I have just 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

Quoting the first one about the [*prisoner dilemma*](https://en.wikipedia.org/wiki/Prisoner%27s_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.

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 ?