— Croyez-vous aux idées dangereuses ?
— Qu’entendez-vous par là ?
— Croyez-vous que certaines idées soient aussi dangereuses pour certains esprits que le poison pour le corps ?
— Mais, oui, peut-être.
Guy de Maupassant, « Divorce » (1888)
“Do you believe in dangerous ideas?”
“What do you mean by that?”
“Do you believe that certain ideas are as dangerous for some minds as poison is for the body?”
“Well, yes, perhaps.”
“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
A beautifully subtle puzzle:
Scrambled Box Tops
Imagine you have three boxes, one containing two black marbles, one containing two white marbles, and the third, one black and one white marble. The boxes are labelled according to their contents — BB, WW, and BW — but someone has switched the labels so that every box is now incorrectly labelled. You are allowed to take one marble at a time out of any box, without looking inside, and by this process of sampling you are to determine the contents of all three boxes. What is the smallest number of drawings needed to do this? — Martin Gardner, Mathematical Puzzles and Diversions (1959), chapter 3, “Nine Problems”, #5.
Bald eagle, Haliaeetus leucocephalus (Linnaeus 1776)
Answer: You can learn the contents of all three boxes by drawing just one marble. The key to the solution is your knowledge that the labels on all three boxes are incorrect. You must draw a marble from the box labelled “black-white”. Assume that the marble drawn is black. You know then that the other marble in the box must be black also, otherwise the label on the box would be correct. Since you have now identified the box containing two black marbles, you can tell at once the contents of the box labelled “white-white”: you know it cannot contain two white marbles, because its label has to be wrong; it cannot contain two black marbles, because you have identified that box; therefore it must contain one black and one white marble. The third box, of course, must then be the one containing two white marbles. You can solve the puzzle by the same reasoning if the marble you draw from the “black-white” box happens to be white instead of black.
“Instead, Mr Comey has rocket-fuelled a venomous contest just when Mr Trump was desperate for a lifeline…” — The Guardian view on the FBI’s Clinton probe: exactly the wrong thing to do
Previously pre-posted…
Overlord In Terms of Core Issues Around Maximal Engagement with Key Notions of the Über-Feral 