Max Equals Min

Consider two integers, $$a$$ and $$b$$ , chosen independently and uniformly at random from the set $S = \{-10, -9, \ldots, 9, 10\}$ . Determine the probability that $\max(0,a) = \min(0,b)$ .

Answer

Condition on $$a$$ . If $a \leq 0$ , then $\max(0,a) = 0$ . For equality, we need $\min(0,b) = 0$ , which requires $b \geq 0$ . Hence,

\Pr(a \leq 0) = \frac{11}{21}, \quad \Pr(b \geq 0) = \frac{11}{21},

and the contribution to our probability is $\tfrac{11}{21} \times \tfrac{11}{21} = \tfrac{121}{441}$ .

If $$a > 0$$ , then $\max(0,a) = a$ , which is positive, while $\min(0,b)$ is always $\leq 0$ . They cannot be equal, so this case contributes $$0$$ .

Therefore, by the total law of probability, the desired probability is

\frac{121}{441}.