solve by intuition or counting

Make your choices in the other order. First arbitrarily (and with equal probability) choose a subset of $50$ people. What is the probability that a person selected randomly and uniformly from within your group of $1000$ is within that subset? Seems pretty clear that probability is $50$ hits out of a universe of $1000$.

Method 1 - The obvious method

Your idea of a "first person" is similar to marking out one person, so I'll call that person A.

Now the chance that A won't be chosen will be given as follows:

In the group of 1000 people, there are 999 ways to choose someone else. Hence, probability from this step will be $\frac{999}{1000}$.

In the remaining group of 999 people, there are 998 ways to choose someone else. Hence probability from this step will be $\frac{998}{999}$.

In the remaining group of 998 people, there are 997 ways to choose someone else. Hence probability from this step will be $\frac{997}{998}$.

...

In the remaining group of 951 people, there are 950 ways to choose someone else. Hence probability from this step will be $\frac{950}{951}$.

This means that the chance that A won't be chosen is $\frac{999}{1000}\cdot\frac{998}{999}\cdot\frac{997}{998}\cdot ...\cdot\frac{950}{1000} = \frac{950}{1000}$.

Hence the chance that A will be chosen is $1-\frac{950}{1000}=\frac{50}{1000}$.

Method 2 - A logical method

Probability-wise, there is nothing distinguishing A from any other person. Hence, every person has an equal probability of being chosen.

Hence, the chance that, in those 50 people, A will be a member, would be $\frac{50}{1000}$.