Tuesday, February 27, 2007

Stupid Math Tricks

Joe Miller relates the correct version of a quotation from J. S. Mill:
The Conservatives, as being by the law of their existence the stupidest party…
Mill later clarified his statement like so:
I never meant to say that the Conservatives are generally stupid. I meant to say that stupid people are generally Conservative.
As a matter of strict logic, these are distinct claims; S --> C does not entail C --> S. However, if what Mill says is correct, then we should be able to make a probabilistic judgment about the intelligence of conservatives. Suppose that stupid people constitute 60% of the population (a generously low number). Suppose that smart people split 50-50 between conservatism and liberalism. And suppose that stupid people split 80-20 between conservatism and liberalism, consistent with Mill’s claim. Now, say you meet a random person who turns out to be conservative, and you know nothing else about him. How likely is he to be stupid? A quick application of Bayes’ Rule tells us
P(stupid | conservative) = [(0.8)(0.6)] / [(0.8)(0.6) + (0.5)(0.4)] = 0.71
And thus we may conclude that a random conservative person – that is, a randomly chosen person who turns out to be conservative – is more than 70% likely to be stupid.

More generally, if s = the fraction of the public that is stupid, a = the fraction of stupid people who are conservative, and b = the fraction of smart people who are conservative, then the probability a randomly chosen person is stupid given that he is conservative is
P(stupid | conservative) = as / [as + b(1 – s)]
Just plug in the numbers you think appropriate. Of course, it’s worth noting that the meaning of ‘conservative’ has changed substantially since Mill’s time. And being neither a conservative nor a liberal (in the modern American sense of that word), I have no dog in this hunt. Stupidity seems well represented in both parties. My interest here is purely mathematical!

3 comments:

ZERO DIVISOR said...

That's about right. 70% of the conservatives would benefit from liberal policies yet because they are retards and easily manipulated by their own party and the right wing media like FOX, they vote Republican thereby fattening the pockets of the 30% of the Republicans who are smart, egotistical, and rich and who don't give a hoot about the 70% of their own party that helped put the criminal evil doers in high office. But I really didn't need any theorem of probability to understand that obvious and sorry fact.

Benjamin said...

Is your math right?

My thinking on Bayes is:
P(stupid|conservative)
=P(conservative|stupid)*P(stupid)
------------------------
P(conservative)
=.8*.6/.5 = .96

Another way to think about it (values in P):
C L Total
Stupid A? B? .6
Not Stupid C? D? .4
Total .5 .5 1

If stupid people are 80% conservative then we get the following fractions of population:
A = Stupid conservatives = .6*.8=.48
B = Stupid Liberals = .6*.2=.12
C = Smart Conservatives = .5-.48=.02
D = Smart Liberals = .5-.12=.38
P(stupid|conservative) =
= % of pop that are stupid conservatives / percent % conservative
=.96

Did I get something wrong?

Glen Whitman said...

My thinking on Bayes is:
P(stupid|conservative)
=P(conservative|stupid)*P(stupid)
------------------------
P(conservative)
=.8*.6/.5 = .96


The problem is you're dividing by the wrong thing, because P(conservative) is not 0.5. Given my initial assumptions, there are more conservatives than liberals. In fact, the population is 68% conservative (80% of the stupid people and 50% of the smart people, with 60% of the population being stupid).

If stupid people are 80% conservative then we get the following fractions of population:
A = Stupid conservatives = .6*.8=.48
B = Stupid Liberals = .6*.2=.12
C = Smart Conservatives = .5-.48=.02
D = Smart Liberals = .5-.12=.38


A and B are right, but C and D are not. They should be:

C = smart conservatives = (.4)(.5) = 0.20
D = smart liberals = (.4)(.5) = 0.20