# Sicherman Dice

Sicherman dice are a particular pair of six-sided dice that, when rolled, produce sums with the same probability distribution as a pair of standard six-sided dice. In other words, these dice do not have the arrangement of 1, 2, 3, 4, 5, and 6 on their six sides, but when rolled together, still produce the same distribution of sums:

• One way to roll a 2
• Two ways to roll a 3
• Three ways to roll a 4
• Four ways to roll a 5
• Five ways to roll a 6
• Six ways to roll a 7
• Five ways to roll a 8
• Four ways to roll a 9
• Three ways to roll a 10
• Two ways to roll a 11
• One way to roll a 12

Sicherman dice are the only alternative arrangement of six-sided dice with positive integers that produce the same probability distribution as a pair of standard six-sided dice. Can you figure out what numbers belong on Sicherman dice?

Hint: One of the dice has no numbers greater than 4.

George Sicherman discovered this arrangement in 1978, and it was popularized by mathematician and legendary puzzle-maker Martin Gardner.

## Click to see solution

One die has 1, 2, 2, 3, 3, 4 and the other has 1, 3, 4, 5, 6, 8.

There’s only way one to produce a 2

In order for the two dice to produce a 2 in exactly one way, there must be exactly one 1 on each die. If there were more than one 1 on one of the die, there would be more than one way to roll a 2. If there were no 1’s on one of the die, there would be no way to roll a 2 (since all the numbers must be positive integers). So far we know the sides are: 1, ?, ?, ?, ?, ? and 1, ?, ?, ?, ?, ?.

Rule out 11, 10, and 9

There can’t be a 11. If there was a 11 on one of the die, the other die could not have numbers greater than 1 (otherwise there would be sums greater than 12). But then there would be multiple ways to roll a 12.

The same logic rules out 10 and 9. If there was a 10, the other die would have all 1’s and 2’s, but that would exceed the number of ways to roll 11 or 12. If there was a 11, the other die would have to have three 1’s, two 2’s, and one 3 to avoid exceeding the number of ways to roll 10, 11, and 12, but we know from earlier there can only be one 1 on each die.

There must be exactly one 8 and one 4

There must be an 8, otherwise there’s no way to roll a 12 (since the hint specified one of the dice has no numbers greater than 4). So there must be exactly one 8 on one die and exactly one 4 on the other. So far we know the sides are (in ascending order): 1, ?, ?, ?, ?, 4 and 1, ?, ?, ?, ?, 8.

Max two ways to roll 11 and two ways to roll 3

Now we know the first die has some combination of 2’s and 3’s for its remaining four numbers. There can’t be more than two 3’s, or else there would be more than two ways to roll a 11. But there also can’t be more than two 2’s, or else there would be more than two ways to roll a 3. So we now we solved the first die: 1, 2, 2, 3, 3, 4 and 1, ?, ?, ?, ?, 8.

Process of Elimination

Now it’s straightforward to just eliminate the sums that are already possible, and deduce the remaining numbers. We’ve eliminated all the ways to roll 2, 3, 11, and 12. We’ve already accounted for two out of the three ways to roll 4 (3 and 3 with 1) and two out of the three ways to roll 10 (2 and 2 with 8).

To make up the one missing way to roll 4, there must be exactly one 3 on the other die (it can’t be a 1 or 2, or there it would add more than one additional way, since there is more than one 3 and one 2 on the first die).

Using the same logic, there must be exactly one 6 on the other die to produce the one missing way to roll 10. Now we have: 1, 2, 2, 3, 3, 4 and 1, 3, ?, ?, 6, 8.

Finally, there needs to be four ways to roll a 5. We currently have three: 4 + 1 and 2 + 3 (twice). To produce one and only one more, there must be exactly one 4 on the other die.

Using the same logic to get four ways to roll a 9, we find that there must be exactly one 5 on the other die. Now we’ve solved for all the numbers: 1, 2, 2, 3, 3, 4 and 1, 3, 4, 5, 6, 8.

Want your own pair of Sicherman dice?

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