I love this problem, which I found on the Internet many years ago. I suppose you could find a solution by Googling, but that’s of course no fair.
A standard pair of six-sided dice induce a probability distribution on the outcomes 1 through 12: The probability of rolling a 1 is 0, of rolling a 2 is 1/36, of rolling a 3 is 1/18, etc. Is there any nonstandard pair of six-sided dice that induces exactly the same probability distribution? If so, how many such pairs are there?
(A non-standard pair of six-sided dice might have, say, the numbers 1,2,2,3,8,9 on one cube and the numbers 2,3,4,4,4,4 on the other.)
What about the same problem for seven-sided dice, or eight-sided, or n-sided?
SPOILER: (PARTIAL) SOLUTION
Well, if you’re allowed to have zeroes on the dice faces, then you could decrease all the numbers on one dice by one, and increase all the numbers on the other dice by one, and then the total of every roll would be the same as it was before, so you’d have the same distribution.
If you allow negative numbers, then you can decrease/increase the faces by n instead of 1, and get infinitely many such pairs of dice.
SPOILER: (REST OF) SOLUTION
On the other hand, suppose you’re not allowed to have zeroes. Then if you have a 1/36 chance of getting a total of 2, that can only come from having a 1 from each dice. So each dice must have exactly one face with a 1.
If you’re required to have a 2/36 chance of getting a total of 3, there must be two possible outcomes that would give you a 3. Without zeroes, that can only come from a 1 on one dice and a 2 on the other, so each dice must have exactly one face with a 2.
Continuing in the same way, you can prove that the only such dice that give you the desired probability distribution are two dice with a 1, 2, 3, 4, 5, and 6.
The flaw in that solution is that there is another way to get a 2/36 chance of getting 3 – two 2’s on one dice and no 2’s on the other.
Yes, we could have 0,1,2,3,4,5 and 2,3,4,5,6,7 on the faces, or infinitely many similar modifications of the standard dice. This surely doesn’t count as solving the puzzle though, so….
I also found a nontrivial example involving convolutions/cyclotomic polynomials. Spoiler alert : divide 211515486 and 232668072 by the first taxicab number to see my solution. This example leads to another infinite family of solutions, of course.
I also found a number of ways using dice of other shapes, eg, a 9 sided die and a 4 sided die with sides 1,2,2,3,3,3,4,4,5 and 1,4,4,7, or with sides 1,2,3,3,4,5,5,6,7 and 1,2,4,5, to replicate a pair of standard six-sided dice.
Here are the solutions for 8 sided with numbers 1.. :
* 1,2,2,3,3,4,4,5 & 1,3,5,5,7,7,9,11
* 1,2,2,3,5,6,6,7 & 1,3,3,5,5,7,7,9
* 1,2,3,3,4,4,5,6 & 1,2,5,5,6,6,9,10
* 1,2,3,4,5,6,7,8 & 1,2,3,4,5,6,7,8
Hmm… all my solutions involve symmetrical dice – for each one there’s a number M such that if you subtract each number from M, you get the same die back.
Challenge : find a pair of dice that solves the generalised puzzle, where at least one of the pair is not symmetrical, or prove that no such pair exists.
@EricK you’re right, all right I’ll have to think about that some more… although I still maintain my trivial solution is still a solution :)
Bennett Haselton’s logic is correct, it just needs to be followed through. He has already established that each die must have exactly one face with one pip. Although if there were no other constraints it would be possible to get the correct probability for 3 by having one die with two faces with two pips on, and one die with two faces with one pip on, that would violate the constraint we have already identified for the arrangement of one-pip faces. The arrangement of the rest of the pips follows. So if zero or negative pip values aren’t allowed then there is only one arrangement of pips that will produce the normal probabilities. This result must also follow for all pairs of n-sided dice.
Hmm, well I am clearly completely wrong :) @EricK
yes, you could have one dice that is a 7,8,9,10,11,12 and another that is 15,4,14,5,13,6. The probability works out the same (Two outcomes, 18 and 11, that have a 1/36 probability of occurring and the rest have a 1/18 chance of occurring.) This is just an example, the same principle applies to other sets of numbers (21,22,23,24,25,26 and 20,27,19,28,18,29)
@mike H, regarding your 8-sided solutions: maybe i’m misunderstanding the problem (or your suggestion), but the max sum must be 12, no?
Before some math head beats me to it, let me make a little contribution. Two dice with six sides each leads to 36 possible outcomes with a specific probability distribution. We can generate the same distribution using a single die with 36 faces:
The number 2 would appear on one side.
The number 3 would appear on two sides….
The number 7 would appear on six sides
The number 8 would appear on five sides….
And the number 12 would appear on the last side.
See? All those years of D&D were not an entire waste.
I enjoyed the little deviation to taxicab numbers – I had not come across these before. The number Mike H requires is the second taxicab number numerically (if you count 2 as Ta(1)), but of course is the first chronologically, since it was this one that gave them all their name.
Forgot to add: We can also generate the same distribution using any die having natural-number multiples of 36 faces, with proportionate representations of the numbers 2, 3, 4, etc.
For anyone who wants a spoiler, there is a Wikipedia article on this subject. See Sicherman dice.
(without looking at comments)…
Infinitely many pairs of dice. Take two standard dice, subtract n from each side of 1 die, add n to each side of the other die, and you are guaranteed to get the same result. Am I missing something? Are negative numbers against the rules?
@wellplacedadjective – my 8-sided die solutions are not for the original 6-sided die problem, but for the extension to 8 sided dice.
Without using the polynomial formula I was able to reason my way to the answer for six-sided dice by scribbling on a half sheet of paper. It’s not too hard to find the answer if you are careful in your reasoning, but it is fascinating that there is one and only one answer.
The key constraints to guide the reasoning for N-sided dice are:
* each die must contain exactly one ‘1’
* between the two dice there must be exactly two ‘2’s
* if the ‘2’s are on the same die, then there must be a total of three ‘3’s
* if each die contains a ‘2’, then there must be a total of two ‘3’s
* the largest number on each die must be unique on the die
* the largest numbers on the two dice must sum to 2N
Because of these rules, I believe the only ways to put ‘1’s, ‘2’s, and ‘3’s on dice are:
(1,2,2,3,3,3,…)&(1,…)
(1,2,2,3,3,…)&(1,3,…)
(1,2,2,3,…)&(1,3,3,…)
(1,2,3,3,…)&(1,2,…)
(1,2,3,…)&(1,2,3,…)
The remaining sides cannot contain a ‘1’, ‘2’, or ‘3’. Does this seem correct?
Using those constraints you can arrive at the answer fairly quickly for N=4 or N=6.
Unless I made some mistakes, the only answer for 3- or 5-sided dice is the ‘standard’ numbering. Is this true for all dice with an odd number of sides?
(If you don’t like dice with an odd number of sides, then you can use a 10-sided die and put each of five numbers on two of the sides to make a ‘5-sided’ die.)
@Thomas Bayes :
these two nine-sided dice work
* 1,2,2,3,3,3,4,4,5 and 1,4,4,7,7,7,10,10,13
and these two 15-sided dice:
* 1,2,2,3,3,3,4,4,4,5,5,5,6,6,7 and 1,4,6,7,9,10,11,12,13,14,15,17,18,20,23
These are the only (non-standard) solutions for 9 and 15 sides.
Oh, foo – I just checked wikipedia, and my trick is already there. I wonder if I should still blog about it?
I think 0 should be allowed on dice, so I would say YES, there are non-standard pairs of dice that have the same distribution. They can be found by subtracting one from each face of one standard die and adding one to each face of another standard die. For example:
Die 1: 0,1,2,3,4,5
Die 2: 2,3,4,5,6,7
Since you just subtracted one from one of the dies, and added one to the other die, the probability distributions of the sums does not change. To formally prove that you would just apply the fact that
X + Y = (X + 1) + (Y – 1).
I don’t think you should allow negative numbers on dice, so I say there is only one non-standard pair that I know of which gives the same distribution as the standard die. There could be other non-standard pairs that I haven’t thought of, but I suspect not (and will try to prove it).