Hitting a point on an expanding grid

Question

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:

1. Is my reasoning correct?
2. The answer is very simple which begs the question whether there is a more direct way to see it.
3. 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.

Answer 1

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 $n= \max(x,y)$