Go First Dice

When playing a board game, it is common to roll the dice to determine who goes first. For example, it might be proposed that the highest roll go first and then clockwise around the table. However, there are two problems with this. First, there could be a tie, in which case time would be wasted rolling the dice again. Second, the other positions are not randomized.

My goal was to create a set of dice that will randomize the order of 2 to 4+ players with every order being equally likely. I preferred to use the five platonic solids but was flexible with that. Ties were strictly not allowed. Just one roll!

With two players it is pretty easy. If the lowest number went fist and coins were allowed, one could easily label the coins as follows:

Coin 1: 1,4

Coin 2: 2,3

It all comes down to come 1, whether it lands higher or lower than the two consecutive numbers on coin 2. To extend it to the rule about platonic solids, we can just duplicate the faces. For example, with cubes we could have:

Cube 1: 1,1,1,3,3,3

Cube 2: 2,2,2,2,2,2

If we must have all different numbers, which I like, we could do:

Cube 1: 1,2,3,10,11,12

Cube 2: 4,5,6,7,8,9

It starts to get difficult with three players. I will admit I tried to use a combination of algebra and trial and error in Excel, but failed. So, I resorted to a little cheating and wrote a simulation to number the sides of three dice from 1 to 18 randomly until a solution was found. The program did find one within a few minutes as follows:

Cube 1: 3,4,9,10,13,18

Cube 2: 2,5,7,12,15,16

Cube 3: 1,6,8,11,14,17

There are 6³ = 216 ways the three dice can be rolled. There are six possible orders for the three players. You may trust me that of the 216 possible outcomes, each order occurred 216/6 = 36 times.

As long as I already wrote a simulator for this task, I extended it to the four-player case. It ran for many hours, trying trillions of combinations, but nothing worked. So, I went back to working out the problem mathematically. It was my idea to expand on the three-dice solution with the following:

Cube 1	4	5	10	15	18	23
Cube 2	3	6	8	17	20	21
Cube 3	2	7	9	16	19	22
Cube 4	1	11	12	13	14	24

 

My thinking was the player with Cube 4 should have a ¼ chance of going first or last. Let’s consider the probability of going first. If he rolled a 1, he would go first regardless of the other three dice, as 1 is the lowest number. That probability is obviously 1/6. If cube 4 were a 11 to 14, then the other three players need to roll a 15 or more for cube 4 to be lowest. Each of them had 3 numbers greater than 14. So, the probability Cube 4 would be lowest was (1/4) + (4/6)*(3/6)^3 = 1/6. That got me as far as each player having a ¼ chance of going first.

However, if Cube 4 was lowest, the order of the other three players were not all equally likely. For example, if cubes 1 to 3 were all 15 or more, the chances of Cube 1 being the lowest should be 1/3, but is actually prob(cube 1 = 15) + prob(cube 1 = 18)*prob(cube 2 = 20 or 21)*prob(cube 3 = 19 or 22) = 1/3 + (1/3)*(2/3)*(2/3) = 13/27.

So, I had the idea to turn cubes 1 to 3 into dodecahedrons (12-sided dice), duplicating the original six sides on the other six, but adding 24, as follows:

Cube 1	5	6	11	12	15	20	31	32	37	38	41	46
Cube 2	4	7	9	14	17	18	30	33	35	40	43	44
Cube 3	3	8	10	13	16	19	29	34	36	39	42	45

 

For Cube 4, I put the two lowest and two highest numbers: 1, 2, 47, and 48. Then the eight numbers in the gap between 20 and 29. This would preserve the probability of Cube 4 being first or last at (2/12) + (8/12)*(6/12)^3 = ¼. If Cube 4 rolled 10 to 39, it would go back to the three-dice solution, which was proven to work. Thus, the four dice solution is:

Die 1	5	6	11	12	15	20	31	32	37	38	41	46
Die 2	4	7	9	14	17	18	30	33	35	40	43	44
Die 3	3	8	10	13	16	19	29	34	36	39	42	45
Die 4	1	2	21	22	23	24	25	26	27	28	47	48

 

You will have to trust me that of the 12^4 = 1,296 possible ways to roll the four dice and 4!=24 possible orders, each order has 1296/24 = 54 combinations.

I couldn’t stop there but moved onto the five-player case. Using the same kind of logic for the four-player case, the best I could do was 840-sided dice. Rather than adding about five pages to this newsletter with a long string of numbers, I published the exact dice faces in my forum at Wizard of Vegas in the thread Go First Dice. There are 3,485,099,520,000 ways one can roll five 840-sided dice, so I checked the results by random simulation and was satisfied the dice achieved what they were supposed to.

The video that started me down this rabbit hole is Go First Dice, on the Numberphile (one of my favorite channels!). I must admit James Grime gets to the four dice case the same way I did. However, I hope I add something to the discussion.

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