Bangu (2010) claims that Bertrand’s paradox rests on a hitherto unrecognized assumption, which assumption is sufficiently dubious to throw the burden of proof back onto ‘objectors to [the principle of indifference]’ (2010: 31). We show that Bangu’s objection to the assumption is ill-founded and that the assumption is provably true.
Bangu discerns a ‘common structure’ in some of Bertrand’s paradoxes:
One begins with a variable x … and then one considers a scaling transformation θ such that x′=θ(x). (Bangu 2010: 31)
Speaking in these terms on Bertrand’s behalf, we construe his objection to the principle of indifference thus: The second premiss might be rejected: we might have an empirical situation in which the transformed variable is a causally dependent variable, in which case there is no reason to suppose the probability distribution of the dependent variable is uniform. However, the principle of indifference is supposed to be an unrestricted principle, so Bertrand’s argument cannot be generally defeated by this point.
At least when θ is a nice function,1 the probability of choosing at random in the interval [c, b] ⊆ [a, b] should be the same as when choosing at random in the interval [θ(c), θ(b)] ⊆ [θ(a), θ(b)].
Yet applying the principle of indifference results in uniform probability distributions on [a, b] and [θ(a), θ(b)], when the probability that x∈[c, b] is not necessarily equal to the probability that x′∈ [θ(c), θ(b)].
Contradiction; therefore reject the principle of indifference.
Bangu rejects the first premiss on the ground that it rests on a false, or at least undefended, assumption:
R: If the argument of a scaling function is random in an interval, then the scaled value is random as well (in the scaled interval). (2010: 33)2
Bangu takes unpredictability ‘to be the standard conception of randomness’ and argues against R on the basis of the following scenario:
Suppose a machine, a random number generator, picks a value x in the interval [a, b]. Randomness here is understood in the predictive sense: there is no way to predict what value this choice will return. The machine records it, but does not communicate it to us. We now ask whether the transformed value is also random in the interval [a′, b′] [=[θ(a), θ(b)]], in the same predictive sense. One might say that this is not so, as there is a crucial difference between the value of x and the transformed value. While we cannot predict the value of x, we can predict the value of the transformed: we find what value has been recorded, and we scale it. So, is the transformed value random in [a′, b′] after all? Or, is the sense of ‘prediction’ not sufficiently well defined to be useful here? (2010: 33)
Bangu’s argument here rests on a premiss that is patently false. In a perfectly straightforward sense, namely prior to the machine selecting, the transformed value is no more predictable than the original. What he calls the transformed value being predictable is the situation of being given the value the machine picked and using that to predict, say, the square by squaring it. But, of course, in that sense of ‘predictable’ the value the machine picked is precisely as predictable! Take the value the machine picked as the prediction of the value the machine picked.
Allow us to spell this out more slowly. Accept that randomness should be understood in terms of unpredictability. When we consider the truth of R, what is at issue is whether if x is not predictable, θ(x) is not predictable. Bangu’s argument seems directed, instead, at whether if x is not predictable but we know x,3θ(x) is not predictable. Now clearly, if θ is what we called a nice function, the unpredictability of x will not carry over to θ(x) once we know x. But that is irrelevant.
The truth of R can be illustrated by an elementary example. Let X be the variable that takes its values from the roll of a fair die. Let Φ(X) be the variable that is zero in the event that X is not 1, and one in the event that X is 1.4 Φ is known. Even though the image of Φ is smaller than its domain, Φ(X) is no more predictable than X. In order to see this, consider a sequence of values of X from such a die roll and the corresponding values of Φ(X): If either of these sequences is predictable then knowing some number of the previous members of the sequence suffices for knowing the next one. Suppose that SX is unpredictable. That means that for all n, knowing 〈x1, … , xn〉 does not suffice for knowing xn+1. Now we consider the predictability of SΦ. We can suppose that we know 〈x1, … , xn〉 and we know 〈Φ(x1), … , Φ(xn)〉, and because SX is unpredictable we do not know xn+1. So for all we know xn+1 could be 1 or 6 and so Φ(xn+1) could be 1 or 0 and hence we do not know Φ(xn+1). So knowing 〈Φ(x1), … ,Φ(xn)〉 does not suffice for knowing Φ(xn+1) and SΦ is unpredictable. Hence if X is unpredictable, Φ(X) is unpredictable, whence if X is random, Φ(X) is random.
SX 1,2,1,1,5,3,3,1,4,1,6,4,2, … xn …
SΦ 1,0,1,1,0,0,0,1,0,1,0,0,0, … Φ(xn) …
In short, knowledge of how to map SX onto SΦ does not render SΦ any more predictable than SX, and that is all that is required for the truth of R in this example. This argument is in fact general. It neither depends on the discreteness of the event spaces nor on Φ. It applies to a wide range of functions. Its conclusion is equivalent to the assumption that Bangu rejects.5
Now one might instead consider ‘special cases’ where θ is a constant function on the random variable X.6 Then it is true that θ(X) is predictable (given the knowledge that θ is a constant function). However, this kind of special case does not matter for Bertrand’s paradox. Bertrand needs only one non-constant function to result in inconsistent probabilities: to refute him requires no such functions to exist. We suggest this is precisely why Bertrand and commentators on his paradox – among them, exceptional philosophers such as Keynes (1921/1963) and van Fraassen (1989) – have not mentioned the assumption that Bangu raises. The assumption is obviously true and for that reason not mentioned.