Thinking about my undergrad days studying math, I wish I more problems were visualized like this

Here's a problem that shows up in math competitions and quant interviews:

Drop 4 points randomly on a circle. What are the chances they all land in the same half?

Try it a few times. Sometimes all four cluster into one semicircle, sometimes they spread out. What probability would you guess?

The obvious (wrong) answer

The circle is symmetric, so place one point anywhere and ask whether the other three land in the same semicircle. Each of those 3 points independently has a $1/2$ chance of landing in that half, so the probability should be $(1/2)^3 = 1/8 = 12.5\%$

Test that reasoning here. Click on the circle to fix your point, then drop 3 more and see whether they land in your semicircle. Repeat it many times and watch the rate:

The rate hovers around 12.5%. The reasoning checks out for a fixed semicircle. But go back to the first demo and drop 4 points a bunch of times. The rate there is closer to 50%, not 12.5%. Something is off.

The semicircle is not fixed

We are not asking "do the points fit in this particular semicircle?" We are asking "do the points fit in any semicircle?" The semicircle gets to move. It can be anchored at whichever point makes it work.

Drop 4 points below and click each one to try anchoring a semicircle there. Notice that at most one anchor ever works:

Fixing the semicircle in advance ignores configurations where the points do cluster together, just not around the chosen anchor. The real question is whether some point is a valid anchor. That is a much more generous condition.

Anchoring at each point

Label the 4 points $1, 2, 3, 4$

$E_i = \text{"all other points lie in the clockwise semicircle starting at point } i\text{"}$

We already know the probability of each individual event. Once point $i$ is the anchor, each of the other $N - 1$

$P(E_i) = \left(\frac{1}{2}\right)^{N-1}$

For 4 points, that is $(1/2)^3 = 1/8$

The points all fit in some semicircle exactly when at least one of these events occurs. We want the probability of the union $E_1 \cup E_2 \cup \cdots \cup E_N$

$N \times \frac{1}{2^{N-1}}$

For $N = 4$

Only one anchor ever works

Go back to the anchor picker and try clicking different points. Whichever anchor's semicircle contains all the others, click the remaining points. Their semicircles always miss somebody.

Place points below and click one to see its semicircle. Watch the gap (the empty arc going counterclockwise from the farthest point back to the anchor):

When $E_i$

Step through this argument on concrete points:

At most one $E_i$

$P(\text{all in some semicircle}) = \sum_{i=1}^{N} P(E_i) = N \cdot \frac{1}{2^{N-1}} = \frac{N}{2^{N-1}}$

For 4 points: $4/8 = 1/2$

Checking the formula

The formula predicts that any two points always fit in a semicircle ($N = 2$

Adjust $N$ and run batches. The simulated rate converges to the prediction:

Smaller arcs

Nothing in the argument required the arc to be a semicircle. If the arc has length $x$ times the circumference (where $x \leq 1/2$

$P(\text{all } N \text{ points in some arc of length } x) = N \cdot x^{N-1}$

Adjust the arc size and number of points:

For example, 5 points all fitting in an arc covering $1/3$ of the circumference: $5 \times (1/3)^4 = 5/81 \approx 6.2\%$

Beyond the circle

The same decomposition works in higher dimensions. For $N$ random points on a sphere, the probability they all lie in a hemisphere is also $N / 2^{N-1}$