With the latest update of Minecraft PE, there was a small change, namely that you can choose between 3 choices: first, to enter your own worlds, secondly to play with friends and third, to join a server. Now my question: Are these servers for free or do you have to make a subscription for it?
Servers are generally the official "relams" of minecraft
or from other providers such as leet.cc also multiplayer worlds
Both costs are understandable
Oh, because there was nothing of payment and only by joining:-)
Of course, there's also a free version that you can't really call "server", but if you want to play with his friends, that works great.
You both need an Xbox account, so you have to sign up for the pocket edititon. Make sure you are friends.
Now one can enter one world and invite the other via the pause menu. The other should receive an invitation and so you can play together over the Internet.
PS: Maybe you have to install the mobile due Xbox app, so that's possible. Try it out xD.
You do not have to pay either, you are welcome to join foreign servers… Are just strangers and not your own…
Oh yes, that's what I really wanted to know:-) So visiting others is free and creating yourself is chargeable?
Thank you but I think if you play next to each other then you do not need Internet:-)
Obtaining yourself is expensive.
I recommend you servers, for these there are quite cheap providers.
https://deluxenode.com/store/minecraft/bedrock
You need a server software there.
official from Mojang https://minecraft.net/...r/bedrock/
Pocketmine (or better: altay, has implemented mobspawning https://github.com/...eam/Altay)
Nukkit https://nukkitx.com
Realms are quite overpriced, but easier to handle.
Ok then it's good I'm only interested in joining other servers I do not have to do it myself:-)
I just mean that in Minecraft PE sometimes servers are listed that you can enter and there's nothing that it costs something:-)
Of course, joining does not cost anything
That's correct, but you must be both in the same Wi-Fi network. Not with this method!
Yes exactly:-)
