# Game Theory in Tennis–Summer

Today, I’m going to explore with you, through the lens of Game Theories, strategies within tennis.

As we know, in a tennis match, players are trying to score points against each other by hitting the ball into a corner of the opponent’s court where s/he is not able to hit it back.

Let’s focus our attention on a pass between two tennis players. The server, player 1,  has two choices: passing towards the receiver’s left (backhand side) or right (forehand side, assuming right-handed). The receiver at net, player 2, also has two choices: leaning left or right to catch the ball. Because each player is differently skilled at backhand and forehand and player’s satisfaction depends on his possibility of scoring/preventing his opponent from scoring, we have the following payoff matrix:

 Player 1(left)/2 (top) Left Right Left 50,50 80,20 Right 90,10 20,80

In this matrix, we cannot see any strictly/weakly dominated strategy or traditional Nash Equilibrium. In fact, this observation coincides with our real world experience: a tennis match is never good if the players always lean to one specific side!

In this tennis game, however, Nash Equilibrium still exists, but in such a Nash Equilibrium both players are playing, what game theory calls, Mixed Strategies. A mixed strategy is a randomization of available strategies that assign each strategy (left/right) a possibility p such that 0‹p‹1 and all p add up to 1. For example, a possible mixed strategy for player 1 is hitting left 20% of the time and hitting right 80% of the time.

The payoff of a mixed strategy is logically given by the sum of the probabilities of individual strategies time their respective expected payoffs against the opponent’s strategy. For instance, if player 1 plays the mixed strategy (0.2, 0.8) and player 2 plays the mixed strategy (0.7,0.3), then the payoff of player 1’s mixed strategy is:

0.2×(0.7×50+0.3×80)+0.8×(0.7×90+0.3×20)=67

So, given all these definitions, how do we find the mixed-strategy equilibrium of this game? We have to infer a step further.

If the two players are at Nash Equilibrium, then both of them are playing their best responses to the other’s strategy. If a mixed strategy is a best response, then individual strategies within the mix must also all be best responses, or else we can create a better response by taking the strategy with a lower payoff out of the mix. Thus, the only way both hitting left and right are best responses is that their expected payoff  are equal.

Let’s notate player 1’s equilibrium mixed strategy as (p, 1-p) and player 2’s equilibrium mixed strategy as (q, 1-q). Given our inference, if we want to calculate player 1’s equilibrium mix, we can look at player 2’s expected payoffs. Because player 2 is playing a mixed-strategy best response, then as we previously concluded, his expected payoff of leaning right and left should be equal. Thus, we have:

Left   50p+10(1-p)=20p+80(1-p)   Right

We can solve this equation: p=0.7, 1-p=0.3. Thus, player 1’s mix strategy at Nash Equilibrium is (0.7,0.3).

Similarly, we can solve for player 2’s Nash Equilibrium mixed strategy with player 1’s payoff.

50q+80(1-q)=90q+20(1-q)      q= 0.6, 1-q=0.4

Thus, the game has a mixed strategy Nash Equilibrium at {(0.7,0.3), (0.6,0.4)}.

This model draws heavily on the concepts I’ve introduced in my previous posts and has slightly more calculations and variables than my previous examples. Yet, despite its being slightly more complex, it is by far the most accurate representation of the choices we face in sports and in life.

See you next week!

## 1 thought on “Game Theory in Tennis–Summer”

1. Susan Waterhouse

Ahh, so now we get to the more realistic situations… of course, in math we often need to simplify situations first, build a model and then ask how can we improve. What are the challenges with the improved mixed strategy? Does it need further refinement? Are there good algorithms for applying this method to situations with more options beyond 2 x 2?

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