I roll 4 dice and record their sum. How many different ways can I get a sum divisible by 3?
The probable sums divisible by 3 are: 6, 9, 12, 15, 18, 21, 24. Each one of these has a different probability. Adding all their probabilities together, I get: 33.33076%
The number of different ways you get these 7 totals, I get: 432 ways.
I got these numbers by expanding the following:
expand [n+n^2+n^3+n^4+n^5+n^6]^4
n^24+4 n^23+10 n^22+20 n^21+35 n^20+56 n^19+80 n^18+104 n^17+125 n^16+140 n^15+146 n^14+140 n^13+125 n^12+104 n^11+80 n^10+56 n^9+35 n^8+20 n^7+10 n^6+4 n^5+n^4
(21 terms).
And adding all the coefficients of n^24, n^21, n^18, n^15, n^12, n^9, and n^6.
I can't see any straight forward way to do this.
The 4 dice would have to add up to 6,9,12,15,18,21 or 24
The only way for the dice to add to 24 is 6,6,6,6 that is one combination.
Add to 21
6,6,6,3
6,6,5,4
6,5,5,5, So there are 3 ways to add to 21
Add to 6
1,1,1,3
1,1,2,2 So there are only 2 ways to add to 6
Add to 9
1,1,1,6
1,1,2,5
1,1,3,4
1,2,2,4
1,2,3,3 that is it I think so that is 5 ways
Add to 12
1,1,4,6,
1,1,5,5
1,2,3,6
1,2,4,5
2,2,2,6
2,2,3,5
2,2,4,4
2,3,3,4
3,3,3,3 that seems to be 9 ways
Add to 15
1,2,6,6
1,3,5,6
1,4,5,5
1,4,4,6
2,2,5,6
2,3,4,6
2,3,5,5
2,4,4,5
3,3,3,6
3,3,4,5
3,4,4,4 there seems to be 11 there
add to 18 (I am going to try and do it differently - in pairs this time)
6+12 1,5,6,6
2,4,6,6
3,3,6,6
7+11 1,6,5,6
2,5,5,6
8+10 6,2,6,4
6,2,5,5
5,3,4,6
5,3,5,5
4,4,4,6
4,4,5,5
9+9 6,3,6,3,
6,3,4,5 That seems to be 11 ways
6 2 ways
9 5 ways
12 9 ways
15 11 ways
18 11 ways
21 3 ways
24 1 way
So if I haven't made any mistakes I get 2+5+9+11+11+3+1 =2+5+9+11+11+3+1
= 42 ways of thowing a multiple of 3
Take the case of getting a total of 6 with rolling of 4 dice. As the coefficient of n^6 =10. There are 10 different ways of getting a total of 6, and they are:
1 1 1 3 1 2 1 2 2 1 1 2 3 1 1 1
1 1 2 2 1 2 2 1 2 1 2 1
1 1 3 1 1 3 1 1 2 2 1 1
The exact probability is: 10 / 6^4 =10 / 1296. Look here at W/A:
http://www.wolframalpha.com/input/?i=4+dice+tossed,+total+6
Click on "Exact form". it gives the above number 10/1296. but they reduce it by dividing it by 2 =5/648.
I understand what you are assuming. But W/A assumes that are DIFFERENT as though you had 4 different-coloured dice. That is how they get 10 different ways out of 1,296 ways of tossing 4 six-sided dice.Please note that the total of 432 ways that I got agrees with the probability that I arrived at, which was 33.33%. And 432 ways out of a total of 1,296 =432/1296 =33.33%.
If you understood what I have stated as my assumption - then there is no 'BUT'
Wolfram alpha has answered a different question as it has made a different assumption.
This is why questions should be stated exactly with no room for reasonable alternate interpretations.
I never suggested that your answer was incorrect although you should have stated the assumption that you were making.
(i.e. that all the dice were uniquely identifiable)