# Bayesian billiards

In his seminal paper "An Essay towards solving a Problem in the Doctrine of Chances", Thomas Bayes introduced a thought experiment involving six balls thrown randomly onto a billiards table. This example is often used in popular science as an introduction to the ideas of Bayesian statistics, as it nicely demonstrates the difference between a frequentist maximum likelihood estimate and a Bayesian point estimate. Having seen this example before, I had also heard a claim that the Bayesian estimate is "slightly better", without having heard a clarification of what "better" means in this context. In this post I'll describe the problem and compare the two approaches on simulated data.

## The problem

Let's begin by stating Bayes' thought experiment. This description is taken from The Art of Statistics by David Spiegelhalter (which is a very enjoyable read):

Suppose a white ball is thrown at random on to a billiards table, its position along the table marked with a line, and then the white ball is removed. A number of red balls are then thrown at random on to the table, and you are told only how many lie to the left and how many to the right of the line. Where do you think the line might be, and what should be your probability of the next red ball falling to the left of the line.

Let's set up some notation so that we can discuss possible solutions. First we'll denote the position of the white ball by $\theta$. The exact length of the table is irrelevant, so we'll assume that $0 \leq \theta \leq 1$, so that $\theta$ represents the proportion of the table to the left of the line. Since the ball is thrown randomly, it is reasonable to assume that

$\theta \sim \mathrm{Unif}(0, 1)$Next suppose that we have $n$ red balls that we throw randomly onto the table. The probability that any one red ball stops to the left of the line is $\theta$, so the number $y$ of red balls that end up to the left of the line can be modelled as

$y \sim \mathrm{Binomial}(n, \theta)$Our goal is to come up with the best possible estimate of $\theta$ given $n$ and $y$.

## The maximum likelihood estimate

Maximum likelihood estimation produces an estimate for the unknown parameters in a model by choosing those parameters that maximise the likelihood. In our case, we want to maximise

$L(\theta) := p(y | \theta) = \binom{n}{y}\theta^y (1 - \theta)^{n-y}$Equivalently, since $\log$ is a monotone function, we can maximise the log-likelihood

$l(\theta) := \log(p(y | \theta)) = y\log\theta + (n-y)\log(1 - \theta) + C$where $C$ is a constant independent of $\theta$. We can easily differentiate $l$ to find a local maximum at

$\hat \theta_{MLE} = \frac{y}{n}$In other words, the maximum likelihood estimate for the position of the line is simply the proportion of red balls that ended up to the left of the line.

## A Bayesian point estimate

We have already specified our prior (uniform distribution on $\theta$) and sampling distribution (binomial distribution on $y$), so all that remains is to calculate the posterior. By Bayes' rule

$p(\theta | y) \propto p(\theta)p(y | \theta) \propto \theta^y (1 - \theta)^{n-y}$We recognise the right hand side as an unnormalised Beta density, and so conclude that

$\theta | y \sim \mathrm{Beta}(y + 1, n - y + 1)$This gives us a distribution over possible positions of the line based on the observed data. In order to turn this into a point estimate, we could require for example that our expected squared loss over the posterior distribution is minimised

$\hat \theta_{Bayes} = \mathrm{argmin}_{\tilde \theta} \mathbb{E}[(\theta - \tilde \theta)^2 | y]$which is in fact the same as simply choosing $\hat \theta_{Bayes}$ to be the posterior mean

$\hat \theta_{Bayes} = \mathbb{E}[\theta | y]$In our case, with the Beta posterior this means that

$\hat \theta_{Bayes} = \frac{y + 1}{n + 2}$This is somewhat similar to the maximum likelihood estimate, but the modified numerator and denominator means that our estimates are always pulled back towards the center of the table. Most notable if $0$ of the red balls landed to the left of the line, the maximum likelihood estimate is that the line is on the left edge of the table, whereas the Bayesian posterior mean puts the line at $1 / 7 \approx 0.143$, which intuitively seems more reasonable.

This difference is a consequence of the fact that the Bayesian estimate has built into it the knowledge that we threw the white ball at random, and moderates the chance of making extreme inferences like drawing the line right on the edge of the table. This property is also more generally true of Bayesian method: priors can help regularise our inferences, particularly when we have limited data.

## Comparing the two estimates

To compare the estimates, let's do some simulation. Using NumPy it's very easy for us to perform this experiment thousands or even millions of times.

```
import numpy as np
def simulate(n_trials=1_000_000, n_red=5):
# throw a white ball and n_red red balls onto n_trials tables
white = np.random.random(size=n_trials)
red = np.random.random(size=(n_trials, n_red))
# count the number of red balls to the left of the white ball
n_red_left = (red < white[:, None]).sum(axis=1)
# construct the two estimates
mle = n_red_left / n_red
posterior_mean = (n_red_left + 1) / (n_red + 2)
return white, mle, posterior_mean
```

We can repeat the experiment a million times, recording the true location of the line, and the maximum likelihood and Bayesian estimates.

`white, mle, posterior_mean = simulate(1_000_000)`

### Brier score

The Brier score is a proper scoring rule that measures the quality of probabilistic forecasts. Suppose we make a forecast $0 \leq \hat \theta \leq 1$ that the next red ball will land to the left of the line. We then observe the outcome $o = 0, 1$. The Brier score for this single forecast is

$(\hat \theta - o)^2$that is, it's the square of the probability we assigned to the true outcome
*not* occuring. If we made a forecast $\hat \theta = 1$, the Brier score would
be $0$ if we are correct and $o = 1$, but $1$ if we were wrong and $o = 0$! If
the true outcome is not certain, maybe a more conservative forecast would be
$\hat \theta = 0.7$ in which case we would get a score of $0.3 ^ 2 = 0.09$ if
$o = 1$ and $0.7^2 = 0.49$ if $o = 0$. In general, we minimise the expected
Brier score if $\hat \theta = P(o = 1)$, meaning our forecast is encouraged to
be well calibrated.

If we make a sequence $\hat \theta_i$ of $N$ forecasts with corresponding outcomes, the Brier score is simply an average over all forecasts

$\frac{1}{N} \sum_{i=1}^N (\hat \theta_i - o_i)^2$Why square the difference rather than take the absolute value? This is to encourage proper calibration of forecasts. That is to say if an outcome occurs 70% of the time, the best forecast should be to assign a probability $0.7$ to that outcome. If the score were the absolute difference $|\hat \theta - o|$ then we are actually incentivised to exaggerate our confidence. If $P(o=1) = 0.7$ then a forecast $\hat \theta \equiv 0.7$ yields a score

$\mathbb{E}[|o - 0.7|] = 0.3 \times 0.7 + 0.7 \times 0.3 = 0.42$whereas

$\mathbb{E}[|o - 1|] = 0.3 \times 1 + 0.7 \times 0 = 0.3$so our expected score is actually lower for the worse forecast!

#### Brier score - results

I calculated the Brier scores for both estimates as follows

```
next_ball = np.random.binomial(1, white)
mle_brier = ((mle - next_ball) ** 2).mean()
bayes_brier = ((posterior_mean - next_ball) ** 2).mean()
print(f"MLE Brier score: {mle_brier:.4f}")
print(f"Bayes Brier score: {bayes_brier:.4f}")
```

which output

```
MLE Brier score: 0.2002
Bayes Brier score: 0.1906
```

## Conclusion

While it did turn out that the Bayesian posterior mean wound up having a lower Brier score, the difference was relatively small, so maybe it's not a compelling reason to choose one approach over the other. It is however a nice illustration of the fact that Bayesian inference is able to incorporate domain knowledge. In this case the fact that we know the white ball was thrown randomly and so would be equally likely to be anywhere on the table. The maximum likelihood estimate doesn't account for this knowledge, and is therefore prone to extreme inferences like estimating the line is right on one of the edges of the table.