Probability chance of rolling a nine question

[amberglo]

Hi, here is my question:

Seventeenth-century Italian gamblers thought that the chance of getting a 9 when they rolled three dice were equal to the chance of getting a 10. Calculate these two probabilities to see if they were right. (I have the two answers, just need to know what the steps are). Thanks!

 

[arildno]

Well, if you know answers but haven't the faintest idea HOW to get them, THAT's what you should think about. To give you the two answers is without meaning.

So:
HOW would you proceed to find this out?

 

[amberglo]

Well, that's why I posted this question because I don't know how to get the answers. If I did, I would not be posting in this forum. I've tried many different ways in figuring out the probabilities, but nothing matches up to the correct answers so if you know how to do this, that would be glorious. :)

 

[arildno]

Well, can you complete these two lists, then?

1,2,6*****1,3,6
1,3,5*****1,4,5
1,4,4*****1,5,4
1,5,3*****1,6,3
1,6,2*****2,2,6
2,1,6*****2,3,5
2,2,5*****2,4,4
2,3,4*****2,5,3
2,4,3*****2,6,2
2,5,2*****3,1,6
2,6,1*****3,2,5
3,1,5*****3,3,4
3,2,4*****3,4,3
... .***** ...
... ***** ...

Do you see what relation these lists have to your problem?

 

[amberglo]

Ahh yes, I know one major thing I didn't do, I was making the list for rolling 2 dice (8,1) (1,8), not three.

 

[arildno]

Eeh?

 

[arildno]

Alternatively, the probabilities are given by:
$$\frac{1}{6^{3}}(\frac{d^{9}}{dx^{9}}(\frac{x-x^{7}}{1-x})^{3})\mid_{(x=0)}$$
$$\frac{1}{6^{3}}(\frac{d^{10}}{dx^{10}}(\frac{x-x^{7}}{1-x})^{3})\mid_{(x=0)}$$