Is it the case that certain probability problems can only be solve by listing events

Question

Just wondering if there’s a more systematic approach for those problems using combinatorics, (combinations or permutations) For instance: The probability that the sum of the numbers of 3 rolled dice is 8.

Thanks

Answer 1

Roll two dices. If the sum is $\le7$, then there is exactly one way to roll the last dice. The probability is then $\frac7{12}\cdot\frac16=\frac{7}{72} $

Answer 2

Yes, in the given example one may use generating functions. Consider the polynomial $$P= \frac 16x^1 + \frac 16x^2 + \frac 16x^3 + \frac 16x^4 + \frac 16x^5 + \frac 16x^6 =\frac 16x\cdot\frac {1-x^6}{1-x}\ , $$ then $P^3$ is a polynomial with monomial degrees $k$ from $3$ to $18$, and the coefficient of $x^k$ is the probability to get the sum $k$. The combinatorial job of computing this coefficients still has to be done, but this is exactly what the question was addressing. (A way to move the problem into combinatorics.) In our special case, $P^3=\frac 1{6^3}x^3\cdot \frac {(1-x^6)^3}{(1-x)^3}$ and it is enough for us to compute $P^3$ modulo $O(x^9)$, so $$ \begin{aligned} P^3+O(x^9) &= \frac 1{6^3}x^3\cdot \left(\frac {(1-x^6)^3}{(1-x)^3} +O(x^6)\right) \\ &= \frac 1{6^3}x^3\cdot \left(\frac {1}{(1-x)^3} +O(x^6)\right) \\ &= \frac 1{6^3}x^3\cdot \frac 12\left( 1\cdot 2 + 2\cdot 3 x + 3\cdot 4 x^2 + 4\cdot 5 x^3 + 5\cdot 6 x^4 + 6\cdot 7 x^5 +O(x^6)\right) \\ &= \frac 1{6^3}\cdot \frac 12\left( 1\cdot 2 x^3 + 2\cdot 3 x^4 + 3\cdot 4 x^5 + 4\cdot 5 x^6 + 5\cdot 6 x^7 + 6\cdot 7 x^8 +O(x^9)\right) \ , \end{aligned} $$ so in our case the probability is (the coefficient of $x^8$): $$\frac 1{6^3}\cdot\frac {6\cdot 7}2=\frac 7{72}\ .$$

---

Note: Here are all coefficients of $P^3$, we have computed those in degrees $3$ to $8$. (And by symmetry those in degrees $18$ to $13$.)

sage: P = (x+x^2+x^3+x^4+x^5+x^6)/6
sage: latex((P^3).expand())

$$ \frac{1}{216} \, x^{18} + \frac{1}{72} \, x^{17} + \frac{1}{36} \, x^{16} + \frac{5}{108} \, x^{15} + \frac{5}{72} \, x^{14} + \frac{7}{72} \, x^{13} + \frac{25}{216} \, x^{12} + \frac{1}{8} \, x^{11} + \frac{1}{8} \, x^{10} + \frac{25}{216} \, x^{9} + \frac{7}{72} \, x^{8} + \frac{5}{72} \, x^{7} + \frac{5}{108} \, x^{6} + \frac{1}{36} \, x^{5} + \frac{1}{72} \, x^{4} + \frac{1}{216} \, x^{3}\ . $$