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# Why Winning in Rock-Paper-Scissors Isn’t Everything

Rock-Paper-Scissors works great for deciding who has to take out the garbage. But have you ever noticed what happens when, instead of playing best of three, you just let the game continue round after round? At first, you play a pattern that gives you the upper hand, but then your opponent quickly catches on and turns things in her favor. As strategies evolve, a point is reached where neither side seems to be able to improve any further. Why does that happen?

#### Quanta Magazine

Original story reprinted with permission from Quanta Magazine, an editorially independent publication of the Simons Foundation whose mission is to enhance public understanding of science by covering research developments and trends in mathematics and the physical and life sciences.

In 1950, the mathematician John Nash proved that in any kind of game with a finite number of players and a finite number of options—like Rock-Paper-Scissors—a mix of strategies always exists where no single player can do any better by changing their own strategy alone. The theory behind such stable strategy profiles, which came to be known as “Nash equilibria,” revolutionized the field of game theory, altering the course of economics and changing the way everything from political treaties to network traffic is studied and analyzed. And it earned Nash the Nobel Prize in 1994.

So, what does a Nash equilibrium look like in Rock-Paper-Scissors? Let’s model the situation with you (Player A) and your opponent (Player B) playing the game over and over. Each round, the winner earns a point, the loser loses a point, and ties count as zero.

Now, suppose Player B adopts the (silly) strategy of choosing Paper every turn. After a few rounds of winning, losing, and tying, you are likely to notice the pattern and adopt a winning counterstrategy by choosing Scissors every turn. Let’s call this strategy profile (Scissors, Paper). If every round unfolds as Scissors vs. Paper, you’ll slice your way to a perfect record.

But Player B soon sees the folly in this strategy profile. Observing your reliance on Scissors, she switches to the strategy of always choosing Rock. This strategy profile (Scissors, Rock) starts winning for Player B. But of course, you now switch to Paper. During these stretches of the game, Players A and B are employing what are known as “pure” strategies—a single strategy that is chosen and repeatedly executed.

Clearly, no equilibrium will be achieved here: For any pure strategy, like “always choose Rock,” a counterstrategy can be adopted, like “always choose Paper,” which will force another change in strategy. You and your opponent will forever be chasing each other around the circle of strategies.

But you can also try a “mixed” strategy. Let’s assume that, instead of choosing only one strategy to play, you can randomly choose one of the pure strategies each round. Instead of “always play Rock,” a mixed strategy could be to “play Rock half the time and Scissors the other half.” Nash proved that, when such mixed strategies are allowed, every game like this must have at least one equilibrium point. Let’s find it.

So, what’s a sensible mixed strategy for Rock-Paper-Scissors? A reasonable intuition would be “choose Rock, Paper or Scissors with equal probability,” denoted as (1/3,1/3,1/3). This means Rock, Paper and Scissors are each chosen with probability 1/3. Is this a good strategy?

Well, suppose your opponent’s strategy is “always choose Rock,” a pure strategy that can be represented as (1,0,0). How will the game play out under the strategy profile (1/3,1/3,1/3) for A and (1,0,0) for B?

In order to get better picture of our game, we’ll construct a table that shows the probability of each of the nine possible outcomes every round: Rock for A, Rock for B; Rock for A, Paper for B; and so on. In the chart below, the top row indicates Player B’s choice, and the leftmost column indicates Player A’s choice.