# Making three fair dice

You are given three unlabelled standard 6-sided dice. Can you write every number from 1 to 18 on the dice, such that in any given throw each die has the same chance of being the highest?

Here is a simpler version of the problem with two dice: Making two fair dice

• More interesting is the case when each die is also equally likely to be the lowest. Jun 4 at 18:26
• There is a lot of research and info about higher-number-of-players and other fancier versions of this problem at Go First Dice; obviously heavy spoiler warnings apply.
– mlc
Jun 4 at 20:57
• Indeed that makes the problem more interesting, but too hard to solve manually. I found the puzzle from here youtu.be/-64UT8yikng Jun 5 at 3:57
• After watching that video, I'm unsure about the question. Do we solve for when all three dice are thrown, or do they also need an even chance to win when only two of them are thrown? Jun 5 at 11:55
• I see now where this differs from the video. This puzzle only asks that the dice have an even change of winning. It doesn't care about second and third place. The video solves for the dice having an even chance of being first or second or third, which is a lot more tricky. Jun 5 at 12:02

Each dice should have 1/3 probability of winning. So let's put [17, 18, 1, 2, 3, 4] one dice.

Now we need to put the remaining numbers 5-16 on remaining dices so that they have equal winning chances. Borrowing from Jafe's answer to the two-dice problem, we should put [5,6,7,14,15,16] in one dice and [8,9,10,11,12,13] on the other.

• I verified your solution and it is correct. Well done! Jun 5 at 3:53

Surprisingly, it's the same solution as applying 12 numbers to 2 dice.

Taking the top half of the numbers, 10 to 18, assign any three of them to each die. Then assign the reciprocal (19 - n) of each number to the same die.

One solution:
18, 17, 16, 3, 2, 1
15, 14, 13, 6, 5, 4
12, 11, 10, 9, 8, 7

Another solution:
18, 14, 10, 9, 5, 1
17, 13, 12, 7, 6, 2
16, 15, 11, 8, 4, 3

This also fits @Daniel Mathias' comment of each die having equal chance of being the lowest.

• your first solution is wrong : the first die has a 50% chance of being highest, and 50% of being the lowest. Second and third dices each have a 25%/25% of being highest/lowest. Jun 5 at 12:41
• You are on the right track, but you need to modify the thinking for the first die. It is not meant to win half of its rolls. Jun 5 at 15:27
• Your second solution is also invalid. The first die, with 1 and 18, wins and loses with probability 76/216. The other two have win/loss probabilities of 70/216. Jun 5 at 19:46
• Yeah, the math I used to calculate winning odds was completely off. 😓 Jun 6 at 21:11