# 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?

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.