Calculate $\mathbb{P}(S\mid D=d, P=p, B=b)$ given $\mathbb{P}(S\mid D=d, P=p)$ and $\mathbb{P}(S\mid D=d, B=b)$ Disclaimer: I am not a mathematician!
I have encountered a problem at work. There are a number of products $P$, a number of stores at a large retailer (branches, $B$) and a number of days ($D$). I need to calculate the probability that a product $p$ will sell ($S$) on a particular day $d$ at a particular branch $b$.
I know the probability that a given product will sell on a given day: $$\mathbb{P}(S\mid D=d, P=p).$$
I also know the probability that a branch makes a sale on a given day: $$\mathbb{P}(S\mid D=d, B=b).$$
How do I calculate the probability of a sale of a product at a branch on a day? i.e. from the given information, how do I get the distribution $\mathbb{P}(S\mid D=d, P=p, B=b)$?
Thank you for your help!
 A: 
One cannot.

The trouble is that the joint distribution of $(S,D,P,B)$ is not fully described by the distributions of $(S,D,P)$ and $(S,D,B)$.
To simplify things, assume that $D$ is independent of $(S,P,D)$, that $P$ and $B$ can take two values, say $0$ and $1$, and are independent and uniform on $\{0,1\}$, and that $S$ has probability $\frac12$ and is independent on $P$ and independent on $B$. Then, $P(S\mid P=x)=P(S\mid B=y)=\frac12$ for every $x$ and $y$ in $\{0,1\}$, hence
$$
\frac12=P(S\mid P=x)=\sum_yP(S,B=y\mid P=x)=\sum_ys_{xy}P(B=y\mid P=x)=\frac12\sum_ys_{xy},
$$
and
$$
\frac12=P(S\mid B=y)=\sum_xP(S,P=x\mid B=y)=\sum_xs_{xy}P(P=x\mid B=y)=\frac12\sum_xs_{xy},
$$
where $s_{xy}=P(S\mid P=x,B=y)$ are the conditional probabilities to be computed. A solution is $s_{00}=s_{11}=1-\sigma$, $s_{01}=s_{10}=\sigma$, for every parameter $\sigma$ in $[0,1]$.
One sees that the conditional probabilities $P(S\mid P=x)$ and $P(S\mid B=y)$ are not sufficient to determine $P(S\mid P=x,B=y)$, and in fact even the conditional probabilities $P(S\mid P=x)$ and $P(S\mid B=y)$ and the distributions $P(P=x)$ and $P(B=y)$ are not sufficient.
Finally, one can recognize the extreme cases $\sigma=0$ as $S=[P=B]$ and $\sigma=1$ as $S=[P\ne B]$, for two i.i.d. symmetric Bernoulli random variables.
A: Imagine branch $b$ does not have, on any day, product $p$ in stock. Then regardless of the day $d$ there is no chance that product $p$ will sell in branch $b$ on day $d$, even though there may be a positive chance that product $p$ sells (elsewhere) on day $d$, and that branch $b$ sells something on day $d$. On the other hand one clearly cannot deduce the impossibility of $b$ selling $p$ from the given positive probabilities. In other words those probabilities simply do not carry the information necessary to compute the probability you are after.
