Recitation: Probability that Three Pieces Form a Triangle

What this lecture covers

This recitation problem asks: if you break a unit-length stick at two independent, uniformly random points, what is the probability that the three resulting pieces can be assembled into a triangle? It is a classic application of setting up two random variables, translating a geometric condition into inequalities, and computing a probability as an area in the plane.

It builds on the joint density and independence ideas from the lecture on multiple continuous random variables, applying them to a self-contained geometric puzzle rather than introducing new theory.

Key ideas

• Triangle inequality as the governing condition: three lengths can form a triangle only if the sum of any two exceeds the third.
• Two break points as independent uniform variables: X and Y, the two break locations, are independent and uniform on [0,1], so their joint density is uniform (equal to 1) over the unit square.
• Translating lengths into inequalities: assuming X < Y, the three piece lengths are X, Y - X, and 1 - Y; the triangle condition on these lengths simplifies to Y > 1/2, X < 1/2, and Y < X + 1/2.
• Probability as area: because the joint density is uniform, the probability of forming a triangle equals the area of the region in the unit square satisfying all three inequalities.
• Case splitting by order: since either break point could come first, the problem is solved once for X < Y and once for X > Y, and symmetry makes the second case identical to the first.

Walkthrough

Setting up the pieces and the triangle condition (0:01)

A stick of length 1 is broken at two independently and uniformly chosen points. The video first works out what condition the three resulting pieces must satisfy to form a triangle: any two pieces combined must be longer than the third, applied to all three pairs.

Reducing to inequalities in X and Y (3:05)

Assuming the first break point X lands to the left of the second, Y, the three lengths are X, Y - X, and 1 - Y. Substituting these into the three triangle-inequality conditions and simplifying algebraically reduces them to Y > 1/2, X < 1/2, and Y < X + 1/2.

Finding the region and its area (6:07)

Since X and Y are independent and uniform, their joint density is uniform over the unit square, so the probability of the event is just the area of the region satisfying the three inequalities. Plotting the three boundary lines shows this region is a small triangle with area 1/8.

Handling the other ordering (9:10)

The case where the second break point falls to the left of the first is handled by swapping the roles of X and Y in the same three conditions. By symmetry, this produces a region of the same shape and area, 1/8, in the other half of the square.

Combining both cases (11:13)

Adding the two symmetric contributions gives a total probability of 1/8 + 1/8 = 1/4 that the three pieces can form a triangle.

Before you watch

• Be comfortable modeling two independent random choices as a joint uniform density on a square.
• Know the triangle inequality (sum of any two sides must exceed the third).
• Reviewing how probabilities correspond to areas under a uniform joint density helps this problem go smoothly.

Check your understanding

1. Why does independence of the two break points make their joint density uniform over the unit square?
2. How do the three lengths of the pieces get expressed in terms of X and Y when X < Y?
3. Why is the probability of forming a triangle equal to the area of a region rather than requiring a full double integral?
4. Why does the case Y < X give the same probability as the case X < Y?