Question 1208306
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Are you saying you have 
a)  two dice each with all 6 sides marked 3.
b)  two dice each with 4 sides marked 4 and 2 sides marked 0
c)  two dice each with 3 sides marked 5 and 3 sides marked 1
d)  two dice each with 2 sides marked 6 and 4 sides marked 2

How do you win?  By getting the highest sum?

3s vs 4s, 4s vs 5s, 5s vs 6s, 6s vs 3s

Who or what are the "3s", "4s", "5s" and "6s"?

Sorry, I'm confused by the "vs." statements.  

I can't tell who is playing whom and with what pair of dice?

For the pair of dice 3,3,3,3,3,3, you can only get 6
P(sum=6) = 1

For the pair of dice 4,4,4,4,0,0 you can only get 8 or 4 or 0
P(sum=8)= P(1st die=4)*P(2nd die=4) = (4/6)(4/6) = (2/3)(2/3) = 4/9
P(sum=0)= P(1st die=0)*P(2nd die=0) = (2/6)(2/6) = (1/3)(1/3) = 1/9
P(sum=4)= 1 - 4/9 - 1/9 = 4/9

For the pair of dice 5,5,5,1,1,1 you can only get 10,6, or 2
P(sum=10)= P(1st die=5)*P(2nd die=5) = (3/6)(3/6) = (1/2)(1/2) = 1/4
P(sum=2)= P(1st die=1)*P(2nd die=1) = (3/6)(3/6) = (1/2)(1/2) = 1/4
P(sum=6)= 1 - 1/4 - 1/4 = 1/2

For the pair of dice 6,6,2,2,2,2 you can only get 12 or 8 or 4
P(sum=12)= P(1st die=6)*P(2nd die=6) = (2/6)(2/6) = (1/3)(1/3) = 1/9
P(sum=8)= P(1st die=2)*P(2nd die=2) = (4/6)(4/6) = (2/3)(2/3) = 4/9
P(sum=4)= 1 - 1/9 - 4/9 = 4/9
 
Sorry I can't help more.  I don't understand 

"3s vs 4s, 4s vs 5s, 5s vs 6s, 6s vs 3s".  

Edwin</pre>