TL;DR

BoardGameArena restricts the creation of tables for some selected games only to premium users. How does this affect tournament arrangements?

# What’s the issue?

First and foremost: BoardGameArena is an amazing site and you’re totally encouraged to go premium. Everybody has its own financials, though, and not all players are premium ones.

The amazing thing about BoardGameArena is that you’re in no way prevented from playing premium games; the only restriction is that you have to wait for some other player to create a table for that game.

This, of course, affects the creation of multi-player tournaments of premium games. How many premium members have to be in these tournaments? How should they be arranged?

Let’s take a closer look!

Let’s see a couple of examples for less-crowded tournaments.

## Two-player games

Let’s start simple: a tiny round-robin tournament of 4 players in two-players games:

1st round: (1, 2) (3, 4)
2nd round: (1, 3) (2, 4)
3rd round: (1, 4) (2, 3)


It’s easy to see that we need three premium users here. As soon as player 1 is not premium, in fact, all the other ones need to be premium to be able to start the first table in the three rounds. So this is it: for $n=2$, the total number of players is $n^2 = 4$ and we need 3 of them to be premium.

## Three-player games

Let’s now consider three-players matches:

round 1: (1, 4, 7) (3, 6, 8) (2, 5, 9)
round 2: (4, 5, 6) (1, 2, 3) (7, 8, 9)
round 3: (3, 4, 9) (1, 5, 8) (2, 6, 7)
round 4: (2, 4, 8) (1, 6, 9) (3, 5, 7)


Let’s first select 1 as a premium user (anybody can do at this point):

round 1: (1, 4, 7)
round 2:           (1, 2, 3)
round 3:           (1, 5, 8)
round 4:           (1, 6, 9)


So, in theory, at this point every other player might be non-premium. But of course there are other matches to cater for. We addressed 4 tables, which is the maximum we could do with one single premium player.

Let’s now select the first match (1, 4, 7), and decide that also the other two players are premium. We are going to surely include additional tables, because we already know that this first table is the only one where any pair of 1, 4, and 7 appear at the same time.

round 1: (1, 4, 7)
round 2: (4, 5, 6) (1, 2, 3) (7, 8, 9)
round 3: (3, 4, 9) (1, 5, 8) (2, 6, 7)
round 4: (2, 4, 8) (1, 6, 9) (3, 5, 7)


As a matter of fact, each of them addresses 3 additional games, again because the only game where they encounter each other is the first one, which we already assigned to 1.

Now, we’re only left to choose exactly one additional player in each of the tables of the first round. This can be easily done by choosing any table with 1 in another round, and take all of the other players as premium. Let’s take (1, 2, 3) from the second round then:

round 1: (1, 4, 7) (3, 6, 8) (2, 5, 9)
round 2: (4, 5, 6) (1, 2, 3) (7, 8, 9)
round 3: (3, 4, 9) (1, 5, 8) (2, 6, 7)
round 4: (2, 4, 8) (1, 6, 9) (3, 5, 7)


Now everything is accounted for, with the following premium users: 1, 2, 3, 4, and 7. It’s 5 players out of the total of 9 - which starts being better than the two-player games example we saw before!

The example for three-players games shows us a way to understand how many premium users we need, and how to select them.

For $n$-players games, we have to account for $n \cdot (n + 1)$ total matches, played by $n^2$ players (assuming they are indexed from 1 on), each playing $n + 1$ matches.

Let’s consider player 1, and any two matches (let’s call them A and B) where this player participates. We claim that making all the players in this matches premium is sufficient to cover the whole tournament.

Making 1 premium, of course, addresses $n + 1$ matches, i.e. all matches where 1 participates, including matches A and B.

At this point, let’s make all other $n - 1$ players in match A premium. By construction, match A is the only one in which any pair of them is in the same game, which means that each of them is now capable of addressing additional $n$ games (i.e. all games for each player, except game A itself). So far, then, we covered $(n + 1) + (n - 1) \cdot n = n^2 + 1$ matches.

Now let’s consider match B. Each, of course, participates in B itself (addressed by 1), as well as exactly one match with each participant in match A that is not 1, because we already counted all those matches; in other terms, they already participate into $n$ matches that are already addressed by some other player. Out of the $n + 1$, then, one is left out, which means that making them premium will address them, for a total of $n - 1$ additional matches addressed (we don’t count match B of course).

So, we covered $n^2 + 1 + n - 1 = n^2 + n = n \cdot (n + 1)$ matches, i.e. all of them.

Summarizing, how many players should be made premium? There are $n$ players in match A and $n$ in match B, with 1 shared between them, so the number we are after is $2n - 1$. I suspect that this is also the minimum number of players, but I’m not seeing the proof at the moment.

Here’s a table of needed premium players for each case:

$n$ $n^2$ $2n - 1$
2 4 3
3 9 5
4 16 7
5 25 9
7 49 11
8 64 13
9 81 15

# Conclusions

Arranging tournaments for premium games is definitely doable. Although generally less open than individual games, where only one premium user is strictly needed, they are reasonably open anyway, even more so as the number of players per game increases.

If you want to take a look at all posts, here’s the list: