Payoff matrices

How can we find the Nash equilibrium in a two-player, uncooperative, symmetric game? First it helps to have a more efficient way to describe games than the lengthy word explanations I gave in the first post. We can summarise a game through its payoff matrix: a table showing the different pairs of choices the players can make and the "payoffs" (scores) in each case.

Payoff matrices as an efficient way to describe a game

Let's explore payoff matrices, starting by considering Rock Paper Scissors. Here is a possible payoff matrix for the game.

 

	Rock	Paper	Scissors
Rock	0,0	0,1	1,0
Paper	1,0	0,0	0,1
Scissors	0,1	1,0	0,0

The row headings give the action I choose, the column headings give the action you choose, and the entries say how many points I get (before the comma) and how many points you get (after the comma). There's a lot of redundant information in there. Why?

Remember that the game is symmetric, so knowing what I score when we choose (Rock, Paper) is enough to know what you get when we choose (Paper, Rock). This means we only need half of the entries: the bottom left entries can be worked out by swapping the points round from the matching top right entry. For example, the (Scissors, Rock) entry, highlighted in red in the table, is 0,1, and we could work this out by reversing the (Rock, Scissors) entry, highlighted in blue, of 1,0. Alternatively, we could keep all the entries in the table, but only include my score, and work out what your score is from that.

 

Let's take the second of these options. Let's also make another slight change. I've been referring to Rock Paper Scissors as a "zero-sum game" because we can't both win or both lose at the same time: the actual number of points scored isn't so important as the difference between my number of wins and your number of wins. To really be a zero-sum game, we need to record this difference rather than the original scores, which yields the following table.

	Rock	Paper	Scissors
Rock	0	-1	1
Paper	1	0	-1
Scissors	-1	1	0

With this change we can summarise the payoffs even more efficiently. Suppose we only saw part of the table above:

	Rock	Paper	Scissors
Rock	?	-1	1
Paper	??	?	-1
Scissors	??	??	?

For each of the red question marks, the two players have chosen the same option, and necessarily draw because the game is symmetric, hence must both score zero points.

Let's consider the blue double question mark for the (Scissors, Rock) entry, in the bottom left. When I play Rock vs your Scissors, I score 1 point, so when you play Scissors vs my Rock you must likewise score 1 point. Because it's a zero-sum game, that means in the latter scenario I must score -1 point, and we can fill this in the bottom left corner.

We can work out the other double question mark entries similarly, so we've been able to work out the full table from only the entries in the top right corner. I'll continue to write the tables in full, because it will be convenient for calculations.

Some more payoff matrices

To demonstrate how efficient these tables are for describing games, here are the three other variants on rock paper scissors I needed whole paragraphs to explain in the first post. 

	Rock	Paper	Scissors	Well
Rock	0	-1	1	-1
Paper	1	0	-1	1
Scissors	-1	1	0	-1
Well	1	-1	1	0

	Rock	Paper	Papyrus	Scissors
Rock	0	-1	-1	1
Paper	1	0	0	-1
Papyrus	1	0	0	-1
Scissors	-1	1	1	0

	Rock	Paper	SuperScissors
Rock	0	-1	1
Paper	1	0	-2
SuperScissors	-1	2	0

Payoff matrices allow you to very quickly spot certain dominated strategies and duplicate strategies. In particular, we can see that in the first example that the row entries for Rock are all smaller than or the same as those for Well, which entails that Rock is dominated by Well. Meanwhile, in the second table, the entries in the Paper row and exactly the same as those for the Papyrus row (and similarly for their columns). 

Exercise

See if you can find the omitted entries of this payoff matrix. Do you think there's a dominated strategy?

	Option 1	Option 2	Option 3	Option 4
Option 1	?	3	-2	1
Option 2	?	?	0	6
Option 3	?	?	?	-4
Option 4	?	?	?	?

Arbitrage betting

We can't always spot duplicate or dominated strategies by directly comparing table rows: as in the last post, sometimes a mixture of strategies can mimic or outperform a strategy (or mimic a different mixture of strategies). The fact that there are multiple ways to make the same bet is something familiar to lots of gamblers: for example, you could:

• Bet directly that Chelsea win
• Make two bets simultaneously:
  • That both Chelsea and Aston Villa win: 
  • That Chelsea win but Aston Villa lose or draw.

It may be that one of these betting methods gives you better odds for Chelsea winning, making than the other a dominated strategy. Sometimes the division can be more complicated than this: I remember exploring Paris with a friend who had to pause to place a bet that Southampton's would win the second half of their match, having previously made bets regarding the overall result and the first half. Finding ways to divide up bets like this can be used to great profit: sometimes you can find a sure win, throguh what is known as arbitrage betting.