Stupidity, conservativism, and math:

Several readers objected to the logic of Prof. Brandon’s argument, and not just to its factual underpinnings or the accuracy of its paraphase of John Stuart Mill. Even if “stupid people are generally conservative,” the readers argue, this doesn’t mean that conservative people are especially likely to be stupid — and, they say, that’s a pretty basic error for a philosophy professor to make.

     I think this is one criticism of Brandon that isn’t quite apt. Assume stupid people are “generally” conservative (in the sense of more than 50% of the stupid people being conservative). Assume also — an unstated premise on Brandon’s part, but a plausible one — that the public at large is not generally conservative (in the sense that fewer than 50% of the general public being conservative). Then it follows that the smart people (or at least the nonstupid ones) will indeed be disproportionately nonconservative, and the conservatives will indeed be disproportionately stupid.

     Let’s start with a concrete example. Imagine a population of 1000 people. 60% of them are stupid, and among the stupid, 60% are conservative (“stupid people are generally conservative”). But among the 1000 people at large, only 40% are conservative (the unstated premise). This means that there are 360 (1000 x 60% x 60%) stupid conservatives, but only 400 (1000 x 40%) conservatives total. Thus, there are only 40 (400 – 360) smart conservatives, out of 400 (1000 – 1000 x 60%) smart people total. The smart people are 90% nonconservative, and the conservative people are 90% stupid.

     More broadly, say that:

     s = the proportion of the public that’s stupid (from 0 to 1).

     f = the fraction of stupid people who are conservative (presumably over 0.5, if “stupid people are generally conservative”).

     c = the proportion of the public that’s conservative; I argue that Brandon’s unstated premise, which is plausible as a matter of current fact, is that this is probably 0.5 or below.

     Then we know stupid conservatives are s x f (as a proportion of the population); smart conservatives are c – s x f; smart people total are 1 – s; and the fraction of conservatives that is smart is thus

          (c – s x f) / (1 – s)

     And if indeed c < f (as it is, given Brandon's stated premise and his likely unstated premise), which is to say a smaller fraction of all people are conservative than the fraction of stupid people that are conservative, then (c - s x f) / (1 - s) < (c - s x c) / (1 - s) = c, which is to say that conservatives are underrepresented among smart people compared to the population as a whole.      This, of course, is a highly stylized model, which rests on all sorts of simplifying assumptions that might not be entirely accurate. But neither are they provably inaccurate. So if you grant Brandon the plausible assumption that the public at large is not generally conservative, accept this simplistic but not implausible model, and accept his factual premise, then his argument does make logical sense — not as an ironclad proof, but as a plausible pragmatic explanation — which is why I didn’t fault him on that score.      I am, of course, happy to fault him on other scores, as I’ve done in other posts.

Comments are closed.

Powered by WordPress. Designed by Woo Themes