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!

January 14, 2008 11:37 pm

(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.)

Think about it for half a second.

If the outcomes from 2 to 12 have the usual probabilities, their total probability will add to 1. Therefore no other outcomes are possible.

2. January 14, 2008 11:39 pm

Good point. 🙂 I was thinking “same relative probabilities”. Yeah, that’s my story and I’m sticking to it. 🙂

January 15, 2008 8:20 am

I also thought this was a very hard problem until I saw it in Paul Zeitz’s book _The Art And Craft of Problem Solving_, in the chapter on generating functions.

The generating function idea is that you represent a die showing 1,2,3,4,5,6 on its sides with the polynomial (x^1 + x^2 + x^3 + x^4 + x^5 + x^6). Then rolling two dice and adding becomes multiplying the polynomials:

(x^1 + x^2 + x^3 + x^4 + x^5 + x^6)(x^1 + x^2 + x^3 + x^4 + x^5 + x^6).

When you multiply it all out, the coefficients, like the 4 in 4x^5, tell you how many ways there are of getting a total of 5.

I don’t know whether the custom on this blog is to go into detail from there or to leave it here for other readers to figure out the rest. In any case, enjoy!

January 15, 2008 11:27 am

The usual way to solve this problem is with generating functions. The probabilities associated with rolling one die correspond to the polynomial $1/6x + 1/6x^2 + \dots + 1/6x^6$ (the coefficient of the $x^n$ term gives the probability of rolling an $n$). The nice thing here is that the probabilities of rolling two dice together can be found simply by multiplying their generating functions! (If you work out an example you can probably see why this is the case.) So the probabilities associated with rolling two dice are given by $1/36 (x + x^2 + \dots + x^6)^2$. The key insight now is that there are other ways this polynomial can be factored…

5. January 15, 2008 8:00 pm

Guess it’s time to learn about generating functions! Thanks for the pointers, Joshua and Brent! I don’t know that I have an established “custom” yet of how deeply to go on solutions, but I do prefer what you both did, which was to get the ball rolling without giving away the whole puzzle. Thanks again. 😀

6. January 17, 2008 9:01 am

Hmm. Is a “half-dot” a dot? If so, I have an alternate solution. 🙂

7. January 17, 2008 9:16 am

Each side must have at least one dot. Negative numbers are not allowed

So… no.