# Bayes and the billiard balls: and what’s the probability the Sun rises tomorrow?

This is again from Ten Great Ideas About Chance.

Some years ago I wrote about measuring how fair, or otherwise, a dime coin might be, applying Bayes’s theorem, which requires us to start with a view on the fairness of the coin (the so-called ‘prior’).

Then, as is typical in many such example cases, I made no assumption about fairness and so thought it was equally probable that the coin was completely biased to heads, to tails, and to everything in between.

Bayes, it seems, made a similar initial assumption and used a very appealing analogy to explain this – with billiard balls.

Take N + 1 billiard balls and run them across a snooker table from left to right – assuming you have applied even slightly different forces to them they will end up in different positions.

So if you pick one of the balls how many are to the right or to the left of the picked ball? If we there are M balls to the right of our picked ball we can also model this as how many heads (M) we get out of N coin tosses. So if we pick the rightmost ball then M=0 , the second from right M=1 and if we pick the leftmost then M=N. But for a given M if we pick our ball at random the odds must be:

$\frac{1}{N + 1}$

Which, of course, gives us a flat line as a prior.

And what about the odds on the Sun rising in the morning?

Laplace’s rule of succession, which flows from Bayes’s rule is:

 $\frac{M+1}{N+2}$



(Where there have been N tests and M successes).

Now, if we assume recorded history began in approximately 6000BC and there have been no recorded instances of the Sun failing to rise in the 8000 years this encompasses: that means there have been 2,920,000 tests, all of which have been successes.

So, very approximately, we can assign the odds of 0.9999996575 to the Sun rising tomorrow or, to put it another way, the bookies might want to offer you 3000000 – 1 against or thereabouts that it won’t happen.