### To read before (following some answers or comments)
- 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](http://stla.github.io/stlapblog/posts/PrisonerDilemma.html) (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:
- [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. 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
$$
\text{Probability of $X$ when $X$ knows $Y$}.
$$