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Consider the well-known Rock - Paper - Scissors (RPS) game

Economics

Consider the well-known Rock - Paper - Scissors (RPS) game. Imagine that winning the game gives a pay-off of 1, while loosing the game gives a pay-off of -1. Draws have a pay-off of 0. (Rock beats scissors, scissors beats paper, paper beats rock. If both players play the same action, the result is a draw.) a) Is RPS a zero sum or non-zero sum game? b) Write down the normal form of the RPS game and show the pay-offs for each player. c) Does the RPS game have a pure strategy Nash equilibrium? What about mixed strategy Nash equilibriums?

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1) Zero-sum game refers to the situation in a game when loss of a player exactly equals the gain of the other player thereby no welfare change happens post an event. In the game of rock-paper-scissors, either player 1 wins and player 2 loses or it is the other way around. There is no possibility of both the players winning and also if the game is a tie then also there is no change in welfare. Hence, it is an example of zero-sum game.

2) Refer to the page attached.

3) There will be no pure strategy equilibrium in this game because if for instance player 1 adopts the strategy of playing of paper in each turn. Soon the other player will notice the pattern and will only play scissors as a counterstrategy to win the game. Therefore, no equilibrium will be achieved in such a case as for every strategy choosen there will be some counter strategy. It will lead to situation where both players will keep chasing each other.

Therefore, we now move to 'mixed' strategy

Suppose the probability of player 1 rock, paper scissor is p1, p2 and p3 respectively and similarly for player 2 is q1, q2, q3. Now we will find the expected value for each strategy. If player 2 chooses to play rock, then expected payoff for player 1 would be: p1 * 0 + p2 * 1 + p3 * (-1) = p2 - p3 [ where p1,p2,p3 are probabilities and 0,1,-1 are payoff as indicated in the page attached. Similarly for the paper strategy of player 2, expected value for player 1 would be -p1 + p3 and for scissor strategy the expected payoff would be p1 - p2. Now if the player 1 chooses to play rock, then his expected payoff is -q2 + q3, if he plays paper then it is q1 - q3 and in the case of scissors, it is q2 - q1.

It must be noted that player 2 will such strategy that minimizes the expected payoff of player 1 and player 1 will choose such strategy so as to maximize its own expected payoff. Therefore, the problem now becomes a LP formulation because we need to minimize (p2 - p3, -p1+p3,p1-p2) and maximize (-q2+q3, q1-q3, q2-q1) subject to constraint that all prob sum to 1. Solving this will us the result that optimal probabilities are : (1/3,1/3,1/3) for both the players.

Alternative way,

Looking at the symmetry of the game, it is quite easy to guess that each option must be such that every strategy has an equal probability. Therefore, for both the players choosing (1/3,1/3,1/3) is the optimal mixed strategy.

please see the attached file for the complete solution.