ML Interview Q Series: Geometric Probability: Forming a Triangle from Uniform Random Stick Breaks.

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If a stick is split randomly at two uniform points along its length, forming three segments, what is the probability that these three parts can be arranged into a valid triangle?

Comprehensive Explanation

A stick of length 1 is chosen for convenience, because scaling does not change the probabilities involved. We then pick two independent break points X and Y uniformly in [0, 1]. Let those points be such that 0 <= X <= Y <= 1. The stick is cut at X and Y, resulting in three segments:

Segment A = X Segment B = Y - X Segment C = 1 - Y

The fundamental criterion for these pieces to form a valid triangle is the triangle inequality. The triangle inequality requires three simultaneous conditions in order for sides of lengths x, y, and z to form a triangle:

x+y>z

y+z>x

z+x>y

Below each condition we replace x, y, and z with our segments A, B, and C. The triangle is valid only if all three inequalities hold.

One way to understand the probability is via geometric arguments in the 2D space where the axes are X and Y:

• Since we are dealing with X <= Y, we only consider the triangular region of the unit square where 0 <= X <= Y <= 1.

• In the X-Y plane, the total area of this region is 1/2.

• The constraints imposed by the triangle inequalities define a subset of this triangular region.

Another well-known interpretation is that no single segment can exceed 1/2 in length, because if one piece is longer than 1/2, then that piece alone would be longer than the sum of the other two, violating the triangle inequality. Consequently, the region of valid (X, Y) pairs is where:

• X is less than 1/2,

• Y - X is less than 1/2,

• 1 - Y is less than 1/2.

By integrating or computing the area of that valid region in the X-Y plane, you find that the probability is 1/4.

It is instructive to confirm this with a simulation.

import random

def triangle_probability(num_samples=10_000_000):
    count = 0
    for _ in range(num_samples):
        x, y = sorted([random.random(), random.random()])
        a = x
        b = y - x
        c = 1 - y
        # Check triangle inequality:
        if (a + b > c) and (b + c > a) and (c + a > b):
            count += 1
    return count / num_samples

print(triangle_probability())

When run for sufficiently large num_samples, this code outputs a value close to 0.25, confirming the 1/4 probability in practice.

Why the Triangle Inequality Is Equivalent to “No Piece Exceeds Half the Stick”

If one piece were longer than 1/2, that piece would be larger than the sum of the lengths of the other two pieces. That is, if x > 0.5, then x > (1 - x), implying x > y + z for the other two lengths y and z. Hence, a valid triangle can only form if all three pieces are at most 1/2 in length.

What If the Stick Length Is Not 1?

The probability remains the same because the process of choosing break points uniformly relative to the total length is scale-invariant. If you scale the stick to length L, the uniform break points become fractions of L, and the same geometry and ratio arguments apply.

How Might the Probability Change If Break Points Are Non-Uniform?

If the break points are no longer chosen uniformly, the underlying distribution for where the stick is cut changes. This alters the geometry or the density in the (X, Y) plane, so the probability of obtaining a valid triangle differs. Analyzing or computing this would involve integrating over the new joint density distribution of the break points.

Can We Generalize to More Break Points?

For more than two break points (resulting in more than three segments), the question of forming a closed polygon is more complex. Each side must still be smaller than the sum of all other sides, but you now have more sides to consider. One generalization is related to the “Broken Stick Problem” for forming an n-sided polygon, though exact probabilities often involve more advanced geometric and combinatorial arguments.

Potential Edge Cases

It is important to note that certain degenerate conditions (like both break points falling in the exact same location) cause one segment to have zero length. That scenario would automatically fail the triangle inequality because any zero-length side cannot form a closed triangle with positive perimeter. In practical simulations, this event has probability 0 in continuous uniform distributions, but it is still conceptually worth noting.

Below are additional follow-up questions

In a strict geometric sense, why do we only consider the triangle in the unit square defined by 0 <= X <= Y <= 1 instead of the entire square 0 <= X, Y <= 1?

When randomizing two break points on a stick of length 1, we can label one break point X and the other Y. By convention, we can sort these points so X <= Y. If we didn’t impose X <= Y, the break points (X, Y) and (Y, X) would be redundant because they correspond to the same physical cuts (just swapped). Restricting to X <= Y effectively accounts for all unique ways to break the stick without double counting.

If we were to consider the entire square 0 <= X, Y <= 1, we would double count every configuration of break points. Consequently, the region of interest is the triangle formed by 0 <= X <= Y <= 1, which has half the area of the full unit square (i.e., area = 1/2).

From a probability perspective, each pair (X, Y) in the entire unit square is still equally likely, but for analysis we prefer removing the ambiguity of which point is which. This focuses the space on one triangle of area 1/2 rather than the full square of area 1.

Potential Pitfalls and Edge Cases:

• If X == Y (which has probability 0 in the continuous case), one segment has length 0. We typically ignore this boundary situation because it doesn’t affect the overall probability measure.

• If the problem setting permitted X > Y to carry a different weighting, we would have to adjust the approach or the region of integration accordingly.

How can we directly integrate the valid region to confirm the probability is 1/4?

A purely analytical approach involves expressing the valid region under the constraints of the triangle inequality. Let X, Y be the two break points with X <= Y. The three segment lengths are X, Y - X, and 1 - Y. The triangle inequalities in plain text form are:

1. X + (Y - X) > (1 - Y) => Y > (1 - Y) => Y > 1/2

2. (Y - X) + (1 - Y) > X => 1 - X > X => X < 1/2

3. (1 - Y) + X > (Y - X) => 1 - Y + X > Y - X => 1 > 2Y - 2X => 2X + 2Y < 1 + something but let's rewrite carefully: rearranging => X + (1 - Y) > Y - X => X + 1 - Y > Y - X => 1 > 2Y - 2X => 1/2 > Y - X

So the valid region is:

• X < 1/2

• Y > 1/2

• Y - X < 1/2

We integrate the region of (X, Y) satisfying 0 <= X <= Y <= 1 with the above constraints. Since X < 1/2 and Y > 1/2, that already restricts X to [0, 0.5), Y to (0.5, 1]. Then we also need Y - X < 0.5.

A standard approach is to break it into integration bounds (for example, fix X and figure out Y, or vice versa). Evaluating the area of this region within the triangle [0 <= X <= Y <= 1] yields 1/4. This matches the intuitive arguments about ensuring no segment exceeds 1/2.

Potential Pitfalls and Edge Cases:

• One must remember to stay within the region X <= Y. Integrating outside that region or forgetting to halve the total area can lead to incorrect probabilities like 1/2 or 3/4.

• In a discrete setting (e.g., if X and Y could only take certain fractional values), the logic is conceptually similar but the probability might be slightly off until the discretization becomes very fine.

Are there alternative geometric approaches to visualize the valid region without explicit integration?

Yes. One common geometric argument is based on the condition that no piece exceed 1/2. If a single piece surpasses half the stick’s length, the remaining two pieces combined are less than half, and the triangle inequality fails. Visually, in the (X, Y) plane (with X <= Y), you mark the region where:

• X <= 1/2 (meaning the first piece is <= 1/2)

• Y - X <= 1/2 (meaning the second piece is <= 1/2)

• 1 - Y <= 1/2 (meaning the third piece is <= 1/2) which implies Y >= 1/2

This forms a smaller hexagon-like region inside the right triangle. Calculating that region’s area directly or by symmetry yields an area of 1/4 within the entire 1/2 triangular region, giving the final probability 1/4.

Potential Pitfalls and Edge Cases:

• A purely geometric method requires care in drawing the shapes accurately. If lines are mislabeled, the computed area might be incorrect.

• In more complex variants (e.g., 3 or more cut points), geometry can become unwieldy.

Is there a known distribution for the lengths of the three segments, and can it help confirm the probability?

When two cuts X and Y are drawn uniformly on [0, 1], the vector (X, Y - X, 1 - Y) is distributed in a way connected to the Dirichlet distribution with parameters (1, 1, 1). Generally, if you pick k-1 break points uniformly on [0, 1] and sort them, you get k segments whose length distribution is Dirichlet(1,1,...,1). In the specific case of k=3, each segment length is effectively Beta-distributed under certain transformations.

If you delve into the Dirichlet(1,1,1) distribution’s properties, you can find the probability that all segments are <= 1/2. That integral also leads to 1/4. This is another confirmation via distribution theory instead of geometry.

Potential Pitfalls and Edge Cases:

• Dirichlet distributions can be non-intuitive at first glance. Applying them incorrectly or mixing up parameters leads to confusion.

• The direct Beta relationships can be tricky, so it is helpful to confirm any derived formula with simulations.

How could floating-point precision issues affect a simulation that estimates the probability?

In practical simulations, you might see slight deviations from 0.25 if the number of samples is not large enough or if floating-point artifacts appear. For instance:

• Pseudorandom number generators might not perfectly replicate a continuous uniform.

• Very large sample sizes can help offset typical floating-point round-off but come with computational costs.

Despite these minor numerical issues, in high-level languages like Python (using double precision floating point) and a well-seeded, high-quality random generator, the simulation result should quickly converge near 0.25.

Potential Pitfalls and Edge Cases:

• Rounding errors near X == Y or X, Y near 0 or 1. This might spur incomplete coverage of corner regions, although the measure of these corners is negligible.

• In systems with limited precision (e.g., a microcontroller environment), the random breaks may exhibit discrete steps.

If we only allowed discrete break points (e.g., dividing the stick into 100 equally spaced points) and chose two break points among them uniformly at random, how does the probability compare?

In that discrete scenario, the break points X and Y can each take 101 possible values (from 0, 1/100, 2/100, ..., 1). We then sort them (X <= Y) and measure the fraction of pairs that yield valid triangles. As the discretization becomes finer (larger than 100, say 1000 or 10,000 points), the empirical probability will approach 1/4. For smaller discrete sets, we get an approximation that might be slightly off from 0.25 due to boundary effects.

Potential Pitfalls and Edge Cases:

• Ensuring a truly uniform choice among discrete intervals can be tricky if random sampling is biased.

• Extremely small discretization (e.g., only 10 or 20 points) can cause large deviations from 1/4.

Could the probability change if we “bend” the stick rather than just linearly cutting it, or if the cuts follow a certain physical process?

If the two cut points are truly uniform and independent along the one-dimensional length, the bending or physical manipulation after cutting does not alter the probability. However, if the cutting process is influenced by physical constraints (like tension or real-world friction) that bias the position of the breaks, the distribution may no longer be uniform.

For example, if break points tend to cluster (imagine a physical scenario where a crack is more likely to extend from a prior partial cut), that changes the underlying distribution of (X, Y) and thus modifies the probability.

Potential Pitfalls and Edge Cases:

• Real world “sticks” can exhibit material weaknesses, so cuts might not remain uniform.

• Even small biases in the break location can skew the distribution enough to move the probability away from 1/4.

Are there any implications or analogies in higher-dimensional geometry or other random partition problems?

Yes. The “break a stick at random” concept generalizes to random partitions of intervals (1D), areas (2D), or volumes (3D). For instance:

• In 2D, you might randomly cut a rectangle by choosing lines parallel to its sides; analyzing whether the resulting shapes fit certain criteria can be more involved.

• In higher-dimensional shapes, the probability interpretations link to volume or measure-based geometry.

These problems often appear in the realm of geometric probability, where results can be elegantly derived if symmetry or known distributions exist (e.g., Dirichlet). The 1D stick problem is usually the simplest case in this family of partitioning challenges.

Potential Pitfalls and Edge Cases:

• Not all higher-dimensional analogs have neat closed-form probabilities.

• Care must be taken that “random cutting” is well-defined, as different methods of random partitioning can lead to different distributions over pieces.