## John Stuart Mill in Math

| Peter Klein |

When I teach Bayes’s Theorem to my graduate students I use the Monty Hall paradox for illustration.

• Priors: $p(\textrm{door 1})=\frac{1}{3}$
• Suppose you choose door 3; Monty reveals door 2.
• What’s $p(\textrm{door 1}|\textrm{reveals 2})$?

It’s $\frac{p(\textrm{reveals 2}|\textrm{door 1})\cdot p(\textrm{door 1})}{p(\textrm{reveals 2}|\textrm{door 1})\cdot p(\textrm{door 1})+p(\textrm{reveals 2}|\symbol{126}\textrm{door 1})\cdot p(\symbol{126}\textrm{door 1})}\\\\=\frac{p(\textrm{reveals 2}|\textrm{door 1})\cdot p(\textrm{door 1})}{p(\textrm{reveals 2}|\textrm{door 1})\cdot p(\textrm{door 1})+p(\textrm{reveals 2}|\textrm{door 3})\cdot p(\textrm{door 3})+p(\textrm{reveals 2}|\textrm{door 2})\cdot p(\textrm{door 2})}\\\\=\frac{1\cdot \frac{1}{3}}{1\cdot \frac{1}{3}+\frac{1}{2}\cdot \frac{1}{3}+0\cdot \frac{1}{3}}\\\\=\frac{2}{3}$

So you should switch doors!

The problem is that few students in their 20s or early 30s have ever seen an episode of Let’s Make a Deal. (I have a college friend whose mother was a contestant on the show, dressed as a giant chicken.)

Glenn Whitman provides another example for illustrating Bayes’s Theorem, based on this quotation from John Stuart Mill: “The Conservatives, as being by the law of their existence the stupidest party. . . . ” Mill subsequently offered this clarification: “I never meant to say that the Conservatives are generally stupid. I meant to say that stupid people are generally Conservative.”

Writes Whitman:

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

$p(\textrm{stupid}|\textrm{conservative})=\frac{(0.8)(0.6)}{(0.8)(0.6) + (0.5)(0.4)}=0.71$

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.

These are hypothetical numbers, of course, and have nothing to do with liberalism or conservatism per se (whatever those terms happen to mean today). But maybe this example makes more sense than the old game-show story.

Entry filed under: - Klein -, Ephemera.

• 1. spostrel  |  5 March 2007 at 8:03 pm

The medical test example is quite popular and gets at the importance of prior information. You have a disease of low prevalence in the population (say 1%) and a test with pretty good accuracy (say 90% against both type I and type II error) and you ask how likely it is that you’re really sick if the test says so. Using these numbers, the answer is .9*.01/(.9*.01+.1*.99) = 47.6% to three digits. So even though the test is conditionally accurate nine out of ten times, you have a less than one-half chance of being sick if it says you are.

• 2. Jeff McNeill » Blog Archive » links for 2007-03-08  |  8 March 2007 at 8:22 am

[…] John Stuart Mill in Math « Organizations and Markets Monty Hall problem (tags: statistics) […]

• 3. michael webster  |  14 June 2007 at 8:06 pm

Let’s suppose that what is behind the winning door is $12. The expected value of the bet before any door is shown is$4.

On the theory presented here, you would pay up to $x for the opportunity to switch, where x is 1/2(12) – 1/3(12) – the extra value to switching-$2.

I can guarantee that if you commit to switching as a strategy, and pay say $1.75 every time, and$4 for each play, you will lose money over the long run -nothwithstanding the correctness of Bayes Theorem.

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