ML Interview Q Series: Stick Break Probability: No Piece Over Half Length via Joint Density Integration.

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A stick is broken into three pieces at two randomly chosen points on the stick. What is the probability that no piece is longer than half the length of the stick?

Short Compact solution

We label the two random break points as X and Y, with X being the position closer to the left endpoint of the stick. Since the stick has length 1, the density function of (X, Y) with 0 < X < Y < 1 is f(x, y) = 2. We need 0 ≤ X ≤ 0.5, Y−X ≤ 0.5, and 1−Y ≤ 0.5. This translates to 0 ≤ X ≤ 0.5 and 0.5 ≤ Y ≤ x + 0.5. Integrating the joint density over this region gives

Hence, the probability that no piece exceeds half the stick's length is 1/4.

Comprehensive Explanation

Setting up the problem

We have a stick of unit length (length = 1). We choose two cut points independently and uniformly on (0, 1) and label them in ascending order as X and Y. So X is always the smaller cut point and Y is the larger cut point. That means 0 < X < Y < 1.

Once these cuts are made, we get three pieces:

1. From the left end to X, which has length X.

2. From X to Y, which has length Y - X.

3. From Y to the right end (which is at 1), so its length is 1 - Y.

We are interested in the probability that none of these three lengths exceeds 1/2. In other words, each piece has length at most 1/2.

Joint density function of X and Y

Because we chose two cut points independently and uniformly on (0, 1), the probability density function (pdf) for unordered break points U1 and U2 is uniform over the unit square. However, once we explicitly order them so that X < Y, we are restricting the domain to the triangle 0 < X < Y < 1. Over that triangle, the joint pdf is f(x, y) = 2. This factor of 2 arises because for each unordered pair (U1, U2) in the unit square, exactly one of them will be the smaller (X) and the other the larger (Y). The area of the region 0 < x < y < 1 is 1/2, and so the pdf must integrate to 1 over that triangle. Hence f(x, y) = 2 in 0 < x < y < 1 and 0 otherwise.

Conditions for no piece exceeding 1/2

We want:

• Piece 1: X ≤ 1/2

• Piece 2: (Y - X) ≤ 1/2

• Piece 3: (1 - Y) ≤ 1/2

Rewriting these:

1. X ≤ 1/2

2. Y - X ≤ 1/2 ⇒ Y ≤ X + 1/2

3. 1 - Y ≤ 1/2 ⇒ Y ≥ 1/2

Therefore, we must have 0 ≤ X ≤ 1/2 and 1/2 ≤ Y ≤ X + 1/2. Of course, we also must keep in mind that Y ≤ 1, but that is automatically ensured for X in [0, 0.5] since X + 0.5 ≤ 1.0 + 0.5 = 1.5, and the upper bound is effectively min(X+0.5, 1). Within 0 ≤ X ≤ 0.5, if X + 0.5 exceeds 1, the region is clipped by Y ≤ 1. But we only integrate in the domain X < Y < 1, so practically the relevant region is:

• X from 0 to 0.5

• Y from 0.5 to min(X + 0.5, 1)

For X ∈ [0, 0.5], X + 0.5 ∈ [0.5, 1]. Hence the region for Y is 0.5 ≤ Y ≤ 0.5 + X and also Y ≤ 1. But 0.5 + X is at most 1 when X ≤ 0.5. So effectively, Y goes from 0.5 to X + 0.5, and that is always within [0.5, 1].

Computing the probability

Because f(x, y) = 2 in the region 0 < x < y < 1, the probability we want is the double integral of 2 over the region specified:

We integrate in X from 0 to 0.5, and for each X, Y goes from 0.5 to 0.5 + X:

Inside the inner integral:

• Integrand = 2

• Integration with respect to y from 0.5 to x+0.5

So the inner integral is 2 * [(x + 0.5) - 0.5] = 2 * x = 2x. Now we integrate that in X from 0 to 0.5:

Thus, the probability is 1/4.

Intuitive interpretation

• Because the cuts are random, sometimes you get one long piece and the other two short pieces.

• The condition “no piece longer than 1/2” is fairly restrictive. It restricts where you can place X and Y so that the longest piece is at most 1/2.

• It turns out that precisely 1/4 of all possible placements of the break points satisfy this constraint.

Possible Follow-up Questions

How would you simulate this in code to confirm?

One approach is to perform a Monte Carlo simulation. We can randomly generate two points in [0, 1], order them as X < Y, and then check whether the resulting three segments all have lengths ≤ 1/2. We accumulate the fraction of trials that satisfy this. Below is a quick Python snippet:

import random

def simulate_no_piece_longer_than_half(num_samples=10_000_000):
    count = 0
    for _ in range(num_samples):
        u1 = random.random()
        u2 = random.random()
        X, Y = min(u1, u2), max(u1, u2)
        # Calculate segment lengths
        seg1 = X
        seg2 = Y - X
        seg3 = 1 - Y
        if seg1 <= 0.5 and seg2 <= 0.5 and seg3 <= 0.5:
            count += 1
    return count / num_samples

prob_estimate = simulate_no_piece_longer_than_half()
print("Monte Carlo estimate:", prob_estimate)

After enough samples, the result should approximate 0.25.

Why is the joint pdf f(x, y) = 2 for 0 < x < y < 1?

When two points U1 and U2 are chosen independently in (0, 1), their joint pdf is 1 over the unit square (0, 1) × (0, 1). But when we condition on X = min(U1, U2) and Y = max(U1, U2), we restrict to the set 0 < X < Y < 1, which is exactly half the unit square (because for any point in the unit square, either U1 < U2 or U1 > U2; the diagonal line U1 = U2 has measure zero and can be ignored). Thus, the area of that triangle is 1/2. Since probability must sum (integrate) to 1, the uniform density over that triangle must be 2.

What if we wanted the probability that all three pieces are exactly in non-decreasing order, or some other arrangement?

If the question were about, for instance, the order of lengths, we would have to account for different conditional constraints. The uniform assumption about the cut points might remain the same, but then you would layer on additional inequalities, such as seg1 ≤ seg2 ≤ seg3, and so on. Each scenario can often be handled by describing the region in terms of X and Y and carefully integrating. The essential approach is the same: translate the desired condition into inequalities involving X and Y, and integrate over that region using the known joint pdf.

Is there a simpler geometric interpretation?

Yes. When picking two points on a segment, you can imagine the unit square (U1, U2). The line U1 = U2 splits that square into two right triangles. Label the lower one as X = U1 < U2 = Y. Then we transform the condition "no piece > 1/2" into constraints on X and Y. This forms a region in that right triangle. The probability is the area of that region times the factor for the pdf. Sometimes, for uniform distributions, these problems reduce to a straightforward area computation in the plane.

Could the result change if the break points weren’t uniformly chosen?

Absolutely. The result 1/4 heavily relies on the uniform distribution assumption. If the break points are distributed in some other way (e.g., a distribution that concentrates near the center), the probability of no piece exceeding 1/2 might be different. The uniform case simply means every point in (0,1) is equally likely for a break.

Are there any boundary effects if the break points could be chosen outside (0,1)?

In the usual statement of this classic “broken stick” type problem, we always assume we’re strictly inside (0,1). Including boundary points at 0 or 1 for the breaks would trivially produce a piece of length 0, and typically that’s excluded. So we keep the breaks strictly in the interior. If we allowed a break at the endpoints, the distribution or the integrals might need slight modifications.

These clarifications illustrate the depth of reasoning behind the original result and show that the final probability 1/4 is a neat outcome of carefully integrating over the appropriate region under uniform assumptions.

Below are additional follow-up questions

If we consider the complementary event where at least one piece is longer than half, is that approach simpler or more complex?

One might try to compute the probability of the complementary event: “At least one segment exceeds length 1/2,” and then subtract that from 1. The direct approach is to see if any one of the three segment lengths is greater than 1/2. In principle, you could consider separate events:

• Segment 1 > 1/2

• Segment 2 > 1/2

• Segment 3 > 1/2

and then apply the inclusion-exclusion principle because these events can overlap (though it is impossible for two segments to both exceed 1/2 simultaneously, but you still need to track that logic). Concretely:

• For Segment 1 > 1/2, that requires X > 1/2.

• For Segment 2 > 1/2, that requires Y - X > 1/2.

• For Segment 3 > 1/2, that requires 1 - Y > 1/2, or Y < 1/2.

You could sum the probabilities of these conditions, but you must carefully note that for a continuous probability distribution, the event “Segment 1 > 1/2 and Segment 2 > 1/2” has probability zero, yet it still merits conceptual inspection. In practice, computing the area of the region that satisfies at least one of these inequalities may be somewhat more cumbersome in geometry. Hence, it can be instructive, but you often end up with a more direct calculation by focusing on “none of the segments exceed 1/2,” which is a more compact set of inequalities.

A potential pitfall here is mixing up or double-counting areas in the plane if you attempt a naive inclusion-exclusion. Another subtlety is remembering that we are working only in the triangle 0 < X < Y < 1 with density 2, rather than in the entire unit square. Real-world parallels might include scenarios where you only measure one segment in some practical test, incorrectly concluding about the total probability of having a large segment—thus leading to biases or incomplete data if you do not account for all three pieces systematically.

How can this be generalized to the case of more than two breaks?

If we place N > 2 random breaks (so that we end up with N+1 pieces), then the probability that no piece exceeds 1/2 will depend on more complex constraints among all break positions. Symbolically, you have ordered positions X1 < X2 < ... < XN in (0,1), each corresponding to one of the random cut points. Then the piece lengths are X1, X2 - X1, X3 - X2, ..., XN - X_{N-1}, and 1 - XN. Ensuring each piece is ≤ 1/2 translates into a set of N+1 inequalities. The region you need to integrate over can be quite complicated.

The main pitfall is that for large N, enumerating or visualizing the feasible region is not trivial (it becomes an intersection of many constraints in an N-dimensional space). Monte Carlo simulation is often easier for large N, although symbolic integration can still be done with a careful approach. A real-world analogy is a manufacturer who repeatedly slices a wire at multiple points: the constraints for each piece might involve different tolerances, and verifying them all in a high-dimensional space can be tricky without systematic simulation tools.

If we want the conditional probability that the middle segment is the smallest one given that no segment exceeds 1/2, how would we proceed?

You could compute this conditional probability by first restricting attention to the event “no segment > 1/2.” Within that subset, you need to calculate the probability that Y - X (the middle segment) is smaller than both X and 1 - Y. Formally, you would write:

P(middle piece is smallest | no piece exceeds 1/2) = [ P(middle piece is smallest AND no piece exceeds 1/2) ] / [ P(no piece exceeds 1/2) ].

You’d look for the region in 0 < X < Y < 1 where X ≤ 1/2, Y - X ≤ 1/2, 1 - Y ≤ 1/2, and also Y - X ≤ X and Y - X ≤ 1 - Y. After setting up those inequalities, you integrate the joint density 2 over that domain in the X-Y plane.

A subtlety arises when you do the comparisons among all segments, because you must double-check that your region definitions do not accidentally omit or double-count certain boundary conditions (like X = Y - X or Y - X = 1 - Y). In practice, carefully enumerating which piece is the smallest can be error-prone if you do not keep clear track of inequalities. In real-world engineering contexts, an equivalent situation might be ensuring that a certain sub-component dimension is the “minimum dimension” while also satisfying an upper-limit constraint on all sub-component sizes.

Would the problem change if the breaks were chosen sequentially, with knowledge of the first break before choosing the second?

Yes, if the second break is somehow dependent on the position of the first break (for instance, the second break point is more likely to be near the center if the first break happened near an end), the distribution of the pair (X, Y) is no longer uniform over 0 < X < Y < 1. The joint distribution f(x, y) must then be derived from that conditional selection. For instance, if you first pick X uniformly, but then place Y in a narrower sub-interval around the center with higher probability, the feasible region for (X, Y) might not even be described by a simple formula like 2 in the triangular region.

A major pitfall would be to assume independence or uniformity in a real process that is sequential and dependent. In practical scenarios—say, repeated breaks in a brittle material—micro-fractures might cluster in certain areas, so subsequent breaks are not uniformly distributed.

How does this problem connect to the “stick-breaking” interpretation in Bayesian nonparametrics (such as in Dirichlet processes)?

In Bayesian nonparametrics, the “stick-breaking” process typically involves a sequence of random fractions Beta-distributed (for example, Beta(1, α)) to successively break off portions of the stick. Here, the distribution of break points is not uniform in the same sense. Instead, each break fraction is drawn from a distribution that depends on a concentration parameter. Thus, the final arrangement of segments has very different distributional properties than the uniform two-cut scenario.

One pitfall is conflating the classical “break a stick at two uniform points” problem with the iterative Beta-based stick-breaking used in Dirichlet processes. They share the metaphor of breaking a stick, but the random mechanism is quite different. In real-world usage, such Bayesian stick-breaking processes are employed to create discrete distributions with a countably infinite number of clusters. That is not the same as physically cutting a single stick at two uniform points.

What are some challenges if we tried to solve this by enumerating lengths rather than by direct integration?

An enumerative approach might say: “Let’s treat X, Y - X, and 1 - Y as random variables and systematically list out all possible ways they can be ≤ 1/2.” But enumerating continuous variables is not possible in a discrete sense. You could attempt to break the unit interval into small increments and approximate the probability mass, but that effectively becomes a numerical approximation to the integral.

A major pitfall is the loss of accuracy or heavy computational burden if the incremental steps are too coarse. This approach also becomes exceedingly cumbersome if there are many constraints or if you increase the number of break points. In real-world problems (e.g., designing a manufacturing process with continuous dimension constraints), you would rarely try to “enumerate” all possible dimension sets. Instead, you turn to integral geometry or simulation-based approximations for continuous data.

How can we adapt the solution if we want the probability that the largest piece is strictly less than r, for some r < 1/2?

You can generalize by replacing 1/2 with r in all inequalities. You would require:

• X ≤ r

• Y - X ≤ r

• 1 - Y ≤ r

Then you examine the region 0 < X < Y < 1 where these constraints hold: X ≤ r, Y ≤ X + r, and Y ≥ 1 - r. You would compute

∫ [over X in (0, r)] ∫ [over Y in (max(X, 1 - r), X + r)] 2 dy dx,

making sure that the region is valid (specifically, for r < 1/2, one must confirm that X + r ≥ 1 - r for at least part of that range). The resulting area can be written as a function of r. From a real-world standpoint, you might imagine you cannot tolerate any piece greater than some fraction r for mechanical reasons, so you are interested in the design feasibility given that maximum dimension requirement.

The primary pitfall is to handle overlapping constraints carefully: for the integral to make sense, you need X + r ≥ 1 - r, which simplifies to X ≥ 1 - 2r. So you must also ensure that 1 - 2r is within (0, r), which is only possible for certain r < 1/3. If r ≥ 1/2, the question trivially has probability 1, so the real interest is r in (0, 1/2). It illustrates that you cannot blindly apply the same integration bounds without checking the geometry of the intervals.

Could randomness near the boundaries cause unexpected phenomena when doing a physical experiment?

Yes. In a physical setup, you might have measurement error or an inability to break precisely at a point extremely close to 0 or 1. If the break can land at 0.00001, that might still effectively produce a tiny piece that is negligible in length, and in practice might break off or crumble, leading to more complicated outcomes. When experiments are performed on real materials, the location of fractures might be subject to stress concentrations, micro-cracks, or structural inhomogeneities, so the assumption of uniform randomness can fail drastically.

Potential pitfalls include:

• Overlooking that real sticks might not break “cleanly” at a single point, leading to an irregular cut geometry.

• Not accounting for systematic biases in how or where fractures tend to initiate (end effects vs. center stress).

• Relying on the theoretical uniform cut model can produce inaccurate predictions if these real-world complications significantly shift the distribution of break locations.

In any engineering or materials-science application, verifying whether the uniform random assumption is even remotely accurate should be a key step—failing to do so can produce misleading estimates of probabilities such as “no piece exceeds 1/2.”