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Third stage: host opens a door. Fourth stage: player makes a final choice. The player wants to win the car, the TV station wants to keep it. This is a zero-sum two-person game. By von Neumann's theorem from game theory, if we allow both parties fully randomized strategies there exists a minimax solution or Nash equilibrium. [9]
Several variants are considered in Game Theory Evolving by Herbert Gintis. [2] In some variants of the problem, the players are allowed to communicate before deciding to go to the bar. However, they are not required to tell the truth. Named after a bar in Santa Fe, New Mexico, the problem was created in 1994 by W. Brian Arthur.
The corresponding asymmetric rendezvous problem has a simple optimal solution: one player stays put and the other player visits a random permutation of the locations. As well as being problems of theoretical interest, rendezvous problems include real-world problems with applications in the fields of synchronization , operating system design ...
In game theory, a solution concept is a formal rule for predicting how a game will be played. These predictions are called "solutions", and describe which strategies will be adopted by players and, therefore, the result of the game. The most commonly used solution concepts are equilibrium concepts, most famously Nash equilibrium.
Separately, game theory has played a role in online algorithms; in particular, the k-server problem, which has in the past been referred to as games with moving costs and request-answer games. [124] Yao's principle is a game-theoretic technique for proving lower bounds on the computational complexity of randomized algorithms , especially online ...
Constant sum: A game is a constant sum game if the sum of the payoffs to every player are the same for every single set of strategies. In these games, one player gains if and only if another player loses. A constant sum game can be converted into a zero sum game by subtracting a fixed value from all payoffs, leaving their relative order unchanged.
A simple solution dishes out one gold to the odd or even pirates up to 2G depending whether M is an even or odd power of 2. Another way to see this is to realize that every pirate M will have the vote of all the pirates from M/2 + 1 to M out of self preservation since their survival is secured only with the survival of the pirate M.
The free money game is an example of a "special" game with an even number of equilibria. In it, two players have to both vote "yes" rather than "no" to get a reward and the votes are simultaneous. There are two pure-strategy Nash equilibria, (yes, yes) and (no, no), and no mixed strategy equilibria, because the strategy "yes" weakly dominates "no".