“Who could associate mathematics with horror?”

John Buchan answered that question in “Space” (1911), long before H.P. Lovecraft wrote masterpieces like “The Call of Cthulhu” (1926) and “Dreams in the Witchhouse” (1933). But Lovecraft’s use of mathematics is central to his genius. So is his recognition of both the importance and the strangeness of mathematics. Weird fiction and maths go together very well.

But weird fiction is about the intrusion or eruption of the Other into the everyday. Maths can teach you that the everyday is already Other. In short, reality is weird — the World is a Witch House. Let’s start with a situation that isn’t obviously weird. Suppose you had three six-sided dice, A, B and C, each with different set of numbers, like this:

Die A = (1, 2, 3, 6, 6, 6)

Die B = (1, 2, 3, 4, 6, 6)

Die C = (1, 2, 3, 4, 5, 6)

If the dice are fair, i.e. each face has an equal chance of appearing, then it’s clear that, on average, die A will beat both die B and die C, while die B will beat die C. The reasoning is simple: if die A beats die B and die B beats die C, then surely die A will beat die C. It’s a transitive relationship: If Jack is taller than Jim and Jim is taller than John, then Jack is taller than John.

Now try another set of dice with different arrangements of digits:

Die A = (1, 2, 2, 5, 6, 6)

Die B = (1, 1, 4, 5, 5, 5)

Die C = (3, 3, 3, 3, 4, 6)

If you roll the dice, on average die A beats die B and die B beats die C. Clearly, then, die A will also beat die C. Or will it? In fact, it doesn’t: the dice are what is called non-transitive. Die A beats die B and die B beats die C, but die C beats die A.

But how does that work? To see a simpler example of non-transitivity, try a simpler set of random-number generators. Suppose you have a triangle with a short rod passing through its centre at right angles to the plane of the triangle. Now imagine numbering the edges of the triangles (1, 2, 3) and throwing it repeatedly so that it spins in the air before landing on a flat surface. It should be obvious that it will come to rest with one edge facing downward and that each edge has a 1/3 chance of landing like that.

In other words, you could use such a spiked triangle as a random-number generator — you could call it a “trie”, plural “trice”. Examine the set of three trice below. You’ll find that they have the same paradoxical property as the second set of six-sided dice above. Trie A beats trie B, trie B beats trie C, but trie C beats trie A:

Trie A = (1, 5, 8)

Trie B = (3, 4, 7)

Trie C = (2, 3, 9)

When you throw two of the trice, there are nine possible outcomes, because each of three edges on one trie can be matched with three possible edges on the other. The results look like this:

Trie A beats Trie B 5/9ths of the time.

Trie B beats Trie C 5/9ths of the time.

Trie C beats Trie A 5/9ths of the time.

To see how this works, here are the results throw-by-throw:

Trie A = (1, 5, 8)

Trie B = (3, 4, 7)

When Trie A rolls 1…

…and Trie B rolls 3, Trie B wins (Trie A has won 0 out of 1)

…and Trie B rolls 4, Trie B wins (0 out of 2)

…and Trie B rolls 7, Trie B wins (0 out of 3)

When Trie A rolls 5…

…and Trie B rolls 3, Trie A wins (1/4)

…and Trie B rolls 4, Trie A wins (2/5)

…and Trie B rolls 7, Trie B wins (2/6)

When Trie A rolls 8…

…and Trie B rolls 3, Trie A wins (3/7)

…and Trie B rolls 4, Trie A wins (4/8)

…and Trie B rolls 7, Trie A wins (5/9)

Trie B = (3, 4, 7)

Trie C = (2, 3, 9)

When Trie B rolls 3…

…and Trie C rolls 2, Trie B wins (Trie B has won 1 out of 1)

…and Trie C rolls 3, it’s a draw (1 out of 2)

…and Trie C rolls 9, Trie C wins (1 out of 3)

When Trie B rolls 4…

…and Trie C rolls 2, Trie B wins (2/4)

…and Trie C rolls 3, Trie B wins (3/5)

…and Trie C rolls 9, Trie C wins (3/6)

When Trie B rolls 7…

…and Trie C rolls 2, Trie B wins (4/7)

…and Trie C rolls 3, Trie B wins (5/8)

…and Trie C rolls 9, Trie C wins (5/9)

Trie C = (2, 3, 9)

Trie A = (1, 5, 8)

When Trie C rolls 2…

…and Trie A rolls 1, Trie C wins (Trie C has won 1 out of 1)

…and Trie A rolls 5, Trie A wins (1 out of 2)

…and Trie A rolls 8, Trie A wins (1 out of 3)

When Trie C rolls 3…

…and Trie A rolls 1, Trie C wins (2/4)

…and Trie A rolls 5, Trie A wins (2/5)

…and Trie A rolls 8, Trie A wins (2/6)

When Trie C rolls 9…

…and Trie A rolls 1, Trie C wins (3/7)

…and Trie A rolls 5, Trie C wins (4/8)

…and Trie A rolls 8, Trie C wins (5/9)

The same reasoning can be applied to the six-sided non-transitive dice, but there are 36 possible outcomes when two of the dice are thrown against each other, so I won’t list them.

Die A = (1, 2, 2, 5, 6, 6)

Die B = (1, 1, 4, 5, 5, 5)

Die C = (3, 3, 3, 3, 4, 6)

Elsewhere other-posted:

• At the Mountains of Mathness

• Simpson’s Paradox — a simple situation with a very weird outcome