Game Theory through Examples

Note to the Teacher: The focus of this chapter is on mixed strategies in zero-sum games. Some version is completely random, where the optimal player will not win against anybody. Another version has two bad moves not occuring in the Nash equilibrium, which may not be obvious to spot. Then the Nash equilibrium player will more likely win against random players and those using these 2 moves. Furthermore this game is also an example for coalitions and cooperation in a 3-players game. Material is an excel sheet Morra.xls.

Morra is a game that traces back to the Romans and is still somewhat popular in Italy. the call "cinque a la mora" used there caused the word "Tschinggalamorä" used in Switzerland for Italians. In the most common version, two players simultaneously both show one to five fingers with one hand, and at the same time shout a number between 2 and 10, their guess of what the sume of the fingers of both players will be. The one guessing right wins a point, except in the case of a draw, where both players guess right, and nobody gets a point. There is also a variant where the winner wins as many points as the correct guess was. We will analyse three smaller versions below.

MORRA(n,=,1) or MORRA(n,=,sum) or MORRA(n,<,1) refer to three versions that are played with a maximum of n fingers allowed to show. The "=,1" version refers to the version as described above. When the "1" is relaced by the word "sum", the payoff is not 1 unit but rather the sum of the number of fingers shown. If the "=" is replaced by "<", it is not necessary to guess the sum exactly, but rather the one whose guess is closer to the sum of the fingers shown wins.

MORRA(3,=,1)

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

The solution of this game is that the player plays every move with equal probability 1/9. Thus, the base version of Morra, at least in the 3-fingers version, is just a game of luck. The random strategy forms a Nash equilibrium with itself. And even worse, the expectation of this optimal strategy against any other mixed strategy is just 0.

MORRA(3,=,sum)

If the winning player wins the amount of money as the sum of fingers shown, calling high might seem to be more profitable. Again the normal form with weakly dominated moves deleted is shown.

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

The Brown procedure needs very long, more than 20000 runs, to close to a mix, like 0%, 5%, 36%, 5%, 24%, 5%, 21%, 4%, 0%. But is this already the solution?

MORRA(3,<,1)

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%. This version obviously radicalizes---...

3-PERSON MORRA

3-PERSON MORRA All three versions can be played also by three players. The one or two winning wins the amount of 1 or the sum of shown fingers from the other players. If they are two winners, they share, meaning each gets 1/2 or half the sum of fingers shown.

Even without analyzing this three-player game, since it is symmetric and zero-sum, we know that the expected payoff for each player is 0. In what followis we will compare this to the expected payoff if two of the players communicate and work together.

3-PERSON MORRA(2,=;,sum) with COALITION

3-PERSON MORRA(2,<,1) 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.

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.