Understanding equiprobable trinomial identity With $f(x_1,x_2,x_3,x_1+x_2+x_3;\,1/3,1/3,1/3):= \frac{(x_1+x_2+x_3)!}{x_1!\,x_2!\,x_3!\, 3^{x_1+x_2+x_3}}$ denoting the probability mass function of the equiprobable trinomial distribution as in wiki/Multinomial_distribution, for any $x\in \mathbb N$ it holds, $$\sum_{k_1,k_2\in \mathbb N_0} f(x-1,x+k_1,x+k_2;\, 1/3,1/3,1/3)=1.$$
Because by rewriting $((x-1)+(x+k_1)+(x+k_2))! = \Gamma(3x+k_1+k_2)=\int_0^\infty e^{-t} t^{3x+k_1+k_2-1} \,dt$ and applying $\sum_{k=0}^\infty \frac{z^k}{\Gamma(s+k+1)}=\frac{e^z}{z^s\,\Gamma(s)} \, \gamma(s,z)$ (cf. wiki/Incomplete_gamma_function) both sums may be calculated separately and afterwards the $t$ may be integrated out again. (cf. math.SE/double-series-equals-gamma-function.) Therefore, the identity is true even for $x\in [1,\infty)$ if the factorials are replaced by gamma functions.
Is there an intuitive reason behind the identity? (Combinatorial/probabilistic/anything other than mere calculation)
Also, is it true that for $0<\alpha<1/2$, it holds,
$$ \lim_{\substack{x\to\infty\\x\in\mathbb N}} \sum_{\substack{k_1,k_2\in \mathbb N_0\\k_1 < \alpha(k_1+k_2)}} f(x-1,x+k_1,x+k_2;\, 1/3,1/3,1/3)=0.$$
This is an open question on math.SE/calculating-parameter-limit-in-some-infinite-double-series.
 A: $\newcommand{\al}{\alpha}
\newcommand{\be}{\beta}$
The conjecture about the limit can written as 
\begin{equation*}
 S:=\sum_{j=0}^\infty\sum_{k=m_j}^\infty a_{k,j}\underset{x\to\infty}\longrightarrow0, \tag{1}
\end{equation*}
where 
\begin{equation*}
 m_j=\lfloor\be j\rfloor+1,\quad \be:=(1-\al)/\al>1,
\end{equation*}
\begin{equation*}
 a_{k,j}:=\frac{(3x+k+j-1)!}{(x-1)!(x+k)!(x+j)!\,3^{3x+k+j-1}}. 
\end{equation*}
This conjecture is false. Indeed, let $x\to\infty$. Let us write $A\ll B$ and/or $B\gg A$ if $B>0$ and $|A|=O(B)$, and let us write $A\asymp B$ if $A\ll B\ll A$. Let 
\begin{equation*}
 r_{k,j}:=\frac{a_{k+1,j}}{a_{k,j}}. 
\end{equation*}
Then, assuming that $0\le k\ll\sqrt x$ and $0\le j\ll\sqrt x$, we have
\begin{equation*}
 r_{k,j}-1=\frac{j-2k-3}{3x+3k+3}\ll\frac1{\sqrt x}, 
\end{equation*}
whence $a_{k,j}\asymp a_{0,j}$, and similarly $a_{0,j}\asymp a_{0,0}$, so that 
\begin{equation*}
 a_{k,j}\asymp a_{0,0}. 
\end{equation*}
In turn, by Stirling's formula, 
\begin{equation*}
 a_{0,0}\asymp1/x. 
\end{equation*}
Thus, 
\begin{equation*}
 S\ge\sum_{\sqrt x\le j\le2\sqrt x}\;\sum_{k=m_j}^{2m_j} a_{k,j}
 \gg\sum_{\sqrt x\le j\le2\sqrt x}\;\sum_{k=m_j}^{2m_j} a_{0,0}\gg1,
\end{equation*}
so that (1) does not hold. 
A: $\newcommand{\Pr}{\mathbb{P}}$
I claim that this limit is in fact not zero. For each $N$, write $\Pr_N$ to be the probability measure on the triple $(X_1,X_2,X_3)$ that is trinomial with $N$ trials and parameters all equal to $1/3$.  Then the sum you are interested in is \begin{align}\sum_{N} \Pr_N(X_1 = x-1, 0 \leq (X_2 - x) \leq \alpha(X_2 + X_3 - x - x) ) = \sum_{N} \Pr_N(X_1 = x-1,0 \leq (X_2 - x) \leq \alpha(N - 3x + 1) ) \,. \end{align}
I'll show that the above sum restricted to $3x - 1 \leq N \leq 3x - 1 + O(\sqrt{x})$ is bounded below.  Note that \begin{align}\Pr_N(X_1 = x- 1,& 0 \leq (X_2 - x) \leq \alpha(N - 3x + 1)) \\
&= \Pr_N(X_1 = x-1) \cdot \Pr_N(0 \leq (X_2 - x) \leq \alpha(N - 3x + 1)\,|\,X_1 = x-1)\,.\end{align}
Note that for $3x-1 \leq N \leq 3x - 1 + O(\sqrt{x})$, we have that $\Pr_N(X_1 = x-1) = \Omega(1/\sqrt{x})$.  You can either compute this precisely with Stirling's formula, or use a local central limit theorem.
Now, note that the law of $X_2$ conditioned on $X_1 = x -1$ is that of a binomial random variable with $N - x + 1$ trials and success probability $1/2$. Writing $M = N - x - 1$, we then have  $$\Pr_N(0 \leq (X_2 - x) \leq \alpha(N - 3x + 1)\,|\,X_1 = x-1)= \Pr(0 \leq (Y - x) \leq \alpha(M - 2x))$$
where $Y$ is the aforementioned binomial random variable. We can rewrite this last probability as $$\Pr\left(-\frac{1}{2}(M - 2x) \leq Y - \mathbb{E} Y \leq -(\frac{1}{2} - \alpha)(M - 2x)\right)\,.$$
By the central limit theorem, this is bounded below for $M \leq 2x + O(\sqrt{x})$. 
