Three dice with no best one, and what a second roll does
Three six-sided dice. Amber carries
4 4 4 4 4 9, blue carries 2 2 2 7 7 7,
green carries 0 5 5 5 5 5. You each pick one and roll;
higher wins. Amber beats blue seven times in twelve. Blue beats green
seven times in twelve. And green beats amber twenty-five times in
thirty-six — so there is no best die, and whoever chooses second
can always pick a winner. That much is a known trick. Here is the part that
is not: take two of your die and two of mine, roll both and add, and
every one of those three verdicts turns round. Blue now beats amber
85 times in 144, green beats blue by the same 85 in 144, and
amber beats green 671 times in 1,296. The cycle is still a cycle and it
runs the other way. From three dice each onwards it stops being a cycle
at all and settles into the ranking you would have guessed from the averages
on the first day — and stays there.
The three dice
Everything on this page is worked out from these eighteen numbers and nothing else. They are not measurements and there is nothing to look up: the faces are the whole input.
The averages are 4⅚, 4½ and 4⅙ — evenly spaced, a third of a pip apart, and in exactly the order the one-roll game contradicts. Amber has the highest average and loses to the die with the lowest.
Every way the roll can come out
Pick a pair. The square below is every possible pair of totals: the rows are what the first die can add up to, going upward, and the columns what the second can, going rightward. Each cell is drawn with the area it deserves, so how likely one of them is to win is not a number beside the picture — it is the coloured area. Slide the dial and both dice get more copies of themselves.
Which way the arrows point
The same dial, and all three verdicts at once. An arrow runs from the die that wins more often than it loses to the die it beats. At one die each the three arrows chase each other round. At two each they still chase each other round, in the opposite direction. After that they stop going round.
All twelve at once
Every pair, at every number of dice from one to twelve, and the height is the share of the games that are not drawn — so the level line across the middle is exactly where a verdict changes hands, with nothing to remember. (The raw chance of a win would not do that. At five dice each amber wins only 44.1% of its games against blue and beats it comfortably all the same, because 27.7% are drawn and blue takes just 28.1%. A line at 44.1 under a rule marked a half is a picture of amber losing.) The two dotted columns are the only lengths under twelve where a draw is possible at all; the section after this one says why they are five and ten.
Amber over blue and blue over green are drawn as one line
with rings on it, not as two lines, because they are not two nearly equal
numbers — they are the same number, exactly, at every one of the
twelve. There is a reason, and it is three lines of algebra, below.
Why two of the three are the same number
Each of these dice is one number plus a coin. Amber is
4 + 5·(a 1-in-6 coin); blue is 2 + 5·(a fair coin);
green is 0 + 5·(a 5-in-6 coin). Roll n of one and
the total is 4n + 5X, 2n + 5Y or 5Z,
where X, Y and Z count heads out of n at those three rates.
Amber beats blue when 4n + 5X > 2n + 5Y, which is
Y − X < 2n/5. Blue beats green when
2n + 5Y > 5Z, which is Z − Y < 2n/5. Now
1/6 and 5/6 are mirror images, so Z has the same distribution as
n − X; and a fair coin is its own mirror, so Y has the same
distribution as n − Y. Substitute both and
Z − Y becomes (n − X) − (n − Y), which is
Y − X. Same event, same chance, at every n. Not a coincidence
and not an approximation.
The third pair goes the same way and ends somewhere better. Green beats
amber when 5Z > 4n + 5X, and with Z = n − X′
that is X + X′ < n/5 — the chance that 2n throws of a
1-in-6 coin give fewer than n/5 heads. At one die each that is
(5/6)² = 25/36, and at two dice each it is
(5/6)⁴ = 625/1296, which is the whole of green's collapse in
one line: green only beats amber by never showing its zero, and asking for
that twice is asking for the fourth power of a thing already less than one.
When a draw is possible at all
Amber's faces are both 4 mod 5. Blue's are both
2 mod 5. Green's are both 0 mod 5. So n dice give
totals of 4n, 2n and 0 modulo five,
whatever they land on, and two of those three agree only when
n is a multiple of five. Below twelve dice that means the only ways
to draw this game at all are with five each or ten each — and at five each a
draw is not a curiosity, it is 27.7 per cent of amber-against-blue
games and 32.3 per cent of green-against-amber ones — which is why
the table above marks the length rather than printing a figure, there being
no single figure to print. Everywhere else the two totals cannot be equal, which is why the
arrows above never have a third state.
What this page checks about itself
Two of these are the instrument pointed at something whose answer was known before the page existed: Efron's four dice, a different set with a different cycle, which the same counter has to find; and three identical dice, where it has to find nothing.