Coalitions in 3Person Games
Mixed strategies in zerosum 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 elimnatedthey are also dominated.
Being zerosum, 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%.
3PERSON MORA
3PERSON 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.
...
3PERSON 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 coalitionwhich weakly dominate each other since they
yield the same payoffs for all movesof the single opponentthere 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 3PERSON ROCKSCISSORSPAPER,
this game gives an advantage to the coalition.
Investigate what happens in general symmetric 3person
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?
...
The zerosum 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 
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