Bayes' Theorem: The False Positive Paradox

Bayes' Theorem

$P(A|B) = \frac{P(B|A) \cdot P(A)}{P(B)}$

The false positive paradox

Setup: Disease prevalence = 1%. Test sensitivity = 99% (true positive rate). Test specificity = 99% (true negative rate).

Question: You test positive. What's the probability you actually have the disease?

Step-by-step (natural frequencies method)

Imagine 10,000 people:

Step 1: 100 have the disease (1%), 9,900 don't.

Step 2: Of 100 with disease: 99 test positive (99% sensitivity), 1 tests negative.

Step 3: Of 9,900 without disease: 99 test positive (1% false positive), 9,801 test negative.

Step 4: Total positive tests: $99 + 99 = 198$.

Step 5: Of 198 positives, only 99 actually have the disease.

$P(\text{disease} | \text{positive}) = \frac{99}{198} = 50\%$

The surprise

Even with a 99% accurate test, a positive result only means a 50% chance of having the disease! This happens because the disease is rare (1%), so the 1% false positive rate applied to the 99% healthy population produces as many false positives as true positives.

Try it in the visualization

The tree diagram shows 10,000 people splitting into branches. The surprise is visual: the "false positive" branch is just as large as the "true positive" branch.