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?

Answer

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.

Attribution
Source : Link , Question Author : Guin_go , Answer Author : Mees de Vries

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