Getting, and keeping players

note beforehand: Most of these ideas can be related to more common marketing tactics. A lot of this comes down to having players *trust* your server.

Keeping players – short term

  • Don’t overload them with information. People are here to play and not to read walls of signs. This will already drop their enthousiasm for a new server.
  • Have players online. Nothing screams run more than an empty server. This is ofcourse a catch-22. But a very important one to think about when for example making rules against afking.
  • Promote welcoming new players. This can absolutely make someone enthousiastic about finding a good community and will improve their chance of staying by a ton. You can start this by just welcoming new players when they join and you or another staff member are online. When other players see this they will copy the behaviour. You can also add certain funny welcoming commands with colors or send players a random “+1 karma” message.
  • Show people that you don’t have a reset coming any time soon. This is one of the most important things you can do for a lot of servers if you want to have long-term players.

Keeping players – long term

This one is harder, and also largely depends on how active your playerbase is.

  • Always. Always have something for your players to do. Something grindable is great for this (think walls for factions or farms), But you can also promote building cities or other bigger builds.
  • Events are a great way to pull many players at once including ones that initially stopped playing
  • Know how to handle griefing. Preferably without staff intervention. If players know beforehand that their building will stay up long-term they will invest more time in it. And in addition to that you won’t have anyone quitting over a griefed project that they put so many hours into. Rollbacking might be an option but sometimes people decide to quit before asking a staff member if something can be rolled back.
  • Make sure the difficulty scales well with playtime. This one is hard to get right but having a good start and then keeping some sort of difficulty when you go higher up in the game makes it entertaining for a longer time. While a lot of people like grinding there should still be challenges for them to tackle that aren’t too easy or too hard. Don’t make challenges that take a month to complete as people are not ready to invest that much time without some reward inbetween.

Set math (notations and P iff Q)

Some math basics for trying to understand https://ocw.tudelft.nl/wp-content/uploads/Algoritmiek_proofs.pdf and/or MIT’s OCW on math for computer science.

Sets

for set A and B

∪ = union, Everything in A AND everything in B (including overlap)
∩ = intersection, Everything in A AND B (just the overlap)

A ∈ B = A is part of B (smaller side of ∈ towards the smaller set)
Q.E.D. = Proof done (normally a block symbol, but wordpres….)
P iff Q = P if and only if Q = P ↔ Q) = P implies Q and vice-versa

∩ intersection (wikipedia)

Set proof (both statements imply eachother)

With set A,B,C:

A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)
the intersection of A with (B andor C)
=
(The intersection between A and B) andor (the intersection between A and C)

This looks correct, but let’s prove it.

Proof.
We show that z ∈ A ∩ (B ∪ C) implies that
z ∈ (A ∩ B) ∪ (A ∩ C) and vice-versa.

First, we show that z ∈ A ∩ (B ∪ C) implies that
z ∈ (A ∩ B) ∪ (A ∩ C):

Assume that z ∈ A ∩ (B ∪ C).
Then z is in A and z is also in B or C. Thus, z is in
either A∩B or A∩C, which implies z ∈ (A∩B)∪(A∩C)

——-

Now, we show that z ∈ (A∩B)∪(A∩C) implies that
z ∈ A ∩ (B ∪ C).

Assume that z ∈ (A ∩ B) ∪ (A ∩ C).
Then z is in both A and B or else z is in both A and
C. Thus, z is in A and z is also in B or C. This implies
that z ∈ A ∩ (B ∪ C).

Q.E.D.


Proof by contradiction

Proof by contra-positive ((P → Q) ↔ (¬Q → ¬P) is a tautology. )

In order to prove a proposition P by contradiction:

  1. Write, “We use proof by contradiction.”
  2. Write, “Suppose P is false.”
  3. Deduce a logical contradiction.
  4. Write, “This is a contradiction. Therefore, P must
    be true.”

Theorem: sqrt(2) is irrational

proving something irrational is hard, so instead assume it to be rational:

sqrt(2) = a/b (in the lowest terms a and b possible)
2 = (a^2)/(b^2)
a^2 = 2(b^2)
which means a^2 is even
thus a is even
thus a^2 is a multiple of 4
thus b^2 is a multiple of 4
thus b is also even

a and b are even which is a contradiction with the lowest term rule
thus a/b is not rational