combinatorics - Two dice with the same probability for each sum?

A friend invites you to play a game. The game is using two standard six-sided dice with faces numbered 1, 2, 3, 4, 5, 6 each.
As usual, the dice are considered distinguishable, i.e. throwing a 1 with dice 1 and a 2 with dice 2 is different than throwing a 1 with dice 2 and a 2 with dice 1 even in both cases the sum is 3. You friend claims that each sum from 2-12 on both dice appears with the same probability

How can you prove if this is true or not?

Answer

I think:

These dice can't exist.

Here's my proof:

Let $a$, $b$, $c$, $d$ be the probabilities of the first die being 1, the second die being 1, the second die being 6, the first die being 6, respectively.

Since each sum from 2-12 has equal probability, each sum has 1/11 probability.

There is only one way to roll a 2 or 12: 1-1 and 6-6, so we know:

$ab = 1/11$
$cd = 1/11$

Now consider the ways to make 7. Among the ways to make 7 are 1-6 and 6-1, along with 4 other ways. So we know:

$ac + bd \leq 1/11$

Substituting $1/(11a)$ for $b$ and $1/(11c)$ for $d$, we get:

$ac + 1/(121ac) \leq 1/11$

Multiplying by $ac$ on both sides (note $ac$ can't be negative so the $\leq$ can't flip):

$a^2c^2 + 1/121 \leq ac/11$

Subtracting $ac/11$ from both sides:

$a^2c^2 - ac/11 + 1/121 \leq 0$

Now at this point you can plug this into some equation solver and find out that there are no real solutions. You can also see this from the quadratic equation. Solving for $a$, the square root part of the quadratic equation is this:

$\sqrt{c^2/121 - 4c^2 * X}$, where $X \geq 1/121$ based on the inequality. Since the value inside the square root is always $\leq 0$, the equation doesn't have a real solution unless $c = 0$, which can't be true because $cd = 1/11$.