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

\begin{array}{rcl} P(W\,\vert\,B_1,I)& = &1 \\
\hspace{0.8cm}\end{displaymath} $\displaystyle \Rightarrow$ $\displaystyle \hspace{0.3cm}
P(B_1\,\vert\,W,I)=\,$   $\displaystyle \mbox{\boldmath$?$}$ (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.

Giulio D'Agostini 2010-09-30