Hitting a point on an expanding grid A question got posted today that intrigued me, sadly the formulation was quite unclear and the question got deleted after a couple of minutes. I will try to formulate it more clearly and hopefully correctly and give my answer. The letters I use try to follow the original poster's notation.
My formulation: Take a sequence of independent random vectors $f_t := (X_t, Y_t) \sim U \{-t,t\}^2$ (i.e. both $X_t$ and $Y_t$ have the discrete uniform distribution on the set $\{-t,..,t\}$)
Take some constant vector $p := (x,y)$. What is the probability that $(X_t, Y_t) = (x,y)$ for at least one $t$, i.e. that the random vector $f_t$ at some point hits $p$.
My questions:


*

*Is my reasoning correct?

*The answer is very simple which begs the question whether there is a more direct way to see it.

*How to calculate the product in the answer? (Answered already)


Thank you.
note: The original formulation used $U\{ -2t,2t\}$ which gives a more difficult answer (qualitatively the same, only vanishes faster), so I reformulated it in this way, because the seeming simplicity of the answer to my question is what interested me.
 A: First, note that 
$$
P(f_t=p) = \left\{\begin{aligned}
&0 && t<\max(x,y)\\
&1/t^2 && t\geq \max(x,y)
\end{aligned}
\right.$$
that is we cannot hit $(x,y)$ if neither of our random variables can attain the value of either $x$ or $y$ and if we can, than the probability is simply $1$ divided by the number of all possible pairs (of which there are $t^2$)
The find the probability of our vectors hitting $p$, we look at the complement, i.e. 
$$P\big(\;\exists t: f_t = p\;\big) = 1-P\big(\;  \forall t: f_t\neq p  \big)$$
Since all $f_t$ are independent, this can be rewritten as
$$P\big(\;  \forall t: f_t\neq p  \big) = P\Big( \bigcap_{t=1}^\infty [f_t \neq p] \Big) \stackrel{\bot}=\prod_{t=1}^\infty P\big([f_t\neq p]\big) = \prod_{t=\max(x,y)}^\infty \frac{t^2-1}{t^2}$$
Where we used the fact that
$$
P(f_t\neq p) = \left\{\begin{aligned}
&1 && t<\max(x,y)\\
&\frac{t^2-1}{t^2} && t\geq \max(x,y)
\end{aligned}
\right.$$
Now, using the reasoning provided in this answer we get the answer
$$\prod_{t=n}^\infty \frac{t^2-1}{t^2} = \frac{n-1}n$$
That is
$$P\big(\;\exists t: f_t = p\;\big)  = P(\text{we hit the point } p) = 1 - \frac{n-1}n = \frac 1n$$
Where $n= \max(x,y)$
