Why Tzameti 13 Is A Zero Sum Game.
This is a zero-sum game because at its end one player wins all, while the other loses everything (including his life). In a zero-sum game one party wins and the other loses. In a zero-sum game if we add together the expected probabilities of the opponents they sum to zero, thus the name.
In zero-sum games, somebody wins. Most games are of this type: soccer, football, basketball, chess, etc. One party wins everything and the other loses everything. There is no partial win in chess or you don't half win a football game now do you? There is a tie, in which nobody wins, but that is not zero-sum in the game theory sense. It is more akin to positive sum. If one team has 1 and another has 1 their sum would be 2, and a positive sum would be obtained thus no winner could be declared at the conclusion. We are not considering that case. Also, remember that the two shooters are not the only players. 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. This is much like the investment model in which investors place money in a trading market (like the stock market) and providers of access to the market, profit by providing access to the investors with fees charged. But, I digress, let's have an illustration with an example from Tzameti 13.
Consider the payoff matrix T of the players, Number 13 (X1) and Number 6 (Y1) pay-off outcomes below:
Y1 Y2
T= X1 │ 850,000 Fr -850,000 Fr│
X2 │ -850,000 Fr850,000 Fr │
In this example, X and Y have 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 E(P) 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. This game is played until there is a winner and this means until a player dies, so 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 the same. This game must always be a zero-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. This can be a zero-sum for the players' lives but not for the proxy betting players. That is to say, both players can kill each other and that would be zero-sum for them. The possibility of each chambering a round with no bullet and firing will only restart the game. If they both kill each presumably the game restarts with the last living game players. Remember there can be no draw.
In the case where the game is played until the death of one player, 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)
Nash Equilibrium Strategy
Nash equilibrium is an algorithm that quantifies opposing player strategies with the following rule:
Given two or more players in a game, their individual strategies could not be unilaterally changed and yield an advantage. It is named after the famous game theorist John Nash, that first described the idea. It is alternately described as no individual player in a game having an incentive to change his/her strategies to yield a positive outcome. It means that if we have a payoff matrix for any number of players, one combination is best compared to all others. That is, it yields an advantage compared to all others. Let's revisit the above matrix and apply Nash Equilibrium. The elements of the Nash Equilibrium matrix will be a 3X3 matrix. By the way, this strategy can apply to more than human players in a game, as I'm sure any reader realizes.
Elements are:
Row/Column X1Y1 =[1,0]
Row/Column X2Y1 =[0,1]
Row/Column X3Y3 =[0,0]
And the matrix is:
Y1 Y2 Y3
X1 │1 00 │
X2 │0 1 0 │
X3 │0 0 0 │
Here we have the first row/column combination X1Y1, X2Y1, X3Y3 is the only winning strategy for row is X1Y1. The only winning strategy for column is X2Y2. All other combinations result in no wins. Thus X1Y1 =[1,0] is a Nash Equilibrium for Row (number 6) and X2Y1 =[0,1] is a Nash Equilibrium for Column (number 13). What this means is that for any choice of other than the 2 described above when one opponent chooses a move, the other opponent can't do better than choosing the option given. Whenever Xx is 1 then the opponent Yx must be 0, i.e., he dies. And vice versa whenever Yx is 1, then Xx is 0. Thus, the optimal strategy is as described above.
Of course, the problem with applying Nash Equilibrium to this game is even though the players have this knowledge they had no choice in the matter. They both knew their guns were loaded with 5 of 6 bullets and they're chances were 50%, but couldn't opt out of playing the duelist round. Thus, this movie really doesn't portray true Nash Equilibrium. But, the outcome estimates above show the possible Nash Equilibrium.
The key point to take away from this analysis is there is no draw in the duelist portion of Tzameti 13, somebody has to die and if both die, the monstrous game would have to be restarted by its own rules! Not a game many would want to play, huh?
9/6/09 Robleh Wais: Updated 4/6/14