# The Bayesian umbrella

At a recent MathsJam, there was a puzzle. This is nothing out of the ordinary. It went something like:

If an absent-minded professor takes his umbrella into a classroom, there’s a probability of $\frac{1}{4}$ that he’ll absent-mindedly leave it there. One day, he sets off with his umbrella, teaches in three classrooms, and comes back to his office… without his umbrella. What’s the probability he left it in the first classroom?

A fragile consensus immediately formed around $\frac{1}{4}$ - trick question! - until we realised we had more information: he’d left his umbrella *somewhere*, so we’d need something more complicated. As usual with this kind of question ((where ‘this kind of question’ means ‘one you have to think about’)) , there’s a brutal slog way to do it; once you’ve done that, you can then see a more elegant way to do it.

If we don’t know about him losing his umbrella somewhere, we can use a probability tree sort of method to say:

- there’s a $\frac{1}{4}$ probability of him leaving it in the first classroom;
- there’s a $\frac{3}{4} \times \frac{1}{4} = \frac{3}{16}$ probability of leaving it in classroom 2;
- there’s a $\frac{3}{4} \times \frac{3}{4} \times \frac{1}{4} = \frac{9}{64}$ probability of leaving it in classroom 3;
- there’s a $\left(\frac{3}{4}\right)^3 = \frac{27}{64}$ probability of keeping hold of it.

Which is to say, over any 64 days, he’d expect to leave it in the first classroom 16 times, in the second 12 times, and in the third 9 times. That means the probability of it being in the first classroom, given that he left it somewhere, is $\frac{16}{16+12+9} = \frac{16}{37}$, about 43%.

A more simple way is to use Bayes’ rule and say it’s the probability of him leaving it somewhere and leaving it in classroom 1 ($\frac{1}{4}$) out of the probability of him leaving it somewhere ($1 - \frac{27}{64}$), giving $\frac{1}{4} \div \frac{37}{64} = \frac{16}{37}$.

## A selection of other posts

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