In this city, there are clearly more green cabs than blue ones – you see roughly one blue cab for every five green ones. The prior odds, in other words, are 1:5 (pronounced “one to five”) in favour of the blue company being involved.

However, an eyewitness identified the cab as blue. Court tests of this witness’ reliability under nighttime circumstances indicate a sensitivity of 80 % and false positive rate of 20 %. This translates to an odds ratio of \(0.8/0.2 = 0.8 × 5 = 4\).

Computing the posterior odds is now trivial: prior was 1:5, and odds ratio is 4, so when we multiply we get posterior odds of 4:5, which means after the witness testifies, we think there’s just under an even chance that the blue company was involved rather than the green one.

This problem is usually presented with green cabs being 85 % of the cabs in the city. If we try to apply Bayes’ rule the way it’s normally phrased, we start with

\[p(H|E) = \frac{p(E|H)p(H)}{p(E)}.\]

We can compute \(p(E) = 0.8 × 0.15 + 0.2 × 0.85 = 0.29\) and then plug in the numbers

\[p(H|E) = \frac{0.8 × 0.15}{0.29} = 0.41.\]

In other words, in this more accurate version of the computation we see that the true chance of it being a blue cab is 41 %. For most practical problems, our quicker approximation of “just under even chance” will be fine.

There is just one of urn A, and two urns B and C. This means the odds in favour of being handed urn A is 1:2.

Assuming you were handed urn A, the probability of picking up a red ball would be \(1/5 = 0.2\).

Assuming you were handed one of the other urns, the probability of picking up a red ball would be \((2 + 3)/10 = 1/2 = 0.5\).

Thus, the odds ratio is \(0.2/0.5 = 0.2 × 2 = 0.4\).

We multiply, and get posterior odds

\[1:2 × 0.4 =\] \[= 0.4 : 2 \approx\] \[\approx 0.5 : 2 =\] \[= 1: 4.\]

In other words, we went from a one-in-three chance of holding urn a to a one-in-five chance. Picking up a red ball was evidence against urn A. Specifically, it about halved our belief in holding urn A.

This is the essence of Bayesian hypothesis testing: each competing hypothesis (or parameter value) is a metaphysical urn and then we evaluate what effect the evidence has on our belief of drawing from each urn. The result is a posterior distribution over urns… or hypotheses.

If you have multiple pieces of independent observations, you can multiply their odds ratios together to get the strength of all the evidence combined. (If the observations are correlated, you have to work a bit harder.)

In other words, if in the taxi cab problem we had a second, similar eyewitness independently saying the same thing, the evidence of both combined would have an odds ratio of \(4×4 = 15\), resulting in posterior odds of 16:5, which means now we would think it three times more likely that the culprit was from the blue company rather than the green one.

This is, in fact, what I mean when I say that “the plural of anecdote is data”: there is no magic threshold at which observations become many enough to become significant. Evidence is nothing but small independent anecdotes whose odds ratios multiply together to become a large number.

In fact, whether you are talking about the product of multiple pieces of evidence, or just one observation, the odds ratio measures the strength of the evidence. A sensitivity of 80 % for the witness in the taxi cab problem might have sounded weak, but the odds ratio tells us that this single observation quadruples our belief in the blue company being guilty. That’s some strong evidence!

(This is even more natural when you work in log-odds instead of odds, because then you just sum together the strength of the evidence expressed in log-odds-difference. The problem is that we generally don’t have intuition for estimating and working with log-odds. This is something I want to practise but I haven’t gotten around to it yet.)

Sometimes it’s hard to estimate the probability of low probability events. For example, how likely do I think it is that it was snowing in Stockholm in the summer of last year? I have no idea.

If someone told me it did snow in Stockholm in the summer of last year, I might double my belief in it – but that doesn’t really help me estimate my prior. However, if six independent people told me the same thing, I might judge that snow a 50–50 event, or having odds of 1. So, going backwards, my prior odds must have been \(1/2^6 \approx 1/50 = 0.02\).

Low odds (roughly \(< 0.15\)) are easy to convert to probabilities because they’re basically the same number. So odds of 0.02 means a probability of 2 %.

Thus, I can conclude that my prior odds for snow the previous summer in Stockholm must have been 2 %. This is higher than I would have thought, but I also trust this number more than my gut reaction.