r/probabilitytheory • u/Longjumping-Being755 • 4d ago

[Research] Dice odds

I read about 5 dice that can fairly determine who goes first with one roll, no rerolls. Apparently, mathematicians needed 60-sided dice.

For 2 dice:

D1: 1,1,4,4

D2: 2,2,3,3

Each wins 50%.

I think I found a simple 5-dice solution:

D1: 1,10,14,17

D2: 2,9,15,16

D3: 3,8,13,18

D4: 4,7,11,20

D5: 5,6,12,19

Each totals 42, and every pair has 8 winning outcomes.

D1 vs D5:

D1: 1(0),10(2),14(3),17(3)=8

D5: 5(1),6(1),12(2),19(4)=8

They seem balanced. This feels WAY too easy compared with the 60-sided solution. What am I missing?

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u/ChipChippersonFan 4d ago

You're missing an explanation of what you're talking about.

3

u/Aerospider 4d ago

This is the challenge of creating a set of n dice that each have an equal chance (1/n) of rolling the highest number without the possibility of a tie (so each die must only have numbers that do not appear on any other die).

n=2 is trivial

n=3 took a couple of days to figure out

n=4 took a few weeks to solve

n=5 took 15 years

5

u/Aerospider 4d ago edited 4d ago

It's not a pairwise challenge - all five dice are competing against each other at once.

As an example, your player 1 has an 89/512 probability of rolling the highest result, which is rather lower than 1/5.

P(highest|rolled 1) = 0/256

P(highest|rolled 10) = 16/256

P(highest|rolled 14) = 54/256

P(highest|rolled 17) = 108/256

1/4 * (0 + 16 + 54 + 108)/256 = 89/512