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](https://en.wikipedia.org/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](https://en.wikipedia.org/wiki/Incomplete_gamma_function#Lower_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](https://math.stackexchange.com/questions/2364283/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](https://math.stackexchange.com/questions/3361616/calculating-parameter-limit-in-some-infinite-double-series).



Understanding equiprobable trinomial identity
$\newcommand{\Pr}{\mathbb{P}}$


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 $M \leq 2x + O(\sqrt{x})$. 


