Propensity vs probability

Back to our toy experiment, I then see no problem saying that box $B_i$ has probability $\pi_i$ to produce white balls, meaning that such a `probability' is a physical property of the box, something that measures its propensity (or bent, tendency, preference)to produce white balls.

It is a matter of fact that, if we have full confidence that a physical  system has propensity $\pi$ to produce event $E$, then we shall use $\pi$ to form the “strength of our conjecture or anticipation” of its occurrence, that is $P(E\,\vert\,\pi,I)=\pi$. But it is often the case in real life that, even if we hypothesize that such a propensity does exist, we are not certain about its value, as it happens with box $B_?$. In this case we have to take into account all possible values of propensity. This is the meaning of Equation (12), which we can rewrite in more general terms as

$$P(E\,\vert\,I) = \sum_i \pi_i\cdot P(\pi_i\,\vert\,I)\,.$$ (13)

We can extend the reasoning to a continuous set of $\pi$, indicated by $p$ for its clear meaning of the parameter of a Bernoulli process, to which we associate then a probability density function (pdf), indicated by $f(p\,\vert\,I)$:

$$P(E\,\vert\,I) = \int_0^1\! p\,f(p\,\vert\,I)\,\mbox{d} p$$ (14)

The special case in which our probability, meant as degree of belief, coincides with a particular value of propensity, is when $P(\pi_i\,\vert\,I)$ is 1 for a particular $i$, or $f(p\,\vert\,I)$ is a Dirac delta-function. This is the difference between boxes $B_?$ and $B_E$. In $B_E$ our degree of belief of $1/2$ on White or Black is directly related to its assumed propensity to give balls of either colors. In $B_?$ a numerically identical degree of belief arises from averaging all possible propensity values (initially equally likely). And therefore the “strength of our conjecture or anticipation” (6) is the same in the two cases. Instead, if we had at the very beginning only the boxes with at least one white ball, the probability of White from $B_?^{(1-5)}$ becomes, applying the above formula, $\sum_{i=1}^{5}(i/5)\times (1/5) = 3/5$.

We are clearly talking about probabilities of propensities, as when we are interested in detector efficiencies, or in branching ratios of unstable particles (or in the proportion of the population in a country that shares a given character or opinion, or the many other cases in which we use a binomial distribution, whose parameter $p$ has, or might be given, the meaning of propensity). But there are other cases in which probability has no propensity interpretation, as in the case of the probability of a box composition, or, more generally, when we make inference on the parameter of a model. This occurs for instance in our toy experiment when we were talking about $P(B_i\,\vert\,I)$, a concept to which no serious scientist objects, as well as he/she has nothing against talking e.g. of $90\%$ probability that the mass of a black hole lies within a given interval of values (with the exception of a minority of highly ideologized guys).