# Coalitions in 3-Person Games

Mixed strategies in zero-sum games, found by linear programming. Material is an excel sheet Mora.xls.

Two players simultaneously show one or two or three fingers and call a number between two and six. The number closer to the number of fingers shown by both players together wins. This game is played in Italy and the call "cinque a la mora" used there caused the word "Tschinggalamorä" used in Switzerland for Italians.

Here is the normal form of the game for the different moves. The obviously wrong moves (1,5), (1,6), (2,2), (2,6), (3,2), and (3,3) have already been elimnated---they are also dominated. Being zero-sum, only the payoff for player A is shown.
 (1,2) (1,3) (1,4) (2,3) (2,4) (2,5) (3,4) (3,5) (3,6) (1,2) 0 1 1 -1 0 1 -1 -1 0 (1,3) -1 0 1 0 1 1 -1 0 1 (1,4) -1 -1 0 -1 0 1 0 1 1 (2,3) 1 0 1 0 -1 0 -1 -1 -1 (2,4) 0 -1 0 1 0 1 0 -1 0 (2,5) -1 -1 -1 0 -1 0 1 0 1 (3,4) 1 1 0 1 0 -1 0 -1 -1 (3,5) 1 0 -1 1 1 0 1 0 -1 (3,6) 0 -1 -1 1 0 -1 1 1 0

It is interesting to note that none of the moves dominates another. The solution is: Players should play move (1,2) in 50% of the cases and move (3,6) in the other 50%.

## 3-PERSON MORA

3-PERSON MORA Three players put \$1 each on the desk. Then they simultaneously show one or two fingers and call a number between three and six. The number closer to the sum of the number of fingers shown by the players together wins the money on the desk. In case of a tie, they share.

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## 3-PERSON MORA with COALITION

Let's assume the coalition plays on the rows, and the single player on the columns. Then the payoff matrix looks like this, where symmetric cases (playing (1,4)(1,3) yields obviously the same as (1,3)(1,4)) are already eliminated:
 (1,3) (1,4) (1,5) (1,6) (2,3) (2,4) (2,5) (2,6) (1,3),(1,3) 0 1 1 1 0 -2 0 1 (1,3),(1,4) -0.5 1 1 1 1 -0.5 1 1 (1,3),(1,5) -0.5 1 1 1 0 -2 0 1 (1,3),(1,6) -0.5 1 1 1 -0.5 -2 -0.5 1 (1,3),(2,3) 0 -2 0 1 0 -2 -2 -2 (1,3),(2,4) 1 -0.5 1 1 1 -0.5 -2 -0.5 (1,3),(2,5) 0 -2 0 1 1 1 -0.5 1 (1,3),(2,6) -0.5 -2 -0.5 1 1 -0.5 -2 -0.5 (1,4),(1,4) -2 0 1 1 1 0 1 1 (1,4),(1,5) -2 -0.5 1 1 1 -0.5 1 1 (1,4),(1,6) -2 -0.5 1 1 1 -0.5 1 1 (1,4),(2,3) 1 -0.5 1 1 1 -0.5 -2 -0.5 (1,4),(2,4) 1 0 1 1 1 0 -2 0 (1,4),(2,5) 1 -0.5 1 1 1 1 -0.5 1 (1,4),(2,6) 1 -0.5 1 1 1 0 -2 0 (1,5),(1,5) -2 -2 0 1 0 -2 0 1 (1,5),(1,6) -2 -2 -0.5 1 -0.5 -2 -0.5 1 (1,5),(2,3) 0 -2 0 1 1 1 -0.5 1 (1,5),(2,4) 1 -0.5 1 1 1 1 -0.5 1 (1,5),(2,5) 0 -2 0 1 1 1 0 1 (1,5),(2,6) -0.5 -2 -0.5 1 1 1 -0.5 1 (1,6),(1,6) -2 -2 -2 0 -2 -2 -2 0 (1,6),(2,3) -0.5 -2 -0.5 1 1 -0.5 -2 -0.5 (1,6),(2,4) 1 -0.5 1 1 1 0 -2 0 (1,6),(2,5) -0.5 -2 -0.5 1 1 1 -0.5 1 (1,6),(2,6) -2 -2 -2 0 1 0 -2 0 (2,3),(2,3) 0 -2 -2 -2 0 -2 -2 -2 (2,3),(2,4) 1 -0.5 -2 -0.5 1 -0.5 -2 -2 (2,3),(2,5) 1 1 -0.5 1 1 1 -0.5 -2/TD> (2,3),(2,6) 1 -0.5 -2 -0.5 1 1 1 -0.5 (2,4),(2,4) 1 0 -2 0 1 0 -2 -2 (2,4),(2,5) 1 1 -0.5 1 1 1 -0.5 -2 (2,4),(2,6) 1 0 -2 0 1 1 1 -0.5 (2,5),(2,5) 1 1 0 1 1 1 0 -2 (2,5),(2,6) 1 1 -0.5 1 1 1 1 -0.5 (2,6),(2,6) 1 0 -2 0 1 1 1 0

After deleting weakly dominated moves for the coalition, we arrive at the following matrix:
 (1,3) (1,4) (1,5) (1,6) (2,3) (2,4) (2,5) (2,6) (1,3),(1,3) 0 1 1 1 0 -2 0 1 (1,3),(1,4) -0.5 1 1 1 1 -0.5 1 1 (1,4),(1,4) -2 0 1 1 1 0 1 1 (1,4),(2,4) 1 0 1 1 1 0 -2 0 (1,4),(2,5) 1 -0.5 1 1 1 1 -0.5 1 (1,5),(2,4) 1 -0.5 1 1 1 1 -0.5 1 (1,5),(2,5) 0 -2 0 1 1 1 0 1 (2,5),(2,5) 1 1 0 1 1 1 0 -2 (2,5),(2,6) 1 1 -0.5 1 1 1 1 -0.5 (2,6),(2,6) 1 0 -2 0 1 1 1 0

Now the moves (1,6) and (2,3) for the isolated player are dominated. This is also obvious, since a sum of 6 cannot be achieved if the player shows 1 finger, and a sum of 3 is impossible with two fingers.

 (1,3) (1,4) (1,5) (2,4) (2,5) (2,6) (1,3),(1,3) 0 1 1 -2 0 1 (1,3),(1,4) -0.5 1 1 -0.5 1 1 (1,4),(1,4) -2 0 1 0 1 1 (1,4),(2,4) 1 0 1 0 -2 0 (1,4),(2,5) 1 -0.5 1 1 -0.5 1 (1,5),(2,4) 1 -0.5 1 1 -0.5 1 (1,5),(2,5) 0 -2 0 1 0 1 (2,5),(2,5) 1 1 0 1 0 -2 (2,5),(2,6) 1 1 -0.5 1 1 -0.5 (2,6),(2,6) 1 0 -2 1 1 0

Except the moves (1,4)(2,5) or (1,5)(2,4) of the coalition---which weakly dominate each other since they yield the same payoffs for all movesof the single opponent---there is no domination left. The solution is: The coalition plays the pair (1,3)(1,4) 1/3 of the time, either (1,4)(2,5) or (1,5)(2,4) in 1/3 of the cases, and (2,5)(2,6) in the remaining 1/3 of the cases. The expected payoff for the coalition is 1/2. Different to 3-PERSON ROCK-SCISSORS-PAPER, this game gives an advantage to the coalition.

Investigate what happens in general symmetric 3-person games, where every player has just two moves. Is a coalition always worthwhile? And if it is, how much would the expected payoffs for the players in the coalition improve?
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### The zero-sum case

 Cindy chooses 1 Beth chooses 1 Beth chooses 2 Ann chooses 1 0,0,0 r, -2r, r Ann chooses 2 -2r, r, r v, v, -2v
 Cindy chooses 2 Beth chooses 1 Beth chooses 2 Ann chooses 1 r, r, -2r -2v, v, v Ann chooses 2 v, -2v, v 0, 0, 0

 Coalition of Beth and Cindy chooses (1,1) (1,2) (2,2) Ann chooses 1 0, 0 r, -r -2v, 2v Ann chooses 2 -2r, 2r v,-v 0, 0

### The general case

 Cindy chooses 1 Beth chooses 1 Beth chooses 2 Ann chooses 1 q, q, q r, s, r Ann chooses 2 s, r, r v, v, u
 Cindy chooses 2 Beth chooses 1 Beth chooses 2 Ann chooses 1 r, r, s u, v, v Ann chooses 2 v, u, v t, t, t

 Coalition of Beth and Cindy chooses (1,1) (1,2) (2,2) Ann chooses 1 q, 2q r, r+s u, 2v Ann chooses 2 s, 2r v,u+v t, 2t