Breaking out of the prisoner’s dilemma

What happens if you play the prisoners’ dilemma against yourself?



The prisoner’s dilemma often features in television programmes (such as ITV’s Golden Balls) where two contestants have to decide whether they want to share or steal a pot of money. They make their choices in secret from one another and then their decisions are simultaneously revealed.

Let the tuple $(a, b)$ mean that you get $a$ and your opponent gets $b$: so $(3, -1)$ represents you winning £3 and your opponent losing £1. We introduce the pay-off matrix for a non-iterated (played only once) prisoner’s dilemma:

You \ Opponent Cooperate Defect
Cooperate (1, 1) (-1, 3)
Defect (3, -1) (0, 0)

Each player is given the opportunity either to defect or to cooperate. In the original set-up with prisoners the option of defecting meant betraying the other and testifying whilst cooperation represented remaining silent.

The prisoner's dilemma

The prisoner’s dilemma

Nash equilibrium and Pareto efficiency

The definition of Nash equilibrium (NE) implies that if all but one players are playing at the Nash equilibrium strategy, the other player is better off playing it too. So in our pay-off matrix above, (Defect, Defect) is a Nash equilibrium strategy since if you know that your opponent defects, you should defect too. (Cooperate, Cooperate) is not NE since if you know that your opponent cooperates, you should defect.

It is often taught in game theory courses that two rational players should decide to defect as each one is better off defecting no matter what the other player does. This yields to the paradoxical situation of the players foregoing a more favourable outcome to both.

Pareto efficiency is defined as a state where it is impossible to make any one player better off without making at least one contestant worse off. In our example it is clear that (Cooperate, Cooperate) is Pareto efficient, but the (Defect, Defect) outcome is not. This leads to one criticism of Nash equilibrium: the outcome of NE isn’t always Pareto efficient.

Many possible remedies for a more satisfactory solution of the prisoner’s dilemma have been advocated. It is common to play the game repeatedly to show that rational players cooperate on the first few rounds but then start defecting.

A colleague and I stumbled upon the prisoner’s dilemma whilst discussing a different problem and we originally had a mild disagreement over what the optimal strategy should be for our scenario. However, after a short conversation we settled it. Here is how we did it.

Let the probability guide us

Assume that you (P1) believe that the opponent (P2) cooperates with probability $p_2$. Your task is to find the probability of cooperating $p_1$, such that the expected pay-off is maximised. I would like to emphasise that $p_2$ represents your confidence in the statement that P2 will defect. $p_2$ is not the proportion of times P2 defects if you played the prisoner’s dilemma many times. This interpretation, also known as Frequentist, would be problematic for a game played only once. Instead, $p_2$ is simply your state of knowledge about P2. Such a Bayesian interpretation will protect us against mind projection fallacy: the fallacy when we confuse our state of knowledge about reality with the reality itself. For example, if we raise $p_2$ to $1$ does that mean the opponent will cooperate? No, we just think that he will. This will not, telepathically, cause him to cooperate.

Short foray into plausibility reasoning: if $A \implies B$ and you observe $B$, does that make $A$ more plausible? Bayes’ theorem says yes. This aspect is ignored in classical game theory, but we will rely on it here.

Finding the Nash equilibrium

Expected pay-off when $p_1$ is independent of $p_2$

Expected pay-off when $p_1$ is independent of $p_2$

The strategy assumes that the two players make their decisions independently: your choice of $p_1$ does not affect what you believe $p_2$ to be. Hence the expected pay-off is found to be
\[ \mathbb{E}[\text{pay-off}] = p_1 p_2 – p_1 (1 – p_2) + 3 (1-p_1) p_2, \]
which reduces to

\[ \mathbb{E}[\text{pay-off}] = 3p_2 – p_1(1 + p_2). \]
The maximum expected pay-off is always achieved at $p_1 = 0$ for any value of $p_2$ between $0$ and $1$. So you never cooperate. The probability set-up is now redundant as your strategy is deterministic. Such a deterministic strategy is also known as pure. We arrived at the NE strategy again: a strategy which is often referred to as rational, and yet I’d hesitate to refer to people who leave the best option on the table, $(1, 1)$, as rational.

What if you played it with a clone of yourself?

Expected pay-off when you play against a clone of yourself

Expected pay-off when you play against a clone of yourself

Yeah, that’s right. Let’s just clone you for the purpose of the game just before it starts and see how this goes. Because why not? Previously $p_2$ was independent of $p_1$, but can we now say that your choice of $p_1$ doesn’t change your belief about $p_2$? The set-up now is as symmetric as it can possibly get and if you think you are going to cooperate with probability $p_1$, is there any reason to suppose that your clone cooperates with $p_2 \neq p_1$? Well… I argue that there isn’t. So let’s say that $p_1 = p_2 = p$. Therefore
\[ \mathbb{E}[\text{pay-off}] = p^2 – p(1 – p) + 3 (1-p) p = -p(p-2).\]
So we have an inverted parabola with roots at 0 and 2 readily maximised at $p = 1$. So you always cooperate with your own clone. The strategy is, again, deterministic. The intuition is also satisfied: if I defect then the clone defects because of the same reasoning and we both get nothing. If I cooperate then my clone cooperates and we are both happy. Seems logical that we both choose to cooperate.

Back to reality

Expected pay-off when your opponent is not quite you

Expected pay-off when your opponent is not quite you

What if your opponent dresses like you, talks like you, acts like you, but is not quite you? Let us quantify the “not quite you” part as follows: the person you are playing with flips a coin in the morning. If it comes up heads he will defect. It’s just one of those days when he’s angry at everyone. He acts just like you would act otherwise. Say the coin has probability $q$ of coming up tails.
p_2 =
0, & \text{with probability $1 – q$} \\
p_1, & \text{with probability $q$.}
We call $q$ the similarity index: it tells you how similar you think your opponent is to you. Hence, ignoring the details, $\mathbb{E}[p_2] = q p_1$ and
\[ \mathbb{E}[\text{pay-off}] = p_1 (3q – 1) \left(1 – \frac{q}{3q – 1} p_1 \right) \,, \]
which is maximised at
\[p_{\max} =
0, & \text{if $q \leq \frac{1}{3}$} \\
\frac{3q – 1}{2 q}, & \text{if $q > \frac{1}{3}$}
where $p_{\max}$ is dependant on the pay-off matrix. Note that the strategy isn’t deterministic for $1 / 3 < q < 1$ and it sits right in-between Nash equilibrium and the one with a clone. Our intuition is again satisfied: the chance of you cooperating is determined by the pay-off matrix and by the similarity of the opponent’s thinking process to yours.

I cannot emphasise enough that even though it seems like P2 telepathically knows about your strategy, there is no telepathy or magic thinking involved. The $q p$ strategy is what you believe P2 will do and it has nothing to do with what he will actually do.

Putting it in perspective

Should your belief about $p_2$ depend on your strategy $p_1$? I don’t see why it shouldn’t. Biological neural networks in our heads run on the same principles in all of us. So if I convince myself that I should defect then I have anthropomorphic evidence staring right at me that the opponent might think to do the same thing. Therefore I should revise my belief about $p_2$. Another way to put it: we are all “imperfect” clones of each other.

Let us set up a chain of propositions that could fit this. Let A(X) = “X proportion of people choose to defect”, B = “I end up defecting” and C = “My opponent ends up defecting”. Now observing B could raise your estimate of X, which in turn would raise your confidence in C. Note that there is no physical causality involved: it would be preposterous to reason that your decision causes your contestant to change his mind. However, there is a logical causality that governs the probability flow in your mind.

Only when you can verifiably be sure that you are playing against a P2 whose strategy is completely independent from yours, for example if you are playing against rolls of a die, should you stick to the Nash equilibrium strategy.

Different pay-off matrices

We mentioned that $p_{\max}$ was dependent on the pay-off matrix, so let us play around with our pay-off matrices to check how our solution performs. Keep (Cooperate, Cooperate) and (Defect, Defect) fixed, but introduce $(b, c)$ instead for (Cooperate, Defect), such that $b > 1$ and $c< 0$.

You \ Opponent Cooperate Defect
Cooperate (1, 1) (c, b)
Defect (b, c) (0, 0)

It can now be shown that
p_{\max} =
0, & \text{if $q \leq – \frac{c}{b}$} \\
\frac{bq + c}{2(b + c – 1) q}, & \text{if $q > – \frac{c}{b}$}
under a technical assumption that $b + c \geq 2$.

You can see that $p_{\max} \to 1 /2$ as $b \to \infty$ for any fixed $q, c$. Makes sense? You and your contestant are aiming for the $b$ outcome, but only one of you can have it. Might as well let randomness decide who is going to get it. And if $c \to -\infty$ you should stop cooperating.

Connection with a cooperative game theory

It is apparent that when one plays against a clone it is equivalent to deciding upon the strategy beforehand (at time 0) and sticking to it throughout (without the ability to communicate any longer). The field of game theory where players are permitted to decide upon a strategy first and are obliged to stick to it is also known as cooperative game theory. In a cooperative prisoner’s dilemma players can easily achieve the Pareto efficient outcome (C, C) as long as there is a third party ready to enforce the contract.

In fact, our “back to reality” scenario is equivalent to playing such a cooperative dilemma only with a person who is somewhat unreliable and feels the same way about you: then you can both pre-agree that maximising $p (3q – 1) (1 – q p /(3q – 1))$ is the right thing to do.

So it turns out that allowing plausibility reasoning about your opponent’s strategy based on your own strategy in a classical game theory setting, as we have done here, leads us to a cooperative game theory. It appears that it is beneficial for everyone to act as if they signed an implicit contract rather than out of a selfish self-interest. Perhaps this could explain a number of social concepts like queueing, voting, not littering, etc.

Using cooperative game theory to explain social phenomena such as queueing.

Using cooperative game theory to explain social phenomena such as queueing.

What if the game isn’t symmetric?

The solution to these kind of dilemmas now easily follows. Imagine that at time 0 you could pre-agree with the other players about a strategy, what would it be? Just pick the one that leads to a Pareto efficient outcome and stick to it. Hence if you play with clones that’s what you should do. We believe that Pareto efficiency implies that the strategies $s^*_1, s^*_2, \dots, s^*_N$ maximise
\[ \min_{1 \leq n \leq N} \mathbb{E}[\text{pay-off for player $n$}] \]
because no player would agree to reduce his expectation for the gain of another player.

Let’s work through the “back to reality example”, but with different trust issues. At time 0, P1 and P2 meet and disclose:

P1: “Let us make an agreement that you cooperate with chance $p_2$, but I estimate $(1 – q_1)$ chance that you will break the agreement and defect regardless.”

P2: “Similarly, I think there is a $(1 – q_2)$ chance of you violating your agreement of cooperating with probability $p_1$.”

So, with the same pay-off matrix as before, the expected pay-off for player 1 is
F_1(p_1, p_2) = -p_1 + 3p_2 – p_1 p_2
and the pay-off for player 2, $F_2(p_1, p_2)$, is simply $F_1(p_2, p_1)$. For each choice of $(q_1, q_2)$, the formula for what values of $(p_1, p_2)$ we achieve the Pareto efficient outcome can be derived to obtain
(p_1^*, p_2^*) = \operatorname{argmax}_{p_1, p_2}
\min \left[ F_1(p_1, q_1 p_2), F_1(p_2, q_2 p_1). \right]

By symmetry, for $q_1 = q_2 = q$ this is achieved at $p_2^* = q p_1^*$. For the asymmetric case solving it numerically yields the following values:

$q_1$, $q_2$ 0.75, 1 0.5, 1 0.25, 1 0.1, 1 0.75, 0.75 0.5, 0.75 0.25, 0.75
$p_1^*$ 0.86 0.66 0.33 0 0.83 0.61 0.25
$p_2^*$ 0.99 0.93 0.67 0 0.83 0.75 0.44

One interesting choice of $(q_1, q_2)$ is $(0.1, 1)$: when the opponent trusts you unconditionally, but knows you don’t trust him. In that case you are both better off defecting.

In the end, using this method we can find the solution that takes into account how similar you think you appear to your opponent and how similar you think your opponent is to you!

Obligatory references and other titbits

Obviously it turned out that what Stephen and I stumbled upon one evening isn’t new. Douglas Hofstadter introduced the concept of superrationality which is equivalent to playing against yourselves. However, to the best of my knowledge, this interpretation of rationality is still not widely accepted in the orthodox literature.

And finally, some relevant facts from Golden Balls for your information:

  • Individual players on average choose “split” 53% of the time.
  • There is little evidence that contestants’ propensity to cooperate depends positively on the likelihood that their opponent will cooperate (i.e. little evidence for conditional cooperation).
  • Contestants are less likely to cooperate if their opponent has tried to vote them off the show in the first two rounds of the game, which is in line with the notion that people have an intrinsic preference for reciprocity.


Artiom Fiodorov is a PhD student at UCL who (evidently) likes writing. Cognitive science, maths and programming are his recurring themes. When he is not writing for Chalkdust or his blogs, he is studying random walks in random environments.

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