Suppose we have a box with balls labeled $1, \dots, n, n \geq 4$. Suppose I choose $B_1$ from the box, and then $B_2$ from it without replacement. I want to compute $P[B_2 > B_1 + 3]$.
To do this, I was considering an argument via the law of total probability. Breaking this down into conditioning on $B_1$ we find that:
$$ P[B_2 > B_1 + 3] = \frac{1}{n} \left( P[B_2 > B_1 + 3 \mid B_1 = 1] + \dots + P[B_2 > B_1 + 3 \mid B_1 = n-4] \right) $$
We can write $P[B_2 > k + 3 | B_2 = k]$ as $\frac{n-(k+3)}{n-1}$. So we can write out our answer as:
$$ \sum_{k=1}^{n-4} \frac{1}{n} \frac{n-(k+3)}{n-1} $$
I'm preparing for my midterm, and I came across this problem that I was able to solve rather dirtily with the Law of Total Probability. I was wondering if there was a nicer argument via random variables. I understand that $B_1$ is a uniform discrete r.v., but how do I define $B_2$, given that the sample space is dependent on what $B_1$ is?
All $N=n(n-1)/2$ possible ordered pairs have the same probability, so we can just count the number of outcomes. I hope this picture will give you a hint how to do it.