This is a positive-sum game because at its end one player wins all, while the other loses everything (including his life). In a zero-sum game nobody wins. Most readers will be familiar with that term from its use in reference to nuclear war. In positive sum games somebody wins. Most games are of this type: soccer, football, basketball, chess, etc.   We can see it is a positive-sum game (sometimes called a win-win game) when we remember that the two shooters are not the only players.  In fact, they are proxy players for the investors providing the money to bet on this game.  When one player wins, his backer does also, and a win-win outcome obtains for this team of players.  We can illustrate this with an example from the Tzameti 13.

 

Consider the payoff matrix T of the players Number 13 (X₁₎ and Number 6 (Y₁₎ pay-off outcomes below:

                   Y₁                                               Y₂

     X₁        850,000 Fr                        -850,000 Fr

     X₂      -850,000 Fr                       850,000 Fr  

 

 

In this example, X and Y have ½ probabilities for each of the following payoffs:

 

Row) X₁Y₁=850,000 Fr

Column) X₁Y₂=-850,000 Fr

Row) X₂Y₁=-850,000 Fr

Column) X₂Y₂=850,000 Fr.

 

The expected average payoff probability will be

 

E(P)=1/2(X₁Y₁)+ 1/2(X₁Y₂) + 1/2(X₂Y₁) + 1/2(X₂Y₂) =0

 

As shown above if this game could be played repeatedly, then the outcome probabilities would sum to zero.  But, since this game is played once, the choice of the players (to fire first or not) doesn’t matter.  Whoever has the good fortune to fire and kill the other will win.  Waiting for the other player to fire or firing first doesn’t help, since the probability is ½.  This game can never be a zero-sum game, and must always be a positive-sum game regardless of each player’s strategy.  In the case where both players fire and kill each other it’s a draw as previously stated.

 

In the case where the game is played only once then we must reformulate the payoff matrix to be:

 

                   Y₁                                               Y₂

     X₁             1                                     0

= 850,000 Fr

     X₂                0                                  1

 

 

This matrix resolves to 2 solutions:

 

1)      X₁Y₁ (1) + 0 = 850,000 Fr Row wins (number 13) and Column loses (number 6 dies)

2)       X₂Y₁ (0)+ (1 *X₂Y₂) = 850,000 Fr Column wins (number 6) and Row loses (number 13 dies)

Return to 13 Tzameti

9/6/09 Ken Wais