solve by intuition or counting Say there are 1000 people, and you arbitrarily choose 50 of them. What's the probability you choose the "first" person?
To solve this type of problem, I always count: $$\frac{1 * \binom{999}{49}}{\binom{1000}{50}}$$
But when I asked my friend, he immediately said 50/1000 without doing a counting argument. How does someone know the answer w/o doing the counting because otherwise I just don't know how to do it. My friend just said "it just is" but that's obviously a useless answer haha.
 A: 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 $1000$.
A: 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 $\frac{50}{1000}$.
