Question 473709
If we roll 4 dice once, what is the chance the sum will be 12?
<pre>
The other tutor didn't even start to solve your problem.  Counting
the number of ways to have sum 12 is NO EASY TASK!  And the denominator
is not 24 at all.  It's 6<sup>4</sup>.

There are these 11 sets of ways disregrading order that the 4
dice can have sum 12.  However the dice are different so we must 
rearrange each of these in all possible distinguishable arrangements:

1.  1 1 4 6   <-- 4!/2! = 12 distinguishable arrangements     
2.  1 1 5 5   <-- 4!/(2!2!) = 6 distinguishable arrangements
3.  1 2 3 6   <-- 4! = 24 distinguishable arrangements
4.  1 2 4 5   <-- 4! = 24 distinguishable arrangements
5.  1 3 3 5   <-- 4!/2! = 12 distinguishable arrangements
6.  1 3 4 4   <-- 4!/2! = 12 distinguishable arrangements
7.  2 2 2 6   <-- 4!/3! = 4 distinguishable arrangements
8.  2 2 3 5   <-- 4!/2! = 12 distinguishable arrangements
9.  2 2 4 4   <-- 4!/(2!2!) = 6 distinguishable arrangements
10.  2 3 3 4   <-- 4!/2! = 12 distinguishable arrangements
11.  3 3 3 3   <-- 4!/4! = 1 distinguishable arrangement

The numerator of the desired probability is

12+6+24+24+12+12+4+12+6+12+1 = 125

The denominator is 6<sup>4</sup> since each die can land 6 ways:

So the answer is {{{125/6^4}}} = {{{125/1296}}}

Edwin</pre>