“In no other branch of mathematics is it so easy for experts to blunder as in probability theory.” — Martin Gardner (1914-2010)
A once very difficult but now very simple problem in probability from Ian Stewart’s Do Dice Play God? (2019):
For three dice [Girolamo] Cardano solved a long-standing conundrum [in the sixteenth century]. Gamblers had long known from experience that when throwing three dice, a total of 10 is more likely than 9. This puzzled them, however, because there are six ways to get a total of 10:
1+4+5; 1+3+6; 2+4+4; 2+2+6; 2+3+5; 3+3+4
But also six ways to get a total of 9:
1+2+6; 1+3+5; 1+4+4; 2+2+5; 2+3+4; 3+3+3
So why does 10 occur more often?
To see the answer, imagine throwing three dice of different colors: red, blue and yellow. How many ways can you get 9 and how many ways can you get 10?
| Roll | Total=9 | Dice #1 (Red) | Dice #2 (Blue) | Dice #3 (Yellow) |
| 01 | 9 = | 1 | 2 | 6 |
| 02 | 9 = | 1 | 3 | 5 |
| 03 | 9 = | 1 | 4 | 4 |
| 04 | 9 = | 1 | 5 | 3 |
| 05 | 9 = | 1 | 6 | 2 |
| 06 | 9 = | 2 | 1 | 6 |
| 07 | 9 = | 2 | 2 | 5 |
| 08 | 9 = | 2 | 3 | 4 |
| 09 | 9 = | 2 | 4 | 3 |
| 10 | 9 = | 2 | 5 | 2 |
| 11 | 9 = | 2 | 6 | 1 |
| 12 | 9 = | 3 | 1 | 5 |
| 13 | 9 = | 3 | 2 | 4 |
| 14 | 9 = | 3 | 3 | 3 |
| 15 | 9 = | 3 | 4 | 2 |
| 16 | 9 = | 3 | 5 | 1 |
| 17 | 9 = | 4 | 1 | 4 |
| 18 | 9 = | 4 | 2 | 3 |
| 19 | 9 = | 4 | 3 | 2 |
| 20 | 9 = | 4 | 4 | 1 |
| 21 | 9 = | 5 | 1 | 3 |
| 22 | 9 = | 5 | 2 | 2 |
| 23 | 9 = | 5 | 3 | 1 |
| 24 | 9 = | 6 | 1 | 2 |
| 25 | 9 = | 6 | 2 | 1 |
| Roll | Total=10 | Dice #1 (Red) | Dice #2 (Blue) | Dice #3 (Yellow) |
| 01 | 10 = | 1 | 3 | 6 |
| 02 | 10 = | 1 | 4 | 5 |
| 03 | 10 = | 1 | 5 | 4 |
| 04 | 10 = | 1 | 6 | 3 |
| 05 | 10 = | 2 | 2 | 6 |
| 06 | 10 = | 2 | 3 | 5 |
| 07 | 10 = | 2 | 4 | 4 |
| 08 | 10 = | 2 | 5 | 3 |
| 09 | 10 = | 2 | 6 | 2 |
| 10 | 10 = | 3 | 1 | 6 |
| 11 | 10 = | 3 | 2 | 5 |
| 12 | 10 = | 3 | 3 | 4 |
| 13 | 10 = | 3 | 4 | 3 |
| 14 | 10 = | 3 | 5 | 2 |
| 15 | 10 = | 3 | 6 | 1 |
| 16 | 10 = | 4 | 1 | 5 |
| 17 | 10 = | 4 | 2 | 4 |
| 18 | 10 = | 4 | 3 | 3 |
| 19 | 10 = | 4 | 4 | 2 |
| 20 | 10 = | 4 | 5 | 1 |
| 21 | 10 = | 5 | 1 | 4 |
| 22 | 10 = | 5 | 2 | 3 |
| 23 | 10 = | 5 | 3 | 2 |
| 24 | 10 = | 5 | 4 | 1 |
| 25 | 10 = | 6 | 1 | 3 |
| 26 | 10 = | 6 | 2 | 2 |
| 27 | 10 = | 6 | 3 | 1 |
Imagine a game with six players, numbered #1 to #6, and one six-sided die. Someone rolls the die and the player who matches the number wins the game. That is, if the die rolls 1, player #1 wins; if the die rolls 2, player #2 wins; and so on. With a fair die, this is a fair game, because each player has exactly a 1/6 chance of winning. You could call it a simultaneous game, because all players are playing at once. It has one rule:
• If the die rolls n, then player #n wins.
Now try a different game with six players and one die. Player #1 rolls the die. If he gets 1, he wins the game. If not, then he leaves the game and player #2 rolls the die. If he gets 2, he wins the game. If not, then he leaves the game and player #3 rolls the die. And so on. You could call this a sequential game, because the players are playing in sequence. It has two rules:
• If player #n rolls n on the die, then he wins.
• If player #n doesn’t roll n, then player n+1 rolls the die.
Is it a fair game? No, definitely not. Player #1 has the best chance of winning. 1/6 or 16.6% of the time he rolls 1 and wins the game. 5/6 of the time, he rolls 2, 3, 4, 5 or 6 and passes the die to player #2. Now player #2 has a 1/6 chance of rolling a 2 and winning. But he has the opportunity to roll the die only 5/6 of the time, so his chance of winning the game is 1/6 * 5/6 = 5/36 = 13.8%. However, if player #2 rolls a 1, 3, 4, 5 or 6, then he loses and player #3 rolls the die. But player #3 has that opportunity only 5/6 * 5/6 = 25/36 of the time. So his chance of winning is 1/6 * 25/36 = 11.57%. And so on.
To put it another way, if the six players play 46656 = 6^6 games under the sequential rules, then on average:
• Player #1 wins 7776 games
• Player #2 wins 6480 games
• Player #3 wins 5400 games
• Player #4 wins 4500 games
• Player #5 wins 3750 games
• Player #6 wins 3125 games
• 15625 games end without a winner.
In other words, player #1 is 20% more likely to win than player #2, 44% more likely than player #3, 72.8% more likely than player #4, 107% more likely than player #5, and 148.8% more likely than player #6. Furthermore, player #2 is 20% more likely to win than player #3, 44% more likely than player #4, 72.8% more likely than player #5, and so on.
But there is a simple way to make the sequential game perfectly fair, so long as it’s played with a fair die. At least, I’ve thought of a simple way, but there might be more than one.
To make the sequential game fair, you add an extra rule:
1. If player #n rolls n on the die, he wins the game.
2. If player #n rolls a number greater than n, he loses and the die passes to player n+1.
3. If player #n rolls a number less than n, then he rolls again.
Let’s run through a possible game to see that it’s fair. Player #1 rolls first. He has a 1/6 chance of rolling a 1 and winning the game. However, 5/6 of the time he loses and passes the die to player #2. If player #2 rolls a 1, he rolls again. In other words, player #2 is effectively playing with a five-sided die, because all rolls of 1 are ignored. Therefore, he has a 1/5 chance of winning the game at that stage.
But hold on: a 1/5 chance of winning is better than a 1/6 chance, which is what player #1 had. So how is the game fair? Well, note the qualifying phrase at the end of the previous paragraph: at that stage. The game doesn’t always reach that stage, because if player #1 rolls a 1, the game is over. Player #2 rolls only if player doesn’t roll 1, which is 5/6 of the time. Therefore player #2’s chance of winning is really 1/5 * 5/6 = 5/30 = 1/6.
However, 4/5 of the time player #2 rolls a 3, 4, 5 or 6 and the die passes to player #3. If player #3 rolls a 1 or 2, he rolls again. In other words, player #3 is effectively playing with a four-sided die, because all rolls of 1 and 2 are ignored. Therefore, he has a 1/4 chance of winning the game at that stage.
A 1/4 chance of winning is better than a 1/5 chance and a 1/6 chance, but the same reasoning applies as before. Player #3 rolls the die only 5/6 * 4/5 = 20/30 = 2/3 of the time, so his chance of winning is really 1/4 * 2/3 = 2/12 = 1/6.
However, 3/4 of the time player #2 rolls a 4, 5 or 6 and the die passes to player #4. If player #4 rolls a 1, 2 or 3, he rolls again. In other words, player #4 is effectively playing with a three-sided die, because all rolls of 1, 2 and 3 are ignored. Therefore, he has a 1/3 chance of winning the game at that stage. 1/3 > 1/4 > 1/5 > 1/6, but the same reasoning applies as before. Player #4 rolls the die only 5/6 * 4/5 * 3/4 = 60/120 = 1/2 of the time, so his chance of winning is really 1/3 * 1/2 = 1/6.
And so on. If the die reaches player #5 and he gets a 1, 2, 3 or 4, then he rolls again. He is effectively rolling with a two-sided die, so his chance of winning is 1/2 * 5/6 * 4/5 * 3/4 * 2/3 = 120/720 = 1/6. If player #5 rolls a 6, he loses and the die passes to player #6. But there’s no need for player #6 to roll the die, because he’s bound to win. He rolls again if he gets a 1, 2, 3, 4 or 5, so eventually he must get a 6 and win the game. If player #5 loses, then player #6 automatically wins.
It’s obvious that this form of the game will get slower as more players drop out, because later players will be rolling again more often. To speed the game up, you can refine the rules like this:
1. If Player #1 rolls a 1, he wins the game. Otherwise…
2. If player #2 rolls a 2, he wins the game. If he rolls a 1, he rolls again. Otherwise…
3. Player #3 rolls twice and adds his scores. If the total is 3, 4 or 5, he wins the game. Otherwise…
4. Player #4 rolls once. If he gets 1 or 2, he wins the game. Otherwise…
5. Player #5 rolls once. If he gets 1, 2 or 3, he wins the game. Otherwise…
6. Player #6 wins the game.
Only player #2 might have to roll more than twice. Player #3 has to roll twice because he needs a way to get a 1/4 chance of winning. If you roll two dice, there are:
• Two ways of getting a total of 3: roll #1 is 1 and roll #2 is 2, or vice versa.
• Three ways of getting a total of 4 = 1+3, 3+1, 2+2.
• Four ways of getting 5 = 1+4, 4+1, 2+3, 3+2.
This means player #3 has 2 + 3 + 4 = 9 ways of winning. But there are thirty-six ways of rolling one die twice. Therefore player #3 has a 9/36 = 1/4 chance of winning. Here are the thirty-six ways of rolling one die twice, with asterisks marking the winning totals for player #3:
01. (1,1)
02. (1,2)*
03. (2,1)*
04. (1,3)*
05. (3,1)*
06. (1,4)*
07. (4,1)*
08. (1,5)
09. (5,1)
10. (1,6)
11. (6,1)
12. (2,2)*
13. (2,3)*
14. (3,2)*
15. (2,4)
16. (4,2)
17. (2,5)
18. (5,2)
19. (2,6)
20. (6,2)
21. (3,3)
22. (3,4)
23. (4,3)
24. (3,5)
25. (5,3)
26. (3,6)
27. (6,3)
28. (4,4)
29. (4,5)
30. (5,4)
31. (4,6)
32. (6,4)
33. (5,5)
34. (5,6)
35. (6,5)
36. (6,6)
“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
You can stop reading now, if you want. Or can you? Are your decisions really your own, or are you and all other human beings merely spectators in the mind-arena, observing but neither influencing nor initiating what goes on there? Are all your apparent choices in your brain, but out of your hands, made by mechanisms beyond, or below, your conscious control?
In short, do you have free will? This is a big topic – one of the biggest. For me, the three most interesting things in the world are the Problem of Consciousness, the Problem of Existence and the Question of Free Will. I call consciousness and existence problems because I think they’re real. They’re actually there to be investigated and explained. I call free will a question because I don’t think it’s real. I don’t believe that human beings can choose freely or that any possible being, natural or supernatural, can do so. And I don’t believe we truly want free will: it’s an excuse for other things and something we gladly reject in certain circumstances.
Continue reading The Brain in Pain…
It’s odd that probability theory is so counter-intuitive to human beings and so late-flowering in mathematics. Men have been gambling for thousands of years, but didn’t develop a good understanding of what happens when dice are rolled or coins are tossed until a few centuries ago. And an intuitive grasp of probability would have been useful long before gambling was invented. Our genes automatically equip us to speak, to walk and to throw, but they don’t equip us to understand by instinct why five-tails-in-a-row makes heads no more likely on the sixth coin-toss than it was on the first.
Dice and gambling tokens from ancient Rome
Or to understand why five-boys-in-a-row makes the birth of a girl next time no more likely than it was during the first pregnancy (at least in theory). Boy/girl, like heads/tails, is a binary choice, so binary numbers are useful for understanding the probabilities of birth or coin-tossing. Questions like these are often asked to test knowledge of elementary probability:
1. Suppose a family have two children and the elder is a boy. What is the probability that both are boys?
2. Suppose a family have two children and at least one is a boy. What is the probability that both are boys?
People sometimes assume that the two questions are equivalent, but binary makes it clear that they’re not. If 1 represents a boy, 0 represents a girl and digit-order represents birth-order, the first question covers these possibilities: 10, 11. So the chance of both children being boys is 1/2 or 50%. The second question covers these possibilities: 10, 01, 11. So the chance of both children being boys is 1/3 = 33·3%. But now examine this question:
3. Suppose a family have two children and only one is called John. What is the probability that both children are boys?
That might seem the equivalent of question 2, but it isn’t. The name “John” doesn’t just identify the child as a boy, it identifies him as a unique boy, distinct from any brother he happens to have. Binary isn’t sufficient any more. So, while boy = 1, John = 2. The possibilities are: 20, 21, 02, 12. The chance of both children being boys is then 1/2 = 50%.
The three questions above are very simple, but I don’t think Archimedes or Euclid ever addressed the mathematics behind them. Perhaps they would have made mistakes if they had. I hope I haven’t, more than two millennia later. Perhaps the difficulty of understanding probability relates to the fact that it involves movement and change. The Greeks developed a highly sophisticated mathematics of static geometry, but did not understand projectiles or falling objects. When mathematicians began understood those in Renaissance Italy, they also began to understand the behaviour of dice, coins and cards. Ideas were on the move then and this new mathematics was obviously related to the rise of science: Galileo (1564-1642) is an important figure in both fields. But the maths and science can be linked with apparently distinct phenomena like Protestantism and classical music. All of these things began to develop in a “band of genius” identified by the American researcher Charles Murray. It runs roughly from Italy through France and Germany to Scotland: from Galileo through Beethoven and Descartes to David Hume.
Map of Europe from Mercator’s Atlas Cosmographicae (1596)
But how far is geography also biology? Having children is a form of gambling: the dice of DNA, shaken in testicle- and ovary-cups, are rolled in a casino run by Mother Nature. Or rather, in a series of casinos where different rules apply: the genetic bets placed in Africa or Europe or Asia haven’t paid off in the same way. In other words, what wins in one place may lose in another. Different environments have favoured different sets of genes with different effects on both bodies and brains. All human beings have many things in common, but saying that we all belong to the same race, the human race, is like saying that we all speak the same language, the human language. It’s a ludicrous and anti-scientific idea, however widely it may be accepted (and enforced) in the modern West.
Languages have fuzzy boundaries. So do races. Languages have dialects and accents, and so, in a sense, do races. The genius that unites Galileo, Beethoven and Hume may have been a particular genetic dialect spoken, as it were, in a particular area of Europe. Or perhaps it’s better to see European genius as a series of overlapping dialects. Testing that idea will involve mathematics and probability theory, and the computers that crunch the data about flesh will run on binary. Apparently disparate things are united by mathematics, but maths unites everything partly because it is everything. Understanding the behaviour of dice in the sixteenth century leads to understanding the behaviour of DNA in the twenty-first.
The next step will be to control the DNA-dice as they roll. China has already begun trying to do that using science first developed in the West. But the West itself is still in the thrall of crypto-religious ideas about equality and environment. These differences have biological causes: the way different races think about genetics, or persuade other races to think about genetics, is related to their genetics. You can’t escape genes any more than you can escape maths. But the latter is a ladder that allows us to see over the old genetic wall and glimpse the possibilities beyond it. The Chinese are trying to climb over the wall using super-computers; the West is still insisting that there’s nothing on the other side. Interesting times are ahead for both flesh and binary.
Appendix
1. Suppose a family have three children and the eldest is a girl. What is the probability that all three are girls?
2. Suppose a family have three children and at least one is a girl. What is the probability that all three are girls?
3. Suppose a family have three children and only one is called Joan. What is the probability that all three are girls?
The possibilities in the first case are: 000, 001, 010, 011. So the chance of three girls is 1/4 = 25%.
The possibilities in the second case are: 000, 001, 010, 011, 100, 101, 110. So the chance of three girls is 1/7 = 14·28%.
The possibilities in the third case are: 200, 201, 210, 211, 020, 021, 120, 121, 002, 012, 102, 112. So the chance of three girls is 3/12 = 1/4 = 25%.
How many ways are there to die? The answer is actually five, if by “die” you mean “roll a die” and by “rolled die” you mean “Platonic polyhedron”. The Platonic polyhedra are the solid shapes in which each polygonal face and each vertex (meeting-point of the edges) are the same. There are surprisingly few. Search as long and as far as you like: you’ll find only five of them in this or any other universe. The standard cubic die is the most familiar: each of its six faces is square and each of its eight vertices is the meeting-point of three edges. The other four Platonic polyhedra are the tetrahedron, with four triangular faces and four vertices; the octahedron, with eight triangular faces and six vertices; the dodecahedron, with twelve pentagonal faces and twenty vertices; and the icosahedron, with twenty triangular faces and twelve vertices. Note the symmetries of face- and vertex-number: the dodecahedron can be created inside the icosahedron, and vice versa. Similarly, the cube, or hexahedron, can be created inside the octahedron, and vice versa. The tetrahedron is self-spawning and pairs itself. Plato wrote about these shapes in his Timaeus (c. 360 B.C.) and based a mathemystical cosmology on them, which is why they are called the Platonic polyhedra.
Tetrahedron
Hexahedron
Octahedron
Dodecahedron
Icosahedron
They make good dice because they have no preferred way to fall: each face has the same relationship with the other faces and the centre of gravity, so no face is likelier to land uppermost. Or downmost, in the case of the tetrahedron, which is why it is the basis of the caltrop. This is a spiked weapon, used for many centuries, that always lands with a sharp point pointing upwards, ready to wound the feet of men and horses or damage tyres and tracks. The other four Platonic polyhedra don’t have a particular role in warfare, as far as I know, but all five might have a role in jurisprudence and might raise an interesting question about probability. Suppose, in some strange Tycholatric, or fortune-worshipping, nation, that one face of each Platonic die represents death. A criminal convicted of a serious offence has to choose one of the five dice. The die is then rolled f times, or as many times as it has faces. If the death-face is rolled, the criminal is executed; if not, he is imprisoned for life.
The question is: Which die should he choose to minimize, or maximize, his chance of getting the death-face? Or doesn’t it matter? After all, for each die, the odds of rolling the death-face are 1/f and the die is rolled f times. Each face of the tetrahedron has a 1/4 chance of being chosen, but the tetrahedron is rolled only four times. For the icosahedron, it’s a much smaller 1/20 chance, but the die is rolled twenty times. Well, it does matter which die is chosen. To see which offers the best odds, you have to raise the odds of not getting the death-face to the power of f, like this:
3/4 x 3/4 x 3/4 x 3/4 = 3/4 ^4 = 27/256 = 0·316…
5/6 ^6 = 15,625 / 46,656 = 0·335…
7/8 ^8 = 5,764,801 / 16,777,216 = 0·344…
11/12 ^12 = 3,138,428,376,721 / 8,916,100,448,256 = 0·352…
19/20 ^20 = 37,589,973,457,545,958,193,355,601 / 104,857,600,000,000,000,000,000,000 = 0·358…
Those represent the odds of avoiding the death-face. Criminals who want to avoid execution should choose the icosahedron. For the odds of rolling the death-face, simply subtract the avoidance-odds from 1, like this:
1 – 3/4 ^4 = 0·684…
1 – 5/6 ^6 = 0·665…
1 – 7/8 ^8 = 0·656…
1 – 11/12 ^12 = 0·648…
1 – 19/20 ^20 = 0·642…
So criminals who prefer execution to life-imprisonment should choose the tetrahedron. If the Tycholatric nation offers freedom to every criminal who rolls the same face of the die f times, then the tetrahedron is also clearly best. The odds of rolling a single specified face f times are 1/f ^f:
1/4 x 1/4 x 1/4 x 1/4 = 1/4^4 = 1 / 256
1/6^6 = 1 / 46,656
1/8^8 = 1 / 16,777,216
1/12^12 = 1 / 8,916,100,448,256
1/20^20 = 1 / 104,857,600,000,000,000,000,000,000
But there are f faces on each polyhedron, so the odds of rolling any face f times are 1/f ^(f-1). On average, of every sixty-four (256/4) criminals who choose to roll the tetrahedron, one will roll the same face four times and be reprieved. If a hundred criminals face the death-penalty each year and all choose to roll the tetrahedron, one criminal will be reprieved roughly every eight months. But if all criminals choose to roll the icosahedron and they have been rolling since the Big Bang, just under fourteen billion years ago, it is very, very, very unlikely that any have yet been reprieved.
Overlord In Terms of Core Issues Around Maximal Engagement with Key Notions of the Über-Feral 