Jul 24, 2017

# How would you expect optimal play to change?

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# Game Theory

INSTRUCTIONS:

Finding the mixed-equilibrium strategies for zero-sum games can yield results that seem perplexing at first. In the children’s game Rock-Paper-Scissors the optimal mixed strategy is for each player to randomly choose each option one-third of the time. But suppose that when rock beats scissors, the winning player scores 2 points, not 1. How would you expect optimal play to change? Interestingly, the players play rock less, not more. Paper is played one-half of the time, and rock and scissors both drop to one-fourth. Why? Create a payoff matrix and use it to verify that the strategies specified result in a mixed-strategy equilibrium for the modified Rock-Scissors- Paper game. Explain the reasons why rock is played less, not more, using game theory concepts. 2. Consider the game of two people approaching one another on a sidewalk. Each chooses right or left. If they make the same choice, they pass one another without a problem and each gets a payoff of 1. If they make opposite choices, they both get payoffs of 0. Find the three Nash equilibria of the game. (One of them is a mixed equilibrium.) Show that the payoff from the mixed equilibrium is only half as good for either player as either of the two pure equilibria.

CONTENT:
Game TheoryStudent:Professor:Course title:Game TheoryIn the Rock-Paper-Scissors game, optimal mixed-strategy is for every player to randomly select each option one-third of the time. However, supposing that when Rock beats Scissors, the winning player scores 2 points, not 1. Optimal play changes. The players Rock less, not more. Paper is played one-half of the time, and Rock and Scissors both drop to one-fourth. In a mixed-strategy, the game is played with probabilities 50%, 25%, or 25% respectively. Adding the aforementioned mixed-strategy – 50% Paper, 25% Rock, 25% Scissors – as an alternative for the player 1, then the anticipated payoffs for the first player against the second player’s pure strategies Paper, Rock, Scissors are 0, 0.25, –0.25 respectively. PaperRockScissorsPaper01-1Rock-101Scissors1-1050-25-25 mix00.25-0.25If player 1, Marylyn, plays the mixed strategy and Monroe plays Paper, then with 50 percent likelihood there would be a tie Paper vs. Paper, with 25 percent chance Monroe would win – her Paper against Marylyn’s Rock –, and with 25 percent chance Marylyn would show Scissors and win against Monroe’s Paper. As such, the anticipated payoff for Marylyn when she plays this mixed strategy against Monroe’s Paper will equal 50%·0 + 25%·(-1) + 25%·1 = 0. Therefore, the values within the 4th row are anticipated values of the matching values within the other rows and similar column, using the probabilities of the mi...

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