# Bayesian Dice#

Click here to run this notebook on Colab

I’ve been enjoying Aubrey Clayton’s new book *Bernoulli’s Fallacy*. Chapter 1, which is about the historical development of competing definitions of probability, is worth the price of admission alone.

One of the examples in the first chapter is a simplified version of a problem posed by Thomas Bayes. The original version, which I wrote about here, involves a billiards (pool) table; Clayton’s version uses dice:

Your friend rolls a six-sided die and secretly records the outcome; this number becomes the target

T. You then put on a blindfold and roll the same six-sided die over and over. You’re unable to see how it lands, so each time your friend […] tells youonlywhether the number you just rolled was greater than, equal to, or less thanT.Suppose in one round of the game we had this sequence of outcomes, with G representing a greater roll, L a lesser roll, and E an equal roll:

G, G, L, E, L, L, L, E, G, L

Based on this data, what is the posterior distribution of *T*?

## Computing likelihoods#

There are two parts of my solution; computing the likelihood of the data under each hypothesis and then using those likelihoods to compute the posterior distribution of *T*.

To compute the likelihoods, I’ll demonstrate one of my favorite idioms, using a meshgrid to apply an operation, like `>`

, to all pairs of values from two sequences.

In this case, the sequences are

`hypos`

: The hypothetical values of*T*, and`outcomes`

: possible outcomes each time we roll the dice

```
hypos = [1,2,3,4,5,6]
outcomes = [1,2,3,4,5,6]
```

If we compute a meshgrid of `outcomes`

and `hypos`

, the result is two arrays.

```
import numpy as np
O, H = np.meshgrid(outcomes, hypos)
```

The first contains the possible outcomes repeated down the columns.

```
O
```

```
array([[1, 2, 3, 4, 5, 6],
[1, 2, 3, 4, 5, 6],
[1, 2, 3, 4, 5, 6],
[1, 2, 3, 4, 5, 6],
[1, 2, 3, 4, 5, 6],
[1, 2, 3, 4, 5, 6]])
```

The second contains the hypotheses repeated across the rows.

```
H
```

```
array([[1, 1, 1, 1, 1, 1],
[2, 2, 2, 2, 2, 2],
[3, 3, 3, 3, 3, 3],
[4, 4, 4, 4, 4, 4],
[5, 5, 5, 5, 5, 5],
[6, 6, 6, 6, 6, 6]])
```

If we apply an operator like `>`

, the result is a Boolean array.

```
O > H
```

```
array([[False, True, True, True, True, True],
[False, False, True, True, True, True],
[False, False, False, True, True, True],
[False, False, False, False, True, True],
[False, False, False, False, False, True],
[False, False, False, False, False, False]])
```

Now we can use `mean`

with `axis=1`

to compute the fraction of `True`

values in each row.

```
(O > H).mean(axis=1)
```

```
array([0.83333333, 0.66666667, 0.5 , 0.33333333, 0.16666667,
0. ])
```

The result is the probability that the outcome is greater than *T*, for each hypothetical value of *T*.
I’ll name this array `gt`

:

```
gt = (O > H).mean(axis=1)
gt
```

```
array([0.83333333, 0.66666667, 0.5 , 0.33333333, 0.16666667,
0. ])
```

The first element of the array is 5/6, which indicates that if *T* is 1, the probability of exceeding it is 5/6.
The second element is 2/3, which indicates that if *T* is 2, the probability of exceeding it is 2/3.
And do on.

Now we can compute the corresponding arrays for less than and equal.

```
lt = (O < H).mean(axis=1)
lt
```

```
array([0. , 0.16666667, 0.33333333, 0.5 , 0.66666667,
0.83333333])
```

```
eq = (O == H).mean(axis=1)
eq
```

```
array([0.16666667, 0.16666667, 0.16666667, 0.16666667, 0.16666667,
0.16666667])
```

In the next section, we’ll use these arrays to do a Bayesian update.

## The Update#

In this example, computing the likelihoods was the hard part. The Bayesian update is easy.
Since *T* was chosen by rolling a fair die, the prior distribution for *T* is uniform.
I’ll use a Pandas `Series`

to represent it.

```
import pandas as pd
pmf = pd.Series(1/6, hypos)
pmf
```

```
1 0.166667
2 0.166667
3 0.166667
4 0.166667
5 0.166667
6 0.166667
dtype: float64
```

Now here’s the sequence of data, encoded using the likelihoods we computed in the previous section.

```
data = [gt, gt, lt, eq, lt, lt, lt, eq, gt, lt]
```

The following loop updates the prior distribution by multiplying by each of the likelihoods.

```
for datum in data:
pmf *= datum
```

Finally, we normalize the posterior.

```
pmf /= pmf.sum()
pmf
```

```
1 0.000000
2 0.016427
3 0.221766
4 0.498973
5 0.262834
6 0.000000
dtype: float64
```

Here’s what it looks like.

```
pmf.plot.bar(xlabel='Target value',
ylabel='PMF',
title='Posterior distribution of $T$');
```

As an aside, you might have noticed that the values in `eq`

are all the same.
So when the value we roll is equal to $T$, we don’t get any new information about *T*.
We could leave the instances of `eq`

out of the data, and we would get the same answer.