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This question arose when playing yahtzee with some friends. Not entirely sure if I'm in the right area, but hope you can help.

How many dice do you need to create a tower whose walls have the same number of eyes?

Through brute force we found that you could do it with 2 dice and 4 dice, so then any multiple or sum of these would be a solution.

But how can you solve such a problem generally?

Or to rephrase it into a hypothesis:

There are no dice towers where every wall has an equal number of eyes and an odd number of dice.

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I have no idea how to tag this, but I am sure that [logic] is not right. – Asaf Karagila Nov 29 '12 at 16:29
@Asaf: dice seemed like a natural choice :-) – joriki Nov 29 '12 at 16:35
@joriki: Now there's choice involved? :-) – Asaf Karagila Nov 29 '12 at 17:51
@Asaf: Wherever you are, there's choice :-) – joriki Nov 29 '12 at 17:59
@joriki: I see I have no choice about that... :-) – Asaf Karagila Nov 29 '12 at 18:00

Your hypothesis is correct. A die has a total of $21$ eyes, $7$ on each pair of opposite sides, so it always has $14$ on the walls no matter how you place it. Since $14=3\cdot4+2$, the number of eyes on the walls is not an integer multiple of $4$ if you have an odd number of dice, so you can't have the same number of eyes on all $4$ walls in this case.

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Each two opposite sides of a die has totally 7 eyes in common (1+6, 2+5, 3+4). If quantity of dice is odd, then sum of numbers of eyes on two opposite walls would be odd.

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