# The prisoner dilemma does not violate classical probability laws

• My background: As a mathematician, probability theory is my field of research; please do not answer/comment just to refer to a course on probability theory.

• I expose below an erroneous application of probability theory. It is summarized at the end, in the Summary section. The two probabilities $\Pr$ and ${\Pr}^\ast$ are explained just before. If you are not a bit specialist in probability theory, please do not downvote if you disagree with this point, unless you show an error in my reasoning at a precise point (after numerous edits, I believe I have enough detailed my explanation so that one can point a precise point).

• The previous comments helped me to write a short summary on my blog (rather oriented in general towards a mathematical audience), about the prisoner dilemma.

I have just come across some papers and slides about quantum cognition, including:

Quoting the first one about the prisoner dilemma:

The literature shows: (1) knowing that one’s partner has defected leads to a higher probability of defection; (2) knowing that one’s partner has cooperated also leads to a higher probability of defection; and, most troubling for Classical Probability theory, (3) not knowing one’s partner’s decision leads to a higher probability of cooperation.

The second one provides some empirical data supporting this claim. The data are some relative frequencies: one deals with the frequentist interpretation of probability.

I disagree with the claim that the law of total probability is violated here. The conditional probabilities are misinterpreted.

Let $A$ and $B$ be the two prisoners. Consider the experiment consisting in asking them to choose between defecting or cooperating, without knowing the choice of the other prisoner.

Then, the conditional probability $P(A \textrm{ defects} \mid B \textrm{ defects})$ is the long-term relative frequency of the event "$A$ defects" among all those experiments for which the event "$B$ defects" occurs.

This has nothing to do with the probability that $A$ defects when $A$ knows that $B$ defects, hereafter denoted by $\Pr^\ast(A \textrm{ defects} \mid B \textrm{ defects})$.

The law of total probability says that $$\Pr(A \textrm{ defects}) = \Pr(A \textrm{ defects} \mid B \textrm{ defects})\Pr(B \textrm{ defects}) + \Pr(A \textrm{ defects} \mid B \textrm{ cooperates})\Pr(B \textrm{ cooperates}),$$ thereby implying that $\Pr(A \textrm{ defects})$, as a weighted average of the two conditional probabilities $\Pr(A \textrm{ defects} \mid B \textrm{ defects})$ and $\Pr(A \textrm{ defects} \mid B \textrm{ cooperates})$, lies between these two conditional probabilities.

The above mentioned papers claim that the law of total probability is violated because $\Pr(A \textrm{ defects})$ does not lie between $\Pr^\ast(A \textrm{ defects} \mid B \textrm{ defects})$ and $\Pr^\ast (A \textrm{ defects} \mid B \textrm{ cooperates})$, where $\Pr^\ast (A \textrm{ defects} \mid B \textrm{ defects})$ is the probability that $A$ defects when $A$ knows that $B$ defects, and, as said before, $${\Pr}^\ast (A \textrm{ defects} \mid B \textrm{ defects}) \neq \Pr(A \textrm{ defects} \mid B \textrm{ defects})$$

So, is it an error, or do I misunderstand the purpose behind the modeling based on quantum probability ?

## EDIT: details on the difference between $\Pr$ and ${\Pr}^\ast$

To explain the difference, I give the way to get an empirical estimate of these probabilites.

### Experiment 1 ($\Pr$)

Ask $A$ and $B$ to perform the prisoner dilemma, without giving any information.

Repeat this experiment a large number of times, independently (with others $A$ and $B$). The estimate of $\Pr(A \textrm{ defects})$ is the relative frequency of the experiments for which $A$ defects. The estimate of $\Pr (A \textrm{ defects} \mid B \textrm{ defects})$ is the relative frequency of the experiments for which "$A$ defects" among all those experiments for which the event "$B$ defects" occurs.

### Experiment 2 ($\Pr^*$)

Ask $A$ and $B$ to perform the prisoner dilemma with $B$ first, and giving the choice of $B$ to $A$.

Then ${\Pr}^\ast (A \textrm{ defects})$ and ${\Pr}^\ast (A \textrm{ defects} \mid B \textrm{ defects})$ are estimated in the same way as before.

The experiment is not the same, in other words this is another probability (${\Pr}^*$) on the probability space.

As you can see in Experiment 1, the conditional probability has nothing to do with the probability that $A$ defects when $A$ knows that $B$ defects. In this experiment, $A$ never knows whether $B$ defects.

Of course, if you follow the above procedure to estimate the empirical probabilities, the law of total probability cannot be violated. This law is not really a principle, this is rather a definition (up to an elementary calculation, this is just the definition of the conditional probability). That makes no sense to say a definition is violated. If it is violated, that's because it has not been correctly used.

# Summary

The law of total probability implies that $\Pr(A \textrm{ defects})$ is a weighted average of the two conditional probabilities $\Pr(A \textrm{ defects} \mid B \textrm{ defects})$ and $\Pr(A \textrm{ defects} \mid B \textrm{ cooperates})$: $$\Pr(A \textrm{ defects}) = wavg\Bigl(\Pr(A \textrm{ defects} \mid B \textrm{ defects}), \Pr(A \textrm{ defects} \mid B \textrm{ coop.})\Bigr)$$ and therefore, it lies between these two conditional probabilities.

Similalry, for the other probability ${\Pr}^\ast$, $${\Pr}^\ast(A \textrm{ defects}) = wavg\Bigl({\Pr}^\ast(A \textrm{ defects} \mid B \textrm{ defects}), {\Pr}^\ast(A \textrm{ defects} \mid B \textrm{ coop.})\Bigr)$$ The so-called violation of the law of total probability is a consequence of the erroneous formula: $$\Pr(A \textrm{ defects}) = wavg\Bigl({\Pr}^\ast(A \textrm{ defects} \mid B \textrm{ defects}), {\Pr}^\ast(A \textrm{ defects} \mid B \textrm{ coop.})\Bigr),$$ "mixing" the two probabilities. Based on this formula, $\Pr(A \textrm{ defects})$ shoud lie between ${\Pr}^\ast(A \textrm{ defects} \mid B \textrm{ defects})$ and ${\Pr}^\ast(A \textrm{ defects} \mid B \textrm{ coop.})$. This is intuitively wrong, and this has been observed to be wrong on empirical data. But this formula is wrong.

As a side note, I think that the misunderstanding could have been caused by the name probability of $X$ knowing $Y$ to call the conditional probability of $X$ given $Y$. This has nothing to do with $X$ knowing something about $Y$: $$\text{Probability of X given Y}$$ does not mean $$\text{Probability of X when X knows Y}.$$

• I have to agree with @matus, below. The authors have not made a mistake: participants completing the Prisoner's Dilemma do not respect the law of total probability, or in other words, A is more likely to defect when they know what choice B made than when they don't, regardless of what that choice B actually makes. – Eoin Oct 3 '15 at 21:53
• As for your claim that we cannot compare the probability space from two different "experiments": either you've uncovered a hitherto unknown fundamental flaw in the logic of experimental psychology (where we compare probabilities from two different conditions all of the time), or you're mistaken. Why on earth can't we compare the two experiments? – Eoin Oct 3 '15 at 21:55
• Finally, is this perhaps one of those strange quirks of frequentist interpretations of probability? Surely from a subjectivist point of view, this is straightforward. – Eoin Oct 3 '15 at 21:57
• Post updated to add this more detailed explanation. – stla Oct 3 '15 at 22:45
• You have misread the paper. They are saying that people empirically do not behave like a rational CP actor would, not that CP has been mathematically violated in the PD (check the paper, it exclusively references people violating the sure thing principle). Gelman discusses the problem of translating between CP and QP more generally on his blog (comments), which I think is what you are actually interested in. – Christian Hummeluhr Oct 5 '15 at 9:15

After having thought more carefully about my own question, I give my own answer.

I would say that in the quote:

The literature shows: (1) knowing that one’s partner has defected leads to a higher probability of defection; (2) knowing that one’s partner has cooperated also leads to a higher probability of defection; and, most troubling for Classical Probability theory, (3) not knowing one’s partner’s decision leads to a higher probability of cooperation.

the word troubling is not the appropriate one.

As explained, the CP translation of this quote is:

1. ${\Pr}^\ast(A \textrm{ defects} \mid B \textrm{ defects}) > 0.5$

2. ${\Pr}^\ast(A \textrm{ defects} \mid B \textrm{ coop.}) > 0.5$

3. $\Pr(A \textrm{ defects}) < 0.5$

And this is not troubling for CP, because $\Pr \neq {\Pr}^\ast$.

Instead of saying this is troubling, I would simply say that providing a relation between (1)-(2) and (3) is not the purpose of CP. And roughly speaking, this is the purpose of quantum cognition, using quantum probability.

In the context of QC, a "conditional probability" such as $P(A \textrm{ defects} \mid B \textrm{ defects})$ should be read as "probability that $A$ defects when $A$ knows that $B$ defects". The "event" at right of the conditional bar "$|$" is something about the cognitive state of $A$. In Pothos & Busemeyer's paper, such a notation is used : $$Prob(happy|employed) = \Vert P_{happy} \psi_{employed} \Vert^2$$ The notation at left is a symbolic one, miming the notation for conditional probability. The notation at right is more rigorous. In quantum probability, $P_{happy}$ is a projection, and $\psi_{employed}$ is a quantum state. Here it would be something like $P_{defects}$ and $\psi_{B \text{ defects}}$, representing the state of $A$ when he knows that $B$ defects.

I claimed before that providing a relation between (1)-(2) and (3) is not the purpose of CP. I should rather say: not in this way. The quantum approach is close to, but not the same as, the idea of a family of probabilities $P_\psi(A \text{ defects})$ indexed by the state $\psi$ of $A$.

Here is another way to interpret the quote in CP, with only one probability $\Pr$. Assume that one firstly selects at random one of the two experiments ("true" prisoner dilemma" or "faked" prisoner dilemma") and then one performs it. Globally, there's only one experiment (a two-stage experiment). The three points in the quote are:

1. $\Pr(A \textrm{ defects} \mid \textrm{$A$knows that$B$defects}) > 0.5$

2. $\Pr(A \textrm{ defects} \mid \textrm{$A$knows that$B$coop.}) > 0.5$

3. $\Pr(A \textrm{ defects}\mid \textrm{$A$knows nothing about$B$}) < 0.5$

and they are not inconsistent with any CP law. Because here one does not know anything about the marginal probability $\Pr(A \textrm{ defects})$.

Consider two random variables $E$ and $F$ with the following states:

$E$: $c$ - A cooperates, $\neg c$ - A defects

$F$: $kc$ - A knows that B cooperated, $k\neg c$ - A knows that B defected

PJ Corr claims that:

1. $p(E=\neg c|F=k\neg c)>0.5$
2. $p(E=\neg c|F=kc)>0.5$
3. $p(E=\neg c)<0.5$

Then the application of the law of the total probability results in a contradiction: $p(E=\neg c|F=k\neg c)\cdot p(F=k\neg c)+p(E=\neg c|F=kc)\cdot p(F=kc)<0.5$

(Note that a weighted average of two variables both greater than $0.5$ must be also greater than $0.5$.)

To defend the probability theory one can extend the state space of $F$ with $\neg k$ - "not knowing one’s partner’s decision".

Or one may add context variables as Corr suggests:

Everyday equivalents of this situation do not, typically, resemble the decision-making dynamics assumed by CP theory: people have psychological processes that it fails to model, or to even consider to be relevant. For example, in this situation, outcome (1) could simply be (self-serving) retaliation; and in the case of (2) taking the easier route, but, we should wonder, what would happen if the partner was a loved one (e.g., your child)? Scenario (3) is a situation of uncertainty and psychological conflict (at least as perceived by many participants).

It is trivial to model these contexts with CP:

1. $p(E=\neg c|F=k\neg c,\mathrm{A\ wants\ a\ selfserving\ retaliation})$
2. $p(E=\neg c|F=kc,\mathrm{A\ loves\ B},\mathrm{B\ loves\ A})$
3. $p(E=\neg c|\mathrm{the\ situation\ is\ uncertain})$

Purpose behind the modeling based on quantum probability: I'm not acquainted with the discussion of quantum probability in cognitive science. However, as with any formalism there is a trade-off between simplicity and explanatory power. One can always devise a cognitive model with high explanatory power by adding contexts and rules for exception handling. Indeed, as evidenced by the recent rise of bayesian cognitive modeling and its critics, probability theory is well suited for such endeavor. As a disadvantage one gets ridiculously complex models and the whole modeling endeavor is difficult to falsify. I can imagine that it is in terms of simplicity that the quantum probability offers a better option than the classical probability.

• You are doing the same mistake. If $A$ knows that $B$ cooperates, this is another experiment than the one when $A$ knows nothing about $B$. You are mixing the conditional probabilities of one experiment with the marginal probabilities of another experiment. – stla Oct 3 '15 at 7:43
• By the way, the law of total probability stems from the definition of the conditional probability and an elementary calculation. This is not really a principle, this is rather a definition (up to the elementary calculation, this is just the definition of the conditional probability). That makes no sense to say a definition is violated. If it is violated, that's because it has not been correctly used. – stla Oct 3 '15 at 7:48
• Please, look up kolmogorov axioms and the corresponding definition of probability. There is no word of "experiment". What you are alluding to is the frequentist interpretation of probability, but we don't know if Corr is subscribing to it. Even in the frequentist interpretation, the only way to distinguish experiments is to redefine the state space and/or the corresponding probabilities. Two experiments with the same state space and identical probabilities are two identical experiments and super-scripting P with a star doesn't change that. – matus Oct 3 '15 at 11:13
• See my udpate. I don't need a course of probability theory, since this is my field of research as a mathematician. – stla Oct 3 '15 at 19:41