Biased dice with strange probabilities

Question

A recent question here, probably soon to be deleted, mentions:

… two weighted d6 dice. … there is a 40% chance of getting a sum of 12, and a uniform distribution for all other possible sums.

This seemed unlikely, but I was not able to rule it out. I would like to find out:

Is there any possible assignment of probabilities to the twelve faces of two cubes, and assignment of a non-negative integer number of pips to the faces, to produce the required probabilities?

To be clear, there must be an $0.4$ probability of rolling a 12, and a $0.06$ probability of rolling each integer from 2 to 11 inclusive.

To my surprise I could not even prove that this was impossible when the cubes had the usual number of pips on each face. (Lulu's answer below gives a simple argument that proves the result for this case.)

I tagged this with generating-functions because that seemed like a fruitful approach, although I have not tried it yet.

Answer 1

In the following we make no assumption on the number of sides of these dice, except that they have more than one, so we may assume that for either die, a number of pips appears only on one side.

Let $a_h, a_l$ be the probability of throwing the highest and the lowest number of pips on the first die, and $b_h, b_l$ the same for the second die. Then we must have $a_hb_h = 0.4$ (with those highest values summing to 12) and $a_lb_l = 0.06$ (with those lowest values summing to 2). Since $a_ha_lb_hb_l = 0.024$, at least one of $a_hb_l$ and $a_lb_h$ must be greater than or equal to $\sqrt{0.024} \approx 0.15 > 0.06$, a contradiction.

Answer 2

No, there isn't (assuming independence of the two dice, and assuming the usual face values)

Let $\{p_i\}$ denote the probabilities on the first die, and $\{\psi_i\}$ the probabilities on the second.

Then $$p_6\times \psi_6=.4\implies \psi_6=\frac {.4}{p_6}$$ and $$p_1\times \psi_1=.06\implies \psi_1=\frac {.06}{p_1}$$

But then the probability of getting a $7$ must be $.06$ so $$p_1\psi_6+p_6\psi_1≤.06\implies x\times .4+\frac {.06}x≤.06$$

where $x=\frac {p_1}{p_6}$.

But $.4x+\frac {.06}x$ has a minimum value of about $.31$ for positive $x$.