# Sprinkle on the sugar, eat the lot

A Christmas Pudding Puzzle

I swear, this one came up in real life!

My partner made a Christmas pudding for the most recent festive season. Delicious, it was.

When it was about half-eaten, I went to microwave a portion. “Hang on,” she said: “there might be a coin in there.”

As is traditional, she had mixed in at least one threepenny bit ((actually, 5p coins, to account for inflation)); as is habitual, she couldn’t remember whether it was one coin or half a dozen.

We could remember that her brother was the only one so far to have found a coin.

So, how likely is it that my portion, let’s say a sixth of the remainder, contains a coin?

### Bring on the Rev!

This is a classic opportunity to apply a bit of Bayesian analysis.

I think Christmas pudding is continuous, even though it is served in portions. It probably wouldn’t make much difference to model things with a binomial rather than a Poisson distribution, but I’m in a Poisson-y sort of mood.

Suppose the number of coins in the whole Christmas pudding is $\Lambda$, with a prior belief that $\Lambda=1$ and $\Lambda=6$ are equally likely.

We have observed one “hit” in half a pudding.

Now we can simply check the probability of that under our two possibilities:

- $p_1 = P\br{X=1|\lambda=\frac{1}{2}} = \frac{1}{2} e^{-\frac{1}{2}} \approx 0.303$
- $p_6 = P\br{X=1|\lambda=3} = 3e^{-3} \approx 0.149$

Our posterior belief is that $\Lambda=1$ with probability $\frac{p_1}{p_1+p_6}\approx 0.670$ and $\Lambda=6$ with probability $0.330$.

This is useful: I can say there’s about a two-in-three chance of the only coin having been found already, and a one-in-three chance of there still being five left to find.

### But what about the next portion?

With this in mind, I can estimate the probability of finding a coin in my pudding!

In case one, there are no more coins.

In case two, there are five coins, and I’m about to eat a sixth of what’s left. Again, this looks a bit Poisson-y; I’d expect $Y \sim Po\br{\frac{5}{6}}$.

The probability of getting no coin is $P(Y=0) = e^{-\frac{5}{6}} \approx 0.435$ - nearly half the time, even if there are five coins still to find, I’d still be coin-free.

That means, my probability of being able to use the microwave without it blowing up would be roughly $0.670 + 0.330 \times 0.435 \approx 0.813$ - a little more than four times out of five, there’s no problem.

So, is it OK to use the microwave? Well, I wouldn’t take those odds. We chopped the pudding up into small bits to check for coins – and it turned out there were none.

An exercise for you, dear reader: what was the probability my *next* portion would be coin-free?