Tetrahedral probability

On both dice, all faces are positive and do not repeat (i.e. No "two 1s" on one die).

P(sum of dice=6)=3/16
P(double 4)=0
P(sum of dice <7)=1/2
P(double 3)=1/16
P(sum of dice >(or equal to)9)=1/4
P(sum of dice=4)=1/8
P(sum of dice=2)=0

What are the numbers on each die?

3 ways to get 6 ................1+5, 2+4, 3+3
8 ways to get <7..................1+2, 1+3, 1+5, 2+2, 2+3, 3+2, 3+3, 4+2
1 way to get 2(3)s..................3+3
4 ways >8................................7+2, 7+3, 7+4, 5+4
2 ways to get 4..........................1+3, 2+2
Only one die has a 1 (1,2,3,4) and (2,3,5,7)

It is easy to show that (1,2,3,4) and (2,3,5,7) solves the problem. I showed how to find the solution without guessing the solution and determined that it was the only solution - in case you don't understand what I did.

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I assume that they are all integers. Let px be the probability mass function of the first dice and py the other.
3/16 = px(1) * py(5) + px(2) * py(4) + px(3) * py(3) + px(4) * py(2) + px(5) * py(1)
0 = px(4) * py(4)
....
0 = px(1) * py(1)

so I just turn this into a set of equations. With enough energy one could probably solve the above system constraining on px(j) >0 => px(j)= 1/4 and that px(1) + ... + px(8) + px( >= 9) = 1 and py(1) + ... + py(8) + py( >= 9) = 1. In stead I went through all possible ways of assigning px(1), px(2), ... , px(>=8) and py(1) , ... py(>=8) and checked if it resulted in the right probabilities, (3/16, 0, 1/2, ..., 0) . There are less than 2^16 ways. I got this solution

dice x = (2,3,5,7)
dixe y = (1,2,3,4)

and there are no other solutions( except the one where x=(1,2,3,4) and y = (2,3,5,7) ).