How rare is the specimen? A Bayesian puzzler.

Further and deeper exploration of paradoxes and challenges of intuition and logic can be found in my recently published book, Probability, Choice and Reason.

An entomologist spots what might be a rare category of beetle, due to the pattern on its back. In the rare category, 98% have the pattern. In the common category, only 5% have the pattern. The rare category accounts for only 0.1% of the population. How likely is the beetle to be rare?

Since only 5 per cent of the common beetles bear the distinctive pattern and 98 per cent of the rare beetles do, intuition would tell you that you have come across a rare insect when you espy the pattern. Bayes’ Theorem tells you something quite different.

To calculate just how likely the beetle is to be rare given that we see the pattern on its back, we apply Bayes’ Theorem.

Posterior probability = ab/ [ab+c (1-a)]

a is the prior probability of the hypothesis (beetle is rare) being true. b is the probability we observe the pattern and the beetle is rare (hypothesis is true). c is the probability we observe the pattern and the beetle is not rare (hypothesis is false).

In this case, a = 0.001 (0.1%); b = 0.98 (98%); c = 0.05 (5%).

So, updated probability = ab/ [ab+c (1-a)] = 0.0192. So there is just a 1.92 per cent chance that the beetle is rare when the entomologist spots the distinctive pattern on its back.

Why the counterintuitive result? Because so few of the population of all beetles are rare, i.e. the prior probability that the beetles is rare is almost vanishingly small and it would take a lot more evidence than that acquired to make a reasonable case for the beetle being rare.

So what is the probability that the beetle is rare given that we observe the distinctive pattern? In other words, what is the probability that the hypothesis (the beetle is rare) is true given the evidence (the pattern). That is 1.92 per cent. What is the probability that we will observe the distinctive pattern if the beetle is rare? In other words, what is the probability of observing the evidence (the pattern) if the hypothesis (the beetle is rare) is true. That is 98 per cent.

To conflate these, to believe these two concepts are the same, is to commit the classic Prosecutor’s Fallacy, i.e. to falsely equate the probability that the defendant is guilty given the observed evidence with the probability of observing the evidence given that the defendant is guilty. It’s a potentially very dangerous fallacy to commit, especially when you happen to be the defendant and the jury has never heard of the Reverend Thomas Bayes.

Appendix

We can also solve the Beetle problem using the traditional notation version of Bayes’ Theorem.

P (HIE) = P (EIH). P (H) / [P (EIH) . P(H) + P (EIH’) . P(H’)]

In this case, P (H) = 0.001 (0.1%); P (EIH) = 0.98 (98%); P (EIH’) = 0.05 (5%).

So, P (HIE) = 0.98 x 0.001/ [0.98 x 0.001 +0.05 x 0.999)] = 0.00098 / 0.00098 + 0.04995 = 0.00098 / 0.05093 = 0.0192. So there is just a 1.92 per cent chance that the beetle is rare when the entomologist spots the distinctive pattern on its back.

Note also that P (HIE) = 0.0192, while P (EIH) = 0.98.

The Prosecutor’s Fallacy is to conflate these two expressions.

Exercise

An entomologist spots what might be a rare category of beetle, due to the pattern on its back. In the rare category, 95% have the pattern. In the common category, only 2% have the pattern. The rare category accounts for only 0.5% of the population. How likely is the beetle to be rare?

References and Links

CS201 – Bayes’ Theorem – Excerpts from Wikipedia

Click to access BayesTheorem.pdf

Jeff Thompson. Bayes’ Theorem. November 20, 2011. https://www.jeffreythompson.org/blog/2011/11/20/bayes-theorem/