Probability that $\exists k\in\mathbb{N}: \sum_{j=1}^{3k}X_j = 2k$ is less than $1$ Let $X_j$, $j\in\mathbb{N}$, be IID random variables taking value in $\{0,1\}$ with each with an equal probability.
Then I want to prove that the probability that there exists $k\in\mathbb{N}$ such that $\sum_{j=1}^{3k}X_j = 2k$ is strictly in between $0$ and $1$.
I was able to show that the probability is larger than $0$ by computing the probability for any fixed $k$ which yields a lower bound.
However, I couldn't figure out how to show the latter.
I'd appreciate any help.
 A: The main idea is that, since $\sum_{j=1}^{3k}X_j$ should be concentrated near its mean $3k/2$, if this sum isn't equal to $2k$ for any small $k$ it should be unlikely to equal $2k$ for any large $k$. We can certainly ensure with positive probability that the sum can differ for small $k$ (e.g. by setting $X_j=0$ for small $j$); it only remains to make explicit the "large $k$ unlikely" step.

Let $N$ be a positive integer. Suppose $X_j=0$ for all $1\leq j\leq 3N$ (this happens with positive probability $1/2^{3N}$), and let $Y_k$ be the event that $\sum_{j=1}^{3(k+N)}X_j=2(k+N)$. Note that
$$\Pr[Y_k]=\frac{\binom{3k}{2k+2N}}{2^{3k}}=\frac{\binom{3k}{k-N}}{2^{3k}}\leq \frac{\binom{3k}k}{2^{3k}}.$$
This probability is pretty small; we have
$$1=\left(\frac 13+\frac 23\right)^{3k}=\sum_{i=0}^{3k}\binom{3k}{i}\left(\frac 13\right)^i\left(\frac 23\right)^{3k-i}\geq \binom{3k}{k}\frac{2^{2k}}{3^{3k}},$$
so
$$\Pr[Y_k]\leq \frac{\left(\frac{3^3}{2^2}\right)^k}{2^{3k}}=\left(\frac{27}{32}\right)^k.$$
In particular, by the union bound, the probability that $Y_k$ occurs for some $k\geq 2N$ is at most
$$\sum_{k=2N}^\infty\left(\frac{27}{32}\right)^k=\frac{32}5\cdot\left(\frac{27}{32}\right)^{2N-1}.$$
Note that $Y_k$ cannot occur for any $k<2N$ since
$$\sum_{j=1}^{3(k+N)}X_j\leq 3k<2(k+N)$$
for such $N$. So, the probability that $\sum_{j=1}^{3\ell}X_j\neq 2\ell$ for all $\ell>0$ is at least the probability that $X_j=0$ for all $1\leq j\leq 3N$ times $1-\frac{32}5\cdot\left(\frac{27}{32}\right)^{2N-1}$. For large $N$, this is positive, as desired.
