Toss dice with 10, 8 and 6 sides. What are the probabilities for obtaining sums of 3, 4, 5, through 12?

Question

I understand how to do this the hard way. Figure out all the possible outcomes, and divide by 480. But is there a shortcut to come up with all of the possible outcomes?

This is a homework question for my 6th grade child, and there must be a simple solution that escapes my mind.

Answer 1

I'm afraid there isn't really anything easier. It shouldn't take too long to do though:
For example, lets say we wanted a sum of 10.
If the '6' die shows a 1, we need a sum of 9, which means (1,8),(2,7)..(8,1), in 8 ways.
If it shows a 2, we need a sum of 8, (1,7) through (7,1), 7 ways.
..
If it shows a 6, we need a sum of 4, (1,3) through (3,1), 3 ways.
So we have 8+7+6+5+4+3 = 33.

For some sums, such as 12, its a tiny bit more complicated, eg if the '6' die shows a 1, we can have anything from (3,8) through (10,1), since the second die can be at most 8. So we have 8 ways there; (2,8) through (9,1) gives 8 ways, (1,8) through (8,1) gives 8 ways, and then we start going back down to 7,6,5, for a total of 8+8+8+7+6+5 = 42.

So it shouldn't take too long to do all the others.

Answer 2

Even I cant think of a simple solution.

But I can conjecture maybe that the sum of probabilities of obtaining 3,4,5...12 = sum of probabilities of obtaining 13,14,15....24 = 1/2

EDIT: I think i goofed up