Distribution of the number of trials until the last arrival in a Bernoulli process before an arbitrary point in time $t$

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I was looking at a problem in the book Introduction to probability by Bertsekas and Tsitsiklis (problem 5 ch. 6) but I got confused by the solution.

Without addressing the details of the problem, suppose that, in a Bernoulli process with parameter $p$, you observe a failure at time $t$. Let $M$ be the number of the last win before time $t$ and $N$ be the number of the next win after time $t$.

Now $X = N-t$ follows a geometric distribution with parameter $p$. The solution argues that $Y = t-M$ is also geometrically distributed with the same parameter "by symmetry and independence of the games". However I'm not sure I understand the last part.

Is it just assumed that $t$ is infinitely large for convenience? If so how could we answer upon the distribution of $Y$ if $t$ was given a specific value?

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Well a couple hours after I posted the question I realised the answer, so I'm answering it here for completeness.

The problem apparently assumes that $t$ is a point after an infinite amount of time of the bernoulli process, so $Y$ is indeed geometrically distributed.

This results in an interesting concept which is known as the The Random Incidence Paradox.

As for when $t$ was given a specific non-infinite value, then the distribution of $Y$ is the like asking the distribution of the number of trials until the first success given that the number is less than $t$, but I'm not going to go into details regarding that.