I have this problem:
If three dice are thrown, what is the probability that the sum of the
numbers obtained is 6?
The only solution that occurs to me is to look for all the possible cases by making the sample space, however, I can not even make the sample space quickly. For example, for the release of two dice, you can quickly make the sample space with this table:
However, for the release of 3 dice I do not know how to do it, and doing it one by one will be a lot of time !, then my questions are:
1) Is there a "trick" to get the sample space, quickly, like that table, but for the release of 3 dice or even more?
2) Is there a more optimal solution than to count all the cases for this exercise?
Thanks in advance.
Best Answer
To summarize the discussion in the comments:
Let $(d_1,d_2,d_3)$ be the result of the dice throws and let $x_i=d_i-1$. Then $$\sum d_i=6\iff \sum x_i=3$$
and of course $0≤x_i≤5$.
The triples of non-negative numbers that sum to $3$ can be enumerated as $(1,1,1)$, $(3,0,0)$ and its permutations, $(2,1,0)$ and its permutations. There are three permutations of $(3,0,0)$ and six of $(2,1,0)$ so we get $10$ possible triples for the $x_i$. Thus there are $10$ possible triples for the $d_i$.
Since each triple is equally probable and there are $216$ triples the answer is $\boxed {\frac {10}{216}}$
Note: the numbers here are very small so it is easy to do the count by explicit enumeration. If, say, we had a large number of dice this would become inconvenient. In those cases, Stars and Bars gives the answer, at least when the desired sum is not greater than the maximal value a single die could show.