I’m not sure how I came across the jeka-math blog, but it contains a lot of higher-level math problems. And a few that seem intriguingly easy, but turn out to be quite tricky, like this one in which the author claims there is another arrangement of dots on 6-sided dice that will produce the usual probabilities of each outcome from 2 through 12.
So far all I’ve managed to conclude is that the sum of the numbers on the 12 sides will have to sum to 42, just as they do on two standard dice. I haven’t decided if we should keep the same numbers, and just jumble them around (that seems most promising) or if we will actually use some higher numbers on one (but not both) of the dice.
(It just occurred to me that he states that the usual probabilities must apply for outcomes from 2 through 12, but doesn’t state that no higher sums are possible, so I asked about that in the comments on his blog.)
Anyhow, nice problem, so I figured I’d alert my readers to it. Enjoy!