# A majority coalition can lose a symmetric zero-sum game

Just a neat little result I found when thinking about Jessica's [recent post](https://agentfoundations.org/item?id=1224).

For n players, let a~i~ be the action of player i and s~i~(a~1~,a~2~,...a~n~) the reward of player i as a function of the actions of all players. Then the game is symmetric if for any permutation p:{1,...n}→{1,...n}

s~i~(a~1~,a~2~,...a~n~) = s~p(i)~(a~p(1)~,a~p(2)~,...a~p(n)~)

The game is zero-sum if the sum of the s~i~ is always zero. Assume players can confer before choosing their actions.

Then it is possible for a majority coalition to strictly lose a zero-sum game, even in a deterministic game where they get to see their opponents' moves before choosing their own.

This seems counter-intuitive. After all, if one coalition has M players, the other has m players, with m<M, and there are no other players, how can the M players lose? Couldn't just m of the M players behave exactly as the smaller coalition, thus getting the same amount in expectation? The problem is the potential loses endured by the remaining M-m players.

For an example, consider the following 5 player colour game (it's a much simplified version of the game I came up with previously, [proposed](/lw/oj4/a_majority_coalition_can_lose_a_symmetric_zerosum/dlwc) by [cousin_it](/user/cousin_it/overview/)). Each player chooses one of two colour, blue or red. Then the players that selected the least commonly chosen colour(s) are the winners; the others are the losers. The losers pay 1 each, and this is split equally between the winners.

Then consider a coalition of three players, the triumvirate. The remaining two players - the duumvirate - choose different colours, red and blue. What can the triumvirate then do? If they all chose the same colour - say blue - then they all lose -1, and the duumvriate loses -1 (from its member that chose blue) and gains 4/1 (for the member that chose red). If they split - say 2 blue, 1 red - then the ones that chose blue lose -1, while the duumveriate loses -1 (from its member that chose blue) and gains 3/2 > 1 (from the member that chose red).

So the duumvriate can always win against the triumvirate.

Of course, it's possible for two members of the triumvirate to create a second duumvirate that will profit from the hapless third member. Feel free to add whatever political metaphor you think this fits.

Larger games
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Variations of this game can make [Jessica's theorem 2](https://agentfoundations.org/item?id=1224) sharp. Let the minority coalition be of size m (the majority coalition is of size M = n-m = qm+r for some unique q and 0≤r<m). The actions are choosing from m colours; apart from that the game is the same as before. And, as before, the members of the minority coalition each choose a different colour.

Then an m-set is a collection of players that each chose a different colour. Split the players into as many disjoint m-sets as possible, with the minority coalition being one of them - say this gives q'+1 m-sets. There are r' remaining players from the majority coalition.

Note that we can consider that any loss from a member of an m-set is spread among the remaining members. That's because all winners are members of m-sets, and players that choose the same colour are interchangeable. So we can assign the loss from the member of an m-sets as being an equivalent gain to the other members. Thus the m-sets only profit from the r' remaining players. And this profit is spread equally among the winners - hence equally among the m-sets.

Thus the majority coalition has a loss of r'/(q'+1), and minimises its loss by minimising r' and maximising q' - hence by setting q'=q and r'=r. Under these circumstances, the minority coalition wins r/(q+1) in total.

Adding 1 to the reward of each player, then dividing all rewards by n, gives the unit-sum game in Jessica's theorem.