Any sum of 2 dice with equal probability The question is the following: Can one create two nonidentical loaded 6-sided dice such that when one throws with both dice and sums their values the probability of any sum (from 2 to 12) is the same. I said nonidentical because its easy to verify that with identical loaded dice its not possible.
Formally: Let's say that $q_{i}$ is the probability that we throw $i$ on the first die and $p_{i}$ is the same for the second die. $p_{i},q_{i} \in [0,1]$ for all $i \in 1\ldots 6$. The question is that with these constraints are there $q_{i}$s and $p_{i}$s that satisfy the following equations:
$ q_{1} \cdot p_{1} = \frac{1}{11}$
$ q_{1} \cdot p_{2} + q_{2} \cdot p_{1} = \frac{1}{11}$
$ q_{1} \cdot p_{3} + q_{2} \cdot p_{2} + q_{3} \cdot p_{1} = \frac{1}{11}$
$ q_{1} \cdot p_{4} + q_{2} \cdot p_{3} + q_{3} \cdot p_{2} + q_{4} \cdot p_{1} = \frac{1}{11}$
$ q_{1} \cdot p_{5} + q_{2} \cdot p_{4} + q_{3} \cdot p_{3} + q_{4} \cdot p_{2} + q_{5} \cdot p_{1} = \frac{1}{11}$
$ q_{1} \cdot p_{6} + q_{2} \cdot p_{5} + q_{3} \cdot p_{4} + q_{4} \cdot p_{3} + q_{5} \cdot p_{2} + q_{6} \cdot p_{1} = \frac{1}{11}$
$ q_{2} \cdot p_{6} + q_{3} \cdot p_{5} + q_{4} \cdot p_{4} + q_{5} \cdot p_{3} + q_{6} \cdot p_{2} = \frac{1}{11}$
$ q_{3} \cdot p_{6} + q_{4} \cdot p_{5} + q_{5} \cdot p_{4} + q_{6} \cdot p_{3} = \frac{1}{11}$
$ q_{4} \cdot p_{6} + q_{5} \cdot p_{5} + q_{6} \cdot p_{4} = \frac{1}{11}$
$ q_{5} \cdot p_{6} + q_{6} \cdot p_{5} = \frac{1}{11}$
$ q_{6} \cdot p_{6} = \frac{1}{11}$

I don't really now how to start with this. Any suggestions are welcome.
 A: Here is an alternate solution, which I ran across while looking through Jim Pitman's undergraduate probability text. (It's problem 3.1.19.)
Let $S$ be the sum of numbers obtained by rolling two dice,, and assume $P(S=2)=P(S=12) = 1/11$. Then
$P(S=7) \ge p_1 q_6 + p_6 q_1 = P(S=2) {q_6 \over q_1} + P(S=12) {q_1 \over q_6}$
and so $P(S=7) \ge 1/11 (q_1/q_6 + q_6/q_1)$. The second factor here is at least two, so $P(S=7) \ge 2/11$. 
A: I believe the following punchline to the generating function argument doesn’t depend on whether the number of sides on each die is even or odd. Plug $z = \zeta = \exp 2 \pi i /11$ into the purported equality $1+z+z^2+\dots+z^{10} = p(z) q(z)$ with $p,q$ quintic polynomials with nonnegative coefficients. The LHS vanishes but the RHS can’t since all terms contributing to $p(\zeta)$ lie in the upper half-plane and similarly for $q$.
A: You can write a polynomial that encodes the probabilities for each die:
$$ P(x) = p_1 x^1 + p_2 x^2 + p_3 x^3 + p_4 x^4 + p_5 x^5 + p_6 x^6 $$
and similarly
$$ Q(x) = q_1 x^1 + q_2 x^2 + q_3 x^3 + q_4 x^4 + q_5 x^5 + q_6 x^6. $$
Then the coefficient of $x^n$ in $P(x) Q(x)$ is exactly the probability that the sum of your two dice is $n$. As Robin Chapman points out, you want to know if it's possible to have
$$ P(x) Q(x) = (x^2 + \cdots + x^{12})/11 $$
where $P$ and $Q$ are both sixth-degree polynomials with positive coefficients and zero constant term.  
For simplicity, I'll let $p(x) = P(x)/x, q(x) = Q(x)/x$.  Then we want
$$ p(x) q(x) = (1 + \cdots + x^{10})/11 $$
where $p$ and $q$ are now fifth-degree polynomials. We can rewrite the right-hand side to get
$$ p(x) q(x) = {(x^{11}-1) \over 11(x-1)} $$
or 
$$ 11 (x-1) p(x) q(x) = x^{11} - 1. $$
The roots of the right-hand side are the eleventh roots of unity.  Therefore the roots of $p$ must be five of the eleventh roots of unity which aren't equal to one,  and the roots of $q$ must be the other five. 
But the coefficients of $p$ and $q$ are real, which means that their roots must occur in complex conjugate pairs.  So $p$ and $q$ must have even degree!  Since five is not even, this is impossible.
(This proof would work if you replace six-sided dice with any even-sided dice. I suspect that what you want is impossible for odd-sided dice, as well, but this particular proof doesn't work.)
A: I've heard this brainteaser before, and usually its phrased that 2-12 must come up equally likely (no comment about other sums).  With this formulation (or interpretation) it becomes possible.  Namely, {0,0,0,6,6,6} and {1,2,3,4,5,6}.  In this case, you can also generate the sum 1, but 2-12 are equally likely (1-12 are equally likely).  Without allowing for other sums I do suspect its impossible (and looks like proofs have been given).  I arrived at this answer by noting we are asking for equal probability for 11 events that come from 36 (6*6) possible outcomes, which immediately seems unlikely.  However equal probability for 12 events from 36 outcomes is far more manageable :)
Phil
A: You can write this as a single polynomial equation
$$p(x)q(x)=\frac1{11}(x^2+x^3+\cdots+x^{12})$$
where $p(x)=p_1x+p_2x^2+\cdots+p_6x^6$ and similarly for $q(x)$.
So this reduces to the question of factorizing $(x^2+\cdots+x^{12})/11$
where the factors satisfy some extra conditions (coefficients positive,
$p(1)=1$ etc.).
This is a standard method (generating functions).
A: You can't even solve this with two-sided dice.  Consider two dice with probabilities p and q of rolling 1, and probabilities (1-p) and (1-q) of rolling 2.  The probability of rolling a sum of 2 is pq, and the probability of rolling a sum of 4 is (1-p)(1-q).  These are equal only if p=(1-q).  Hence they are equal to one third only if p satisfies the quadratic equation p(1-p) = 1/3.  Since this has no real roots, it cannot be done.  This logic extends to multisided dice.
