Geometric Probability Puzzles

Two friends agree to meet between noon and 1pm, each arriving at a uniformly random time, and each waiting 15 minutes for the other before leaving. What's the probability they actually meet?

There's no dice roll or coin flip here — the outcome is a point chosen uniformly from a continuous range, so you can't count favorable cases over total cases the way you would with a deck of cards. Instead you compare areas (or lengths, or volumes): the fraction of the whole space that counts as a "success" is the answer. This is geometric probability, and interviewers reach for it specifically to see whether you default to counting when you should be drawing a picture.

The trick: turn the problem into a shape

Let $x$ be the first friend's arrival time and $y$ be the second's, both uniform on $[0, 60]$ minutes. Every possible pair of arrival times is a point in a 60×60 square — that square is the sample space, and its area represents total probability 1. They meet if $|x - y| \le 15$, which is a diagonal band across the square. The probability of meeting is just:

$$P(\text{meet}) = \frac{\text{area of the band}}{\text{area of the square}} .$$

In plain English: instead of listing outcomes, you shade the region of the square where the condition holds and compare its area to the whole square.

Worked example 1: the meeting problem, solved

The square has area $60 \times 60 = 3{,}600$. The region where they don't meet is two corner triangles, one where $x - y > 15$ and one where $y - x > 15$. Each triangle has legs of length $60 - 15 = 45$, so each has area $\tfrac{1}{2}(45)(45) = 1{,}012.5$. Together the two triangles cover $2{,}025$. The meeting band is the rest:

$$3{,}600 - 2{,}025 = 1{,}575, \qquad P(\text{meet}) = \frac{1{,}575}{3{,}600} = \frac{7}{16} \approx 0.4375 .$$

So despite both people trying in good faith, they miss each other more than half the time — a result purely of geometry, not of anyone being unreliable.

Worked example 2: a broken stick

Snap a stick of length 1 at a uniformly random point $x \in [0,1]$, then snap the longer piece again at a uniformly random point. What's the probability the three pieces can form a triangle? The classic version of this question — break the stick at two random points $x, y$ simultaneously — reduces to the same square-and-region trick: plot $(x,y)$ in the unit square, and the "forms a triangle" condition (no piece longer than $1/2$) carves out a smaller triangle of area $1/4$ inside the unit square. The answer is $P = 1/4$: three random breaks form a valid triangle only a quarter of the time, because it's easy for one piece to accidentally end up more than half the stick.

What this means in practice

Geometric probability is really a special case of a general habit: when outcomes come from a continuous, uniform range, translate the problem into "what fraction of this shape satisfies the condition" rather than trying to enumerate cases. In interviews it shows up as meeting-time puzzles, random-point-on-a-segment puzzles, and darts-on-a-target puzzles, and the whole skill being tested is recognizing that a square, triangle, or circle is the right sample space to draw before you touch any algebra.