math bork 2nd try

Ill pretend I understand all of that

In the end we have the following types of systems (apart from probability system, TrueSkill and ANN):

Team Elo System given by a function g to transform to playerstrength and h for the team size dependency:

 playerstrength = g(elo)
 teamstrength = (sum of team's playerstrength)*h(n)
 team elo = f(teamstrength) 

n is the number of players in a team (can be different for any team) and N the number of teams. Win probabilities are calculated with my team FFA generalization of the elo system.

The current system uses g(elo)=elo and h(n)=1/n (but a wrong FFA calculation). Systems that I developed earlier use h(n)=1/sqrt(n) or h(n)=0.5+0.5^n. Sprung 's system uses g(elo)=B^(elo-eloShift) and h(n)=1/n.

Teamstrength System given by the same but without the calculation of team elo. Win probabilities are distributed proportional to teamstrengths. Shifts in the elo scale should not change the result. From this invariance g(elo)=B^(elo-eloShift) can already be concluded and from this follows the equivalence to the team elo system for N=2 but not for more teams. This is still true for any h.

So g(elo)=B^(elo-eloShift) with h(n)=1/sqrt(n) should be an interesting system. I don't know which of both probability calculations should be used for N>2.

Your argumentation would be correct in a normal strategy game, where teamstrength is the sum of playerstrengths. In ZK however there are extra coms. Thus teamstrength is the average of playerstrengths. 51.5% for team 3 is for no cooperation between team 1 and 2. If they cooperate, team 3's win chance will be higher (68.0%) because of extra coms.
For h=1/(number of the teams' coms) all cases are considered here.

[Spoiler]
Furthermore you could argue that probabilities should be proportional to squares of teamstrengths to better reflect team cooperations in FFA. But in order to leave at least 1v1 unaffected, you need to apply sqrt first :

 playerstrength = B^((elo-eloShift)/2)
 teamstrength = (sum of team's playerstrength)/n
 team elo = log_B(teamstrength^2)+eloShift 

Instead of squares you can use a more general convex function phi (a strictly monotonic increasing diffeomorphism) with inversion theta, which is concave:

 playerstrength = theta(g(elo))
 teamstrength = (sum of team's playerstrength)*h(n)
 team elo = f(phi(teamstrength))

Then you can calculate probabilities either with team elo or proportional to phi(teamstrength). As g is convex and theta concave, the better rating of teams with higher elo deviation is partially compensated.

Yes indeed. Still averages are better than sums here. So in a 2v3, the players in the smaller team would be weighted 1/2 and in the bigger one 1/3 times. Generally the players of a team with n players could be weighted v*(1/n-1)+1 times, where v=0 would be the sum and v=1 the average. So what value do you suggest for the "extra com advantage" v? My evaluation of game outcomes with current balancer (v=1) has shown that the bigger team wins exactly as often as the smaller one, but maybe that's only because of self-balancing prophecy.

In earlier theories I also considered that extra com players have a higher influence in their team with an "extra com weighting" w (from 0 to 1), so that a player's weighting is 1+w*(number of extra coms). But since coms' incomes are distributed equally, a smaller w is appropriate now. (Initial BP is still unequal.) Generally you could just multiply both weightings.

quote: This literally gives you a 1/3 chance to lose against two Null AIs!

That's a very good observation and quite solid proof that the proposed formula doesn't work.

The main problem is that any team balancing algorithm needs to make assumptions about how team composition affects team strength/elo. This is basically the choice of f and g in the OP, although it would be easier to define it as a function p(elo vector of team) that produces team strength/elo, which is essentially a norm, like the p-norm Brackman mentioned (see again the OP, or glimpse over this <- just look at pictures and formulas).

Your example with the teamstrength change looks correct (although you did mess up the signs for the expected outcomes), but the question still stands how to map/combine individual strength/elo to team strength/elo.

Some observations:
- Team strength (as used by you) must be in the interval (0; ∞]. It cannot be < 0 (negative probability?), but it can be arbitrarily large, because for every team with strength p1, one can imagine a stronger team that will beat the first one 2 out of 3 times and thus must have a strength p2 = 2*p1. It could theoretically be zero, but not in practical applications.

- According to Brackman p needs to (essentially) satisfy the triangle inequality to fit the known data: Adding a "variation" elo vector (like [-300; 100; 200]) to a team of equally strong players (like [1500, 1500, 1500]) increases their strength (so p([1200; 1600; 1700]) > p([1500; 1500; 1500])). (Note that this is not inherent to the math used! It would be possible that serious anti-synergies in teams with high elo variance would cause the opposite).

- If all elo values increased by the same flat amount, the ratios of the team strengths would need to stay the same. This pretty much implies an exponential approach (which conveniently also satisfies the first point).

- It is still unclear how different team sizes (= vector lengths) should affect p. Is one player with 5 coms stronger than 5 players with one com each (all of equal elo)? I reckon this largely depends on the players, and it's conceivable that for high numbers of extra coms, multiple players get an edge through micro advantage, but it's hard to judge how the force concentration (and increased coordination) balance that out in the general case.

- All of this has many parallels to the 1v1 probabilities, because that's essentially just a special case of the above.

I'd handle the uneven teams case by not giving an extra comm and just equalizing the strength. Extra comm is a bad way of handling inequal teams (some person suddenly has a different start state without announcement, getting twice the BP but not income to compensate, and no ability to place them separately). It's not a large issue because the matchmaker won't be allowed to create such games anyway.

I'm sorry, but I still don't quite get where the sqrt(average size of all teams) is coming from. I can see why one would want to use some adjustment to adapt to elo-imbalances in games with high player numbers more, but why this exact term?

Just to gain an understanding:
For two teams of 4 players, that would yield an exponent of 2. Using exponentiation laws, this effectively means that the "simple" team elo (I mean the compounded strength without the sqrt() exponent, transformed back into elo via the normal formula) would be doubled for each team before arriving at the actual team strength, so it magnifies the elo differences between the teams.

In effect, your formula suggests that (for non-1v1) a higher elo/strength team is underpredicted (and a lower one is overpredicted) by the "simple" team strength calculation. Did you base this on your data examinations?

quote: It is based on the assumption that probability predictions should be more distinct with higher player numbers due to the law of large numbers.

I'd like to point out that h(n) = 1 also does that :P