Thrice Dice Twice

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

Leave and Let Dice

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)

It’s Only Rot’n’Roll…

It’s Only Rot’n’Roll

A Porphyropolyhedric Tribute to Clark Ashton Smith

Banal, mundane, and dreary. Something needs to be done about the writing of Clark Ashton Smith — and I’ve tried to do it. The problem seems to me that the writing of CAS has been Roman in the gloamin’: that is, its twilight mystery, touched with Grecian glamor, plods across the page in the Roman alphabet, which is highly functional, but aesthetically unadventurous. Has any edition of CAS in English tried to match the beauty and complexity of the text with the beauty and complexity of a font? Not to my knowledge. Calligraphy, in the wider sense, is peripheral, at best, to English literature and and even the hyperlogomania of a book like Finnegans Wake takes place on a highly restricted graphological stage. Imagine what Joyce could have done with other alphabets, other ideographies, to stir into his mad meadish Sternen-stew of polyglossemanticity! And imagine CAS printed, or hand-written, in a script that reflects something of the beauty and complexity of his language. The beauty and fluidity of Georgian or Arabic would suit his tales of Zothique, for example; the complexity and density of Devanagari or Tamil would suit his tales of Hyperborea: but best of all would be a script invented specifically for CAS.

I haven’t supplied that, but I’ve tried to point the way with what I call a CAS-Whole, or porphyropolyhedric tribute to Clark Ashton Smith. It consists of a dodecahedron of paper and purple matches that uses four invented scripts to capture the opening lines of five of CAS’s best stories. In Plato’s cosmology, four of the regular (or Platonic) polyhedrons — the tetrahedron, the hexahedron, the octahedron, and the icosahedron — represent the four elements of which the universe is composed. The final regular polyhedron, the dodecahedron, represents the universe as a whole.[1] Hence, “CAS-Whole”. The purple matches — creating a porphyro-polyhedron — recall CAS’s words in The Black Book: “Strange pleasures are known to him who flaunts the immarcesible purple of poetry before the color-blind.”[2]

The dodecahedron itself, consisting of twelve regular dodecahedrons, is replete with the golden ratio, long regarded as of special significance in aesthetics.[3] One face is entirely black and might be called panglossic, representing all possible scripts in all possible languages; another, on the opposite side of the CAS-Whole, is entirely white and might be called an’glossic, representing silence and the blank page. Between the two, in a kind of “Goldilocks zone” between too much meaning and too little, are ten faces enscribed in four invented scripts with the opening words, in English, of five of CAS’s stories. Eight faces use a single, unadulterated script of the four, spiralling to the centre; two faces combine the four scripts. Given that the scripts are used for standard English, the stories can all be deciphered with a little effort and ingenuity. We are used, when reading in our mother tongues, to understanding with little effort and ingenuity, so the CAS-Whole might be regarded as a reminder of something we should not so carelessly take for granted. Furthermore, like all the Platonic solids, the dodecahedron can serve as a die, so the CAS-Whole reflects those central CASean themes of chance and fortune. Due to my ineptitude and impatience, not all of the faces are good regular pentagons, but that too can be woven into the symbolism of the CAS-Whole. The dodecahedron is not perfect, but I am not CAS and perfect dodecahedra do not occur in nature. Nor will the die roll true: fortune is biased.[4] Critics have pointed out that almost all CAS’s stories about death, so I hope that, imperfect as it is, one might say of the CAS-Whole: “It’s only rot’n’roll — but I like it.”

Notes

1. “There still remained a fifth construction, which God used for embroidering the constellations on the whole heaven.” Timaeus, c. 360 B.C. See http://www.ellopos.net/elpenor/physis/plato-timaeus/triangles.asp?pg=3

2. The Black Book of Clark Ashton Smith, Arkham House, 1979. See http://www.eldritchdark.com/writings/bibliography/writings/nonfiction/35/the-black-book-of-clark-ashton-smith

3. For more on the golden ratio, or golden section, please see http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/phi.html

4. A biased coin can be thrown “fair”, using a simple technique that can be adapted to a biased dodecahedron. Suppose a coin is much likelier to land heads than tails (or vice versa). Simply toss it twice. If it lands HH or TT, toss again. Otherwise, use the first of the two throws: simple probability will prove that even on a biased coin, HT is as likely as TH. Similarly, for a a biased dodecahedron, roll it twelve times. If any face repeats during the twelve rolls, roll twelve times again. When you have a sequence of twelve different faces, choose the first face. Based on my (far from reliable) caculations, there are 8,916,100,448,256 ways to roll a dodecahedral die twelve times, of which 479,001,600 contain no repeating number. One would therefore have to roll the die 18,614 times, on average, to produce a sequence in which no number repeats.