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I roll 4 dice and record their sum. How many different ways can I get a sum divisible by 3?

 Oct 15, 2016

Best Answer 

 #3
avatar+118667 
+5

Please note that I have assumed all of the dice are identical :)

 Oct 15, 2016
 #1
avatar
+5

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.

 Oct 15, 2016
 #2
avatar+118667 
+5

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

 Oct 15, 2016
 #3
avatar+118667 
+5
Best Answer

Please note that I have assumed all of the dice are identical :)

Melody Oct 15, 2016
 #4
avatar
+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.

 Oct 15, 2016
 #5
avatar+118667 
0

No this is not true with my assumption that I have stated.

 

The dice are identical so 3,1,1,1 is identical to 1,1,1,3 and 1,3,1,1 and 1,1,3,1

This is NOT 4 ways this is only 1 way. 

Melody  Oct 15, 2016
 #6
avatar
+2

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%.

 Oct 15, 2016
 #7
avatar+118667 
0

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)

Melody  Oct 15, 2016
 #8
avatar
+3

See the probabilities and number of ways of obtaining a certain sum in dice tossing here:

http://mathworld.wolfram.com/Dice.html   

Starting with Equation (1).

 Oct 15, 2016

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