Alternative way of finding the Distribution of $Y_{3} = (X_{1}X_{2}X_{3})^{1/3}$ Suppose that $X_{1}, X_{2}, X_{3}$ are i.i.d from $p_{x}$. Determine the distribution of $Y_{3} = (X_{1}X_{2}X_{3})^{1/3}$.
$p_{X}(x) = \begin{cases}
1/2 & \text{if } x = 1,\newline
1/4 & \text{if } x = 2,\newline
1/4 & \text{if } x = 3,\newline
0 & \text{otherwise}.\newline
\end{cases}$
I can solve this problem by multiplying all the different combinations of values of x and finding the corresponding probability of Y. But there will be a lot of combinations and I feel like there has to be a better way of solving this problem.
Thank you.

The simplest method I can think of in this case exploits the symmetry (equal probabilities of 2 and 3) to evaluate the distribution of $$\log Y = \left(\log X_1 + \log X_2 +\log X_3\right)/3$$ by expanding the generating function of the common distribution $f(t) = 1/2 + \left(t^{\log 2} + t^{\log 3}\right)/4.$  Writing $a$ and $b$ for the two logarithms and expanding via the multinomial theorem we obtain the expansion
$$f_{3\log Y}(t) = (3f(t))^3 = \frac{1}{8} \sum_{i+j+k=3} \binom{3}{i;j;k} 2^{-j-k}\, t^{ja + kb} = \frac{6}{8} \sum_{i=0}^3 \frac{1}{i! 2^{3-i}}\sum_{j+k=i}\frac{t^ja+kb}{j!k!}.$$
The inner sum is symmetric in $(j,k),$ so that we only need to inspect the values for the cases $(j,k) = (0,0); (1,0); (2,0), (1,1);$ and $(3,0), (2,1).$
It's convenient to use the computer for this.  For instance, Wolfram Alpha will display the coefficients by evaluating Expand[(1/2 + (t^a+t^b)/4)^3].



This tells us, for instance, that the chance $3\log Y = 2\log 2 + \log 3$ equals the chance $3\log Y = 2\log 3 + \log 2$ and each of these chances is the coefficient of $t^{2a+b},$ equal to $3/64.$  Equivalently, this is the chance $Y^3 = \exp(2\log 2 + \log 3) = 2^2(3) = 12$ and it's the chance $Y^3 = \exp(2\log 3 + \log 2)=18.$
