Suppose there are $n$ people playing rock, paper, scissors. The chance of any one of these choices is equal. A game is decisive when only two of three choices are picked - so for example if everybody picks scissors, the game is not decisive. If the players are equally split between rock and paper, then all papers win.
My first inclination is to answer $(\frac{2}{3})^n-(\frac{1}{3})^n$, since this should be equivalent to picking only two choices, then subtracting the chance of all players only picking a single option.
But you can also represent this as a multinomial distribution with the PMF of the form: $P(X=x, Y=y, Z=z) = \frac{n!}{x!y!z!}(\frac{1}{3})^n$, where $n=x+y+z$. The chance that a game is decisive will be the scenario where any one of the three outcomes are zero, and the other two nonzero. So by symmetry, we have: $P(decisive) = 3 \times \sum_{k=1}^{n-1}\frac{n!}{0!k!(n-k)!}(\frac{1}{3})^3$. If we reduce this we get $\frac{2^n-2}{3^{n-1}}$.
I'm wondering where the discrepency came from. Obviously, the answers are very similar in form, but why is it that a difference exists at all between these two approaches? Did I miss something on the naive approach?
Any help would be greatly appreciated, thank you!
Your naive approach is slightly off, I think. Let's start by finding the probability that there is a decisive game with everyone picking rock or paper. If we represent outcomes as strings of R's and P's (e.g. RRPRPPP for $n = 7$), it's easy to see the probability ought to be $$(2^n - 2)/3^n = (2/3)^n - 2 \times (1/3)^n,$$ since we want everyone to pick either R or P, but for neither R nor P to be the only choice. By symmetry, the probability we want is then $$3 \times \frac{2^n - 2}{3^n} = \frac{2^n - 2}{3^{n-1}},$$ which I think is what you meant to write for the result of your sum.