Is it possible to get all possible sums with the same probability if I throw two unfair dice together? I throw 2 unfair dice, suppose that $p_i$ is the probability that the first die can give an $i$ if I throw it, for $i =1,2,3,..6$ and $q_i$ the probability that the second die can give an $i$.
If I throw the dice together, is it possible to get all possible sums $2,3,4,...12$ with the same probability?
Here's what I've tried so far, the probability that I get a $2$ if I throw both dice is $p_1q_1$, the probability that I get $3$ is $p_1q_2+p_2q_1$, and generally the probability that I get $n$ is $$\sum_{i+j=n} p_iq_j$$ where $i=1,2,...6$, $j=1,2,...6$. 
So now in order for all possible sums to appear with the same probability, it must be true that $$p_1q_1=p_1q_2+p_2q_1$$
$$p_1q_2+p_2q_1=p_1q_3+p_2q_2+p_3q_1$$
$$........$$ has a solution, this is where I am stuck I can't find a way to prove that the system above has a solution, can you help?
 A: This is a classical problem. Without changing the problem, we can let the digits on the dice be $0, \ldots, 5$ instead of $1, \ldots, 6$ to make our notation easier. Now we make two polynomials:
$$
P(x) = \sum_{i=0}^5 p_ix^i,\qquad Q(x) = \sum_{i=0}^5q_ix^i.
$$
Now we can succinctly phrase your condition on $p_i, q_i$: it is satisfied if and only if
$$
P(x)Q(x) = \frac1{11}\sum_{i=0}^{10} x^i.
$$
Let's multiply both sides by $11 \times (x-1)$, and you get
$$
11(x-1)P(x)Q(x) = x^{11} - 1.
$$
The 11 zeroes of the polynomial on the right are the 11th roots of unity, which means those are also the zeroes of the polynomial on the left. The term $(x-1)$ takes care of one of the zeroes, and since $P, Q$ are both of degree 5, that means that they each have to have 5 of the other 10 zeroes.
But now note: besides $1$, all of the 11th roots of unity are complex numbers, while $P, Q$ are real polynomials. If a complex number is the root of a real polynomial, then so is its complex conjugate. That means that $P, Q$ must each have an even number of complex zeroes, but we just showed that they also have to have 5 each.
We have reached a contradiction: such $P, Q$, and thus such distributions $p_i, q_i$, do not exist.
A: Let's assume that this is possible. We can derive a contradiction from this assumption.
The probability of rolling a total of $2$ must be $1/11$, and the probability of rolling a total of $12$ must also be $1/11$, so
$$
\begin{align}
p_1 q_1 &= 1/11,\ \text{and}\\
p_6 q_6 &= 1/11.
\end{align}
$$
The probability of rolling a total of $7$ must also be $1/11$. At the same time, the probability of rolling a total of $7$ is greater than or equal to $p_1 q_6 + p_6 q_1$. So,
$$p_1 q_6 + p_6 q_1 \le 1/11.$$
The numbers $p_1$, $p_6$, $q_1$, and $q_6$ are all probabilities, so they can't be negative. Also, if any one of them were $0$, then either $p_1 q_1$ or $p_6 q_6$ would be $0$, which we already know is not the case. So, all four of these numbers are positive. This means that $p_1 q_6$ and $p_6 q_1$ are both positive, and so
$$
\begin{align}
p_1 q_6 &< 1/11,\ \text{and}\\
p_6 q_1 &< 1/11.
\end{align}
$$
If we combine these inequalities with the equations above, we find that
$$
\begin{align}
p_1 q_6 &< p_1 q_1,\\
p_6 q_1 &< p_1 q_1,\\
p_1 q_6 &< p_6 q_6,\ \text{and}\\
p_6 q_1 &< p_6 q_6.
\end{align}
$$
Since the numbers $p_1$, $p_6$, $q_1$ and $q_6$ are all positive, we may cancel them when they appear on both sides of one of these inequalities. Doing that, we conclude that
$$
\begin{align}
q_6 &< q_1,\\
p_6 &< p_1,\\
p_1 &< p_6,\ \text{and}\\
q_1 &< q_6.
\end{align}
$$
But this is impossible.
A: Let take a simpler system. There are 2 outcomes for 2 coins $\{C_1,C_2\}$ - $\{1,2\}$ with probabilities $p_1,p_2$ for $C_1$ and $q_1,q_2$ for $C_2$.
Given condition: probability after 2 coin throws of the sum being $\{2,3,4\}$ is the same.
Question: Can you find some $p_1,p_2,q_1,q_2$ that satisfies the condition?

In this the given condition is $p_1q_1 = p_1q_2 + p_2q_1 = p_2q_2 -(G)$. But we also have the implicit conditions: $p_1 + p_2 = 1-(eq1)$, $q_1+q_2 = 1-(eq2)$. Consider the first and third expressions of the given condition (G).
\begin{align*}
p_1 q_1 &= p_2q_2\\
p_1 q_1 &= (1-p_1)(1-q_1)\\
p_1 q_1 &= 1-p_1-q_1+p_1q_1\\
p_1+q_1 &= 1\tag{eq3}\\
p_2+q_2 &= 1\tag{eq4}\\
\end{align*}
Comparing the equations (eq1) and (eq3) and (eq1) and (eq4), we get $p_2 = q_1$ and $p_1 = q_2$.
Finally from the first two equalities of the given condition, we have:
\begin{align*}
p_1q_1 &= p_1q_2 + p_2q_1\\
p_1p_2 &= p_1p_1 + p_2p_2\\
-p_1p_2 &= p_1^2 + p_2^2 - 2p_1p_2\\
-p_1p_2 &= (p_1 - p_2)^2\\
\end{align*}
Now we have a contradiction as the left side is $-ve$ and the right side is $+ve$

Since we can't find a solution for this simpler system, with just 2 equalities, it is quite unlikely that a solution for the more complicated die system with $\binom{6}{2}$ equalities exists.
Note: This is of course not a proof that no solution exists for the die system.
A: The probability generating function of the dice are $P(x) = \sum_{i=1}^6 p_i x^i$ and $Q(x) = \sum_{i=1}^6 q_i x^i$. The probability generating function for their sum is $R(x) = P(x)Q(x)$. You want all possible sums $2, \ldots, 12$ to have the same probability, or equivalently you want $R(x) = \frac{1}{11} x^2 (1+x+\cdots+x^{10})$.
Hence an equivalent way to state your problem is: can we factor $1+x+\cdots+x^{10} = p(x)q(x)$ where $p, q \in \mathbb{R}_{\geq 0}[x]$ have degree 5?
We can factor $1+x+\cdots+x^{10} = \frac{1-x^{11}}{1-x}$ over the complex numbers as $\prod_{k=1}^{10} (x-\exp(2\pi i k/11))$. Grouping together complex conjugate pairs using
\begin{align*}
  (x-\exp(2\pi i k/m))(x-\exp(-2\pi i k/m))
   &= x^2 - (\exp(2\pi i k/m) + \exp(-2\pi i k/m))x + 1 \\
   &= x^2 - 2\cos(2\pi k/m) x + 1
\end{align*}
gives
\begin{align*}
  1 + x + \cdots + x^{10} = \prod_{k=1}^5 (x^2 - 2\cos(2\pi k/11)x + 1).
\end{align*}
But that says $1 + x + \cdots + x^{10}$ has no real factors of degree $5$ whatsoever, so there is no solution.
Suppose you weaken the requirements and don't insist the dice both have values in $\{1, \ldots, 6\}$. The subsets of factors with non-negative coefficients are $\{\}$,$\{3\}$,$\{4\}$,$\{5\}$,$\{2,4\}$,$\{2,5\}$,$\{3,4\}$,$\{3,5\}$,$\{4,5\}$,$\{2,3,4\}$,$\{2,3,5\}$,$\{2,4,5\}$,$\{3,4,5\}$,$\{2,3,4,5\}$,$\{1,2,3,4,5\}$. Since $1$ is only in one of these, namely $\{1, 2, 3, 4, 5\}$, the only complementary pair is the trivial one, $\{\}, \{1,2,3,4,5\}$ and there are no interesting solutions to the weaker version either.
A: If you want to generate numbers 1 through 12 with uniform probability, it is possible by re-labeling the faces of fair dice.  One die has the faces labeled 1 through 6.  The other has the faces labeled -2, 0, 2, 4, 6, 8.  If the total is not in the 1 through 12 range, roll again.  This is an acceptance-rejection technique.
Also, there is a theorem in probability theory than any desired probability can be constructed by a sequence of Bernoulli trials (coin flips).  Simulate coin flips by having the faces of a die be 0, 0, 0, 1, 1, 1, another die with faces 0, 0, 0, 2, 2, 2, a third die with faces 0 and 4, and a fourth die with faces 0 and 8.   You will roll 0 through 15 with equal probability, and roll again if your number is not in the 1 through 12 range.  Another acceptance-rejection method, with dice simulating coins.
Forgive my "engineering" approach; I can't help it.
