Let's return to the example of Rock Paper Scissors. Common sense tells us that choosing each option a third of the time should be the "best" strategy. Can we pin this down mathematically by proving it is the Nash equilibrium strategy?
Nash strategies are "unexploitable"
A key to calculating Nash strategies is the observation that they are unexploitable. Let's define a "least exploitable" strategy as any strategy which on average ties with or beats every other possible strategy, and remember the definition of a Nash strategy as one of a pair of strategies for which neither player can benefit by changing strategy without their opponent also changing. [Remember as well that we talk about a Nash strategy – rather than a pair of Nash strategies – because if \((A,B)\) denotes a pair of strategies forming a Nash equilibrium, then \( (A,A) \) also forms a Nash equilibrium, at least in the context of a symmetric zero-sum game.] These two concepts of an optimal strategy coincide.Nash strategies are least exploitable strategies, and vice-versa, for two-player symmetric zero-sum gamesSee if you can explain why this is true; I'll give a rigorous explanation in the next post. Note that this result lets us prove a claim from the second post: that choosing randomly between Nash strategies gives another Nash strategy. It's easy to show this for least exploitable strategies, and those coincide with Nash strategies by the above.
Verifying a strategy is Nash
By least exploitability, your average score when choosing Rock against me if I'm playing the Nash strategy is at most 0. This gives us the equation\[\begin{split} &\quad\text{average score you get against my strategy when choosing Rock every time } \\ &\small{\left(= \left\{ \begin{split} &\text{your reward when I choose Rock}\times \text{proportion of the time I choose Rock} \\ +& \text{your reward when I choose Paper}\times \text{proportion of the time I choose Paper} \\ +&\text{your reward when I choose Scissors}\times \text{proportion of the time I choose Scissors}\end{split}\right\} \right)} \\ &\quad \text{is at most } 0.\end{split}\]
Remember the payoff matrix from last time, which gives the scores in each case:
| | Rock | Paper | Scissors |
|---|---|---|---|
| Rock | 0 | -1 | 1 |
| Paper | 1 | 0 | -1 |
| Scissors | -1 | 1 | 0 |
Let's represent the proportions of each option which make up the Nash strategy using some letters so we can shorten the equation. Supposing the Nash strategy involves choosing Rock a proportion \(r\) of the time, Paper a proportion \(p\) of the time and Scissors a proportion \(s\) of the time, the above equation reduces to \begin{align*} 0r+1s+(-1)p&\leq 0, \text{ i.e.} \\ s-p &\leq 0. \end{align*} Similarly if I'm using the Nash strategy then I tie with or beat you if you choose Paper every time, or choose Scissors every time, which gives us the following equations: \[ r-s \leq 0, \quad p-r\leq 0. \] We can verify these hold with \(r=p=s=1/3\), confirming (as we already strongly suspected) that this is the Nash equilibrium! [As an aside, technically to show that a strategy is least exploitable we also need to show it doesn't lose to any mixture of Rock, Paper and Scissors. But this follows from it not losing to any of Rock, Paper or Scissors individually.]
Calculating a Nash strategy when you don't know it in advance
What if we hadn't already known what the Nash strategy was, so we couldn't just check that it satisfied the equations? Without knowing the answer in advance or guessing it, we want a method to find \(r,s,p\) such that \[ s-p\leq 0,\quad r-s\leq 0,\quad p-r\leq 0.\] A trick we can do in this case is add up the three left sides: we get \(s-p+r-s+p-r=0.\) If we take three numbers, none of which is bigger than zero, add them together and get zero, it follows that none of them can have been negative. We therefore in fact have \[s-p=0,~r-s=0,~p-r=0,\] so that \(r=p=s.\) Since \(r+p+s=1\) (these are the only three options, so the proportions of the time we choose them add up to \(1\)) we see that \(r=p=s=1/3.\)This hints at a general trick. By definition a least exploitable, and therefore Nash, strategy satisfies that the average payoff of each fixed opposing strategy is at most zero. But in the case of Rock-Paper-Scissors, in fact the payoffs were equal to zero. Equations with equals signs are much easier to solve than ones with less than or equals signs, so if we can always replace \(\leq \) by \(=\), it will help a lot with calculations. Can we do that? For many games the answer is yes.
Suppose a Nash strategy for a game is a completely mixed strategy, i.e. every option is chosen at least some proportion of the time. Then the reward for playing any fixed strategy against this Nash strategy (which is at most zero because Nash strategies are least exploitable) is in fact exactly equal to zero.
Why? Again, I'll leave this for you to think about for now, and give a justification in the next post. In the mean time, let's find the Nash strategy for Rock-Paper-SuperScissors, as promised. The equations we get by using that the payoff of each pure strategy against the Nash strategy is zero are: \[\text{Rock: }-p+s=0,\quad \text{Paper: }r-2s=0,\quad \text{SuperScissors: }-r+2p=0 \] The first two equations yield \(p=s=r/2\). Then, because the proportions must add to 1, the Nash strategy is \((p,s,r)=(1/4,1/4,1/2)\), i.e. you should choose Paper and SuperScissors a quarter of the time each, and Rock half the time. That's right: you choose scissors less than in normal RPS, despite the double reward!
The Nash strategy for cops and robbers
I found this result surprising the first time I saw it. We've shown that increasing the reward for winning with scissors, instead of encouraging you to pick scissors more often, should actually encourage you to choose rock more often. Intuitively speaking, this is to prevent your opponent from benefitting too much from the super-scissors.The same kind of calculations lead to some remarkable conclusions in modelling real world situations. Consider a simplified game theory model for crime, in which a thief wants to go out robbing if the police won't be patrolling that night, but doesn't want to go thieving if the police will be out and about; meanwhile, the police want to patrol if there'll be a thief out to catch but not otherwise. [Of course, like in Rock-Paper-Scissors, both the police and the thief must choose between their options without knowing what the other will do.] The same kind of calculations as for Rock-Paper-(Scissors/SuperScissors) above show that increasing the punishment for crime (which is like decreasing the reward for making the choice to commit a crime, if you get caught and hence "lose" the game) does not decrease rates of crime, but rather decreases the effort police go to to catch criminals: see the results here (in particular theorems 2 and 3). Similarly, increasing the rewards for catching a thief will decrease the crime rate, despite not increasing the proportion of the time that police spend patrolling.
This sounds insane! To get some intuition for why it can be true, think of changing rewards slowly and people reacting in stages. At first, increasing the reward for catching criminals encourages more patrolling, which makes it more likely that a robber gets caught. But if the police go out too often the robbers, who were willing to accept the old risk of being caught, will no longer go robbing because of the increased risk. At the old Nash equilibrium the risk and rewards were perfectly balanced, and since the jail setences and value of goods to steal haven't changed, the only way them to get balanced again is for the old rate of patrols to come back. So the higher rate of patrolling will die away, because otherwise there will never be robbers out for the police to catch, so they would never get to cash in on the bigger bonuses anyway. The extra pay isn't encouraging to the police to work harder, but it is at least encouraging them to work just as hard despite getting to catch crooks less often.
As a final point, let me emphasise that this is only a mathematical abstraction, and that what works in the model may not work as well in the real world. Regardless of whether it works in practice or not, pay based on numbers of arrests can lead to resentment among policed communities, who feel exploited and unfairly treated based on financial incentives, as the numerous articles complaining about "Collars for Dollars" show, even if these criticisms might not always be fair. The broader claim of the mathematical model – that increasing the probability of capture (for example by hiring more police) is much more effective at reducing crime than increasing the harshness of punishments – has been supported by evidence in practice, not just the theory.
