This week, I shall recommence our discussion on Nash Equilibrium and explore the insights of a popular game called “Rock-Paper-Scissors”.
“Rock paper scissors match.”
“Alright! Rock beats paper!”
“I thought paper covered rock.”
“Nah, rock flies right through paper.”
“Well, what beats rock?”
“Nothing beats rock.”
– Kramer and Mickey in Seinfeld, Ep: The Stand-In
This week, I shall recommence our discussion on Nash Equilibrium and explore the insights of a popular game called “Rock-Paper-Scissors”. What do we know about this game? Consider one Saturday night where you and your friend cannot see eye to eye on what movie to watch. Rather than dragging out the argument where both of you defend your choices until one caves, you decide to settle this dispute by leaving the decision entirely up to Fate. What do you do next? You either choose to toss a coin, or roll a die, OR, you can play this game with your friend.
How does one play this game?
Both of you have to simultaneously gesture your choice by forming one of the three shapes with your hand (see above), and the winner is decided based on the following rule; “Scissors cut Paper, Paper covers Rock, and Rock crushes Scissors”.
Remember how I rewrote the story behind Prisoner’s dilemma in the form of a payoff matrix? I will draw the exact same matrix for this game as well. Assign payoff of 1 to the player who wins and payoff of -1 to the player who loses. That is to say, if player 1 plays Scissors and player 2 plays Rock, then player 2 wins and gets 1 while player 2 gets -1 (Rock beats Scissors remember?). Further, in case of a tie, both the players get nothing as the dispute remains unresolved. The Payoff matrix below gives a concise picture of this game, illustrating all the possible nine outcomes.
Notice what the payoffs are telling us: that one player’s loss is other player’s gain and vice-versa. These kind of games are known as Zero-sum games and are used to explain the distributional gains or losses. Consider a game where a pie is shared between two players. It can be seen that the game has zero-sum properties because if one player takes a larger piece, there is less left for the other.
Let us ask ourselves the following question: Is there is strategy in this game that ensures a win for a player, in all likelihood? In other words, is there is a strategy that player 1 can adopt, that will beat player 2’s strategy, no matter what it is, and of course vice-versa? As per the terminology used in Game theory, what I am asking is if there exists a dominant strategy for each player?
As you can see from the payoff matrix, the answer is NO. If player 2 chooses Rock, the best response for player 1 is to choose Paper. If player 2 chooses Paper, the best response for player 1 is to choose Scissors. And lastly, if player 2 chooses Scissors, then the best response for player 1 is to choose Rock. Therefore, there exists no dominant strategy for either of the players, which will beat all possible strategies of his opponent.
Does this game have a Nash-equilibrium? What is a Nash Equilibrium?
“A pure-strategy Nash equilibrium is a strategy profile with the property that no single player can obtain a higher payoff by deviating unilaterally from this profile.”
This is the formal definition of the equilibrium. How do you understand it? Recall the arguments from the Prisoner’s Dilemma game. I told you that the Nash Equilibrium in that game is (Confess, Confess) with payoffs (-10, -10). What it means is precisely this: It is not possible for either of the prisoners to earn a higher payoff by unilaterally deviating to ‘Not Confess’.
If one of them deviates and chooses to remain silent instead while the other prisoner follows his equilibrium strategy to confess, it would result in a prison sentence of 30 years instead of 10 for the deviating prisoner. And this is exactly what we have shown in the previous blog by illustrating that confessing to the crime is the dominant strategy for both the players.
Coming back to our current game, we have established that Rock-Paper-Scissors does not have a dominant strategy for either of the players. How do you use that information to incur that there is no Nash-equilibrium? Quite simple! If Player 2’s strategy is Rock, Player 1 should choose Paper, but if Player 1 chooses Paper, it is profitable for Player 2 to deviate and choose Scissors instead. When player 2 chooses Scissors, Player 1 would want to deviate and choose Rock, and so forth. Thus, we can see that there is no Nash Equilibrium for this game owing to the cyclical manner in which strategies triumph over each other.
Let us discuss the version of this game played by Kramer and Mickey, of the popular show, Seinfeld. Mickey alters the rules of the game by making Kramer believe that Rock “flies through” Paper, beating it. After this discourse, both of them switch to playing rock and no matter how many rounds they repeat the game, it would only end in a tie! I have drawn the payoff matrix for this version as the following;
In this version of the game, the entries in two cells are now altered. The cell in the first row and second column is altered to read as (1, -1) since Rock beats Paper. Similarly, the cell in the second row and first column is altered to read as (-1, 1). Can you see how playing Rock becomes the dominant strategy in this game and (Rock, Rock) becomes the Nash Equilibrium? Check it yourself to see that for neither of the players, a profitable deviation from the top-left cell is possible. Therefore, both the players will always end up in a tie! Further, this game has no use in resolving a dispute as it no longer captures the role of Fate in deciding the winner.
There are some straightforward extensions of this game where both the players play for multiple rounds (mostly three or five) and the player who has won the maximum rounds is declared the winner. Then, there are some not so straightforward extensions. In The Big Bang Theory, Sheldon and Raj are at a disagreement over what show to watch on TV. To resolve this, Sheldon proposes that they play an alternative version he calls as “Rock-Paper-Scissors-Lizard-Spock”.
The rules of the game are:
“Scissors cuts Paper, Paper covers Rock, Rock crushes Lizard, Lizard poisons Spock, Spock smashes Scissors, Scissors decapitates Lizard, Lizard eats Paper, Paper disproves Spock, Spock vaporizes Rock, Rock crushes Scissors”
I draw the payoff matrix of this game as the following:
Again this game has no Nash Equilibrium. The Rock-Paper-Scissors interplay remains the same as the classical game. Only two more alternative actions have been added, that of Lizard and Spock. The link established using them is again cyclical in nature allowing no strategy to dominate the others. This extended version manages to preserve the randomness of the outcome of the game and keeps it as a game of chance.
[This is just a disclaimer for those who are familiar with some concepts in Game Theory. Don’t get confused when I write that the Nash Equilibrium of the game does not exist. I am just referring to Pure-strategy Nash Equilibrium when I say that. The mixed strategy Nash Eqilibrium of Rock-Paper-Scissors game exists and is to play all three strategies with equal probability of 1/3. Anymore discussion than this is beyond the scope of this entry.]