESST*: t no si es verda
*BLESST*: pero juli me dijo q no ahi como ase nd por es un bug
*BLESST*: de ahi y q tiene q arreglar es scooby
*BLESST*: todo por f7
[UVA]Acrope: comboy el dice que jonas esta alli sin ser invitado
[UVA]Acrope: lo siento tengo que traducir todo esto
[UVA]Acrope: hice una publicacion en el foro sobre este tema
*BLESST*: jajaja si vi jajaja veras es q yo era dueño de NY y jonas me dijo q le deje entra para ayudar ase territorios ajjaja y no se q paso pero nos expulsaron de NY y al dia siguien jonas se a dueño de NY
*BLESST*: aja eso queria sabes como le expulsaba de NY por q me dejo entra a NY bajo unas condiciones jajajjaj
*BLESST*: y no me parece justo por q yo y se la mayoria de territorios ajjajaj
*BLESST*: pero nd juli me dijo q nos expulsaron por ahi un bug q esta de arreglaralgo asi
*BLESST*: y q eso tiene q arreglar scooby
[UVA]Acrope: gracias, hablare con scooby y arreglare esto
*BLESST*: aja
*BLESST*: asi q no ahi proble ajjaja es culpa del bug jajajja

thanks for your attention! also, was wanting to ask about having mod panel access back. when issues arise, there's not a lot I can do to resolve the issue with my current level.

- [UVA]Acrope

[[UVA]Trix]

Hello Acrope!

Well, what happens here is that they always ask me to remove such a player from the gang because that player doesnt belong there. All admins can remove members from any gang.

Now when they ask me if I can remove such a player, I refuse to do it. They should fix that problem between themselves, I no longer interfere.

Its also they said that there is a bug and all the gangs open by themselves, and that is when any player joins and then that person removes everyone and doesnt let anyone join.

You just have to ignore when they tell you that any gang was created by them and that there are other players added there who have not been invited. You never know if they are telling you the truth or not, therefore I already refused to help them with that.

[[UVA]Scooby]

Hi, are you guys talking about the gangs in Territories?

If so:
The gangs don't open on their own, someone will have opened it who is already in the gang.
If there is a bug and this is happening, no one has told me.
when i get some time i'll remove the votekick ability and set it so that the first member is 'admin' and will be the only one who can
remove players.
I know this doesnt help in this situation but no one has done anything wrong here.

If not:

If you are talking about him joining the server with the NY tag, then we can look into getting you clan tag protected.

Please confirm which

[[UVA]Acrope]

It was in regards to gang in territories. I'm under the same assumption of just let them sort it out themselves.

giving the gang creator admin over kicking or adding players would seem to sort the issue on its own. I was approached with the issue and told him I would at least ask about figuring out who the original player was.

hope to see you in game soon Scooby

[[UVA]Scooby]

it takes 3 members to vote the 4th member out, If they dont want you in the gang, if you're idle, if you don't play enough or for what ever reason,
you can be removed by the other 3 if they want you out, so why would you want to rejoin unless they had a good reason to remove you in the first place.

if the gang name was [NY] then it would be fair enough, but in this case, its the same for everyone.

Ill change the script when i can.