One in thirteen - Bayesian reasoning illustrated with a toy model

Let us leave aside Columbo's cameras for a while and begin with a different, simpler, stereotyped situation easier to analyze.

Imagine there are two types of boxes, $B_1$, that only contain white balls ($W$), and $B_2$, that contain one white balls and twelve black (incidentally, just to be precise, although the detail is absolutely irrelevant, we have to infer from Columbo's words, ``You didn't touch any of these twelve cameras. You picked up that one'', the cameras were thirteen).

You take at random a box and extract a ball. The resulting color is white. You might be interested to evaluate the probability that the box is of type $B_1$, in the sense of stating in a quantitative way how much you believe this hypothesis. In formal terms we are interested in $P(B_1\,\vert\,W,I)$, knowing that $P(W\,\vert\,B_1,I)=1$ and $P(W\,\vert\,B_2,I)=1/13$, a problem that can be sketched as

$$\left\{\!\! \begin{array}{rcl} P(W\,\vert\,B_1,I)& = &1 \\ P(W\,\vert\,B_2,I)&=&{1}/{13} \end{array}\right. \hspace{0.8cm} \Rightarrow \hspace{0.3cm} P(B_1\,\vert\,W,I)=\, \mbox{\boldmath$?$}$$ (1)

[Here `$\vert$' stands for `given', or `conditioned by'; $I$ is the general (`background') status of information under which this probability is assessed; `$W,I$' or `$B_i,I$' after `$\vert$' indicates that both conditions are relevant for the evaluation of the probability.]

A typical mistake at this point is to confuse $P(B_1\,\vert\,W,I)$ with $P(W\,\vert\,B_1,I)$, or, more often, $P(B_2\,\vert\,W,I)$ with $P(W\,\vert\,B_2,I)$, as largely discussed in Ref. [1]. Hence we need to learn how to turn properly $P(W\,\vert\,B_1,I)$ into $P(B_1\,\vert\,W,I)$ using the rules of probability theory.