I have come across some papers and slides about *quantum cognition*, including: - [Cold and hot cognition: Quantum probability theory and realistic psychological modeling](https://ppw.kuleuven.be/okp/_pdf/Lee2013QMOCA.pdf), by P. J. Corr - [Applications of quantum probability theory to dynamic decision making](https://community.apan.org/afosr/m/cognition_decision__computational_intelligence_program/121111/download.aspx), by Busemeyer, Balakrishnan and Wang These texts and other ones claim that *the [*prisoner's dilemma*](https://en.wikipedia.org/wiki/Prisoner%27s_dilemma) violates the law of total probability.* Quoting Corr's paper: > 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 slides by Busemeyer & al. provide some empirical data supporting this claim. The above quote has an obvious flaw, as I explain now. I want to emphasize this is not an attempt to discredit quantum cognition. This story of violation of the TP law is older than quantum cognition, but this is where I found it. I asked on a stackexchange network about psychology whether this flaw were known, and then what is the purpose of quantum cognition about the prisoner's dilemma. I got some reactions of persons falling into the flaw. That showed me that it is at least not known by everybody, and this helped me to achieve, I hope, a better explanation, which is the content of this blog post. Consider ***two experiments***: - *The "true" prisoner's dilemma*: each of the prisoners $A$ and $B$ has the choice to cooperate or defect, without knowing the other prisoner's choice. - *The "faked" prisoner's dilemma*: each of the prisoners $A$ and $B$ has the choice to cooperate or defect, but $B$ is asked first, and $A$ knows $B$'s choice. Consider the two probabilities $\Pr$ and $\Pr^\ast$ respectively corresponding to these two experiments. The above quote claims that the literature shows: 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 it claims that this is troubling for Classical Probability theory. This is not troubling. But this is troubling if you assume $\Pr=\Pr^\ast$. And I believe this is the flaw in the reasoning. Indeed, the TP law shows that $\Pr(A \textrm{ defects})$ is a weighted average of $\Pr(A \textrm{ defects} \mid B \textrm{ defects})$ and $\Pr(A \textrm{ defects} \mid B \textrm{ coop.})$. Assuming $\Pr=\Pr^\ast$ implies that $\Pr(A \textrm{ defects})$, which is lower than $0.5$, would be a weighted average of two numbers higher than $0.5$, which is impossible. Of course the two experiments are totally different, and there is no reason to assume $\Pr=\Pr^\ast$. The so-called violation of the TP *actually shows* that $\Pr\neq\Pr^\ast$. I am afraid this misunderstanding of the conditional probability could have an origin in the name which is sometimes used to call it : $\Pr(A \textrm{ defects} \mid B \textrm{ defects})$ is the *probability that* $A \textrm{ defects}$ ***knowing that*** $B \textrm{ defects}$. And it looks like it can be misinterpreted as: the *probability that* $A \textrm{ defects}$ ***when $A$ knows that*** $B \textrm{ defects}$. And $A$ never knows whether $B$ defects in the "true" prisoner dilemma. I think I am starting to have an idea about the purpose of quantum cognition, for this example of the prisoner's dilemma. Roughly speaking, it aims to have a mathematical axiomatic system for the "cognitive state" of $A$, with a rule describing the behavior of this cognitive state when $A$ learns the choice of $B$. Quantum probability could provide such an axiomatic, according to psychologists working in that field. It would provide a mathematical model for the state of $A$ under ignorance of $B$'s choice and also how this state would move when $A$ learns $B$'s choice. This is something different than the purpose of modeling the issue of one of the two above experiments, which is achieved by classical probability theory. ```{r, include=FALSE} knitr::knit_exit() ``` _________ 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 $$ \text{Probability of $X$ when $X$ knows $Y$}. $$