Consecutive rolls of a die whose face count increases by $1$ with each roll. The game starts with a dice with (let's say) $1000$ faces. Every roll, the die magically gets another face. So, after $1$ roll it has $1001$ faces, after another $1002$, and so on. Now the question is:

What is the probability of rolling at least one $1$ in the first $x$ rolls?

 A: Hint If the die at first has $\require{cancel}F$ faces, then


*

*Since there are $F - 1$ faces that have a label other than $1$, the probability of not rolling $1$ on the first roll is
$$\frac{F - 1}{F} .$$

*For the second roll, there are $F + 1$ faces and $F + 1 - 1 = F$ faces that have a label other than $1$. So, the probability of not rolling a $1$ on the second roll is
$$\frac{F}{F + 1} .$$
Since rolling $1$ on the first roll and rolling $1$ on the second are independent events, the probability of not rolling $1$ on either of the first two rolls is the product of the two above probabilities, namely,
$$\frac{F - 1}{\cancel{F}} \cdot \frac{\cancel{F}}{F + 1} = \frac{F - 1}{F + 1},$$

*For the third roll, there are $F + 2$ faces and $F + 1$ of these have a label other than $1$. So, the probability of not rolling $1$ on the third roll is $$\frac{F + 1}{F + 2} ...$$

A: Compute the complement of the propability to not get a 1, i.e.
$$p(x)=1-\prod_{k=1}^x(1-\frac{1}{999+k})= \frac{x}{999+x}$$
