Calculating probability for unconventional 6 sided die Let's consider this situation. We have $3$ different $6$-sided dice. The first die has five blank sides + one '$1$' side. The second die has four blank sides + two '$1$' sides. The third die has $4$ blank sides + one '$1$' side + one '$2$' side.
In an other form :
1st die : $1,0,0,0,0,0$
2nd die : $1,1,0,0,0,0$
3rd die : $1,2,0,0,0,0$
I am trying to calculate the different probabilities of each possible sum outcome (sum of $0,1,2,3$ and $4$). 
Would it be fair to assume that what follows is true?
The probability of having a sum of $1$ on the $3$ dice is equal to (1st dice, 2nd dice, third dice) :
$\frac{1}{6} (1) \times \frac{4}{6} (0) \times \frac{4}{6} (0)  +  \frac{5}{6} (0) \times \frac{2}{6} (1) \times \frac{4}{6} (0) + \frac{5}{6} (0) \times \frac{4}{6} (0) \times \frac{1}{6} (1) = \frac{76}{216}$ or $35.19\%$
For the sum of $1$, I assumed that to have this sum, only one die can have '$1$' and the other two '$0$' so I find the probability for each set of dice (chance of '$1$' $\times$ chance of not '$1$' $\times$ chance of not '$1$'). I understand that this will be different for the sum of $2$ since the dice combination will be different (sometimes only $1$ dice is needed while the other time two dice will be needed) but I would like to confirm that my reasoning is accurate. 
I am terrible with the notions of probability and it's been a few days now that this has been in my head and I wasn't able to find an answer to my assumption anywhere on the internet since I don't know how to really word it. 
Thank you for your time. 
Jeph
edit : Corrected the sum of the probability, miscalculated, put $4\times 4$ as $8$ instead of $16$.
 A: There are 216 possible combinations of 3 six sided dice (6x6x6).
probability of rolling a 
0 is 80/216 = 37%
1 is 76/216 = 35%
2 is 42/216 = 19%
3 is 16/216 = 7%
4 is 2/216  = 1%
This can be calculated by simply writing a table of the results. it simplifies to only 12 combinations because of the duplication of the sides.
Combinations are 
111 (x2)
112 (x2)
110 (x8)
101 (x4)
102 (x4)
100 (x16)
011 (x10)
012 (x10)
010 (x40)
001 (x20)
002 (x20)
000 (x80)
This means that there are 80 different combinations that produce a 000. That is because dice 1 has 5 sides with 0, dice 2 has 4 sides with 0 and dice 3 has 4 sides with 0: 5 x 4 x 4.
So you can go through this calculation for all the combinations to produce the above table.
Then you simply add up all the combinations that produce the result. For example a result of 1 could be 100 (occurs 16 times), 010 (occurs 40 times) or 001 (occurs 20 times). So there are 76 combinations (16 + 40 + 10) that give a total of 1. chance of rolling a 1 is 76/216 = 35%.
A: Yes, your approach to the probability of a sum of $1$ is correct.  Computing the chance of $2$ is a little harder, because you have to add $0-0-2$ to the three ways of getting a zero and two ones.  When you are done, check that the probabilities all sum to $1$.
