Triangle From Breaking the Longer Piece
You break a unit stick at a uniformly random point, creating two pieces. You then take the longer piece and break it again at a uniformly random point along its length.
You now have three pieces. What is the probability that these three pieces can form a triangle?
Hints
- The triangle inequality for three lengths summing to 1 simplifies to: each piece must be less than /2$.
- Condition on the first break point $U$ and use the symmetry around $U = 1/2$. For $U < 1/2$, the longer piece is
- U$, and you need the second break $V$ to satisfy/2 - U < V < 1/2$.\int_0^{1/2} U/(1-U)\,dU$ via the substitution $w = 1 - U$.- Compute $P(\text{triangle} \mid U) = U/(1-U)$ for $U \in (0, 1/2)$, then evaluate
Worked Solution
How to Think About It: The triangle inequality says three lengths form a triangle if and only if each piece is strictly less than the sum of the other two. Since the pieces sum to 1, this simplifies to: each piece must be less than
/2$. The key subtlety here is that we always break the *longer* piece, not a random piece. This biases the procedure toward more balanced splits, so we should expect a higher triangle probability than the classic "two random breaks" problem (which gives/4$). Intuitively, breaking the longer piece is a kind of self-correcting mechanism -- it prevents one piece from being too dominant.Quick Estimate: After the first break, the longer piece has expected length $3/4$. Breaking it uniformly gives two sub-pieces with expected lengths $3/8$ each, plus the short piece with expected length
/4$. All three expected lengths are well below/2$, suggesting the probability is fairly high -- maybe $0.6$ to $0.8$. The classic two-uniform-breaks problem gives/4$, but that allows the case where you break a short piece (making things worse). Breaking the longer piece should do much better.Approach: By symmetry, we can condition on whether $U < 1/2$ or $U > 1/2$ for the first break. By symmetry of $U$ around
/2$, both cases contribute equally, so we only need to compute one and double it (or equivalently, just compute one case since they give the same integral).Formal Solution:
Break the stick at $U \sim \text{Unif}(0,1)$. WLOG condition on $U < 1/2$ (the other case is symmetric). The pieces are $U$ and
-U$, with-U$ being the longer piece.Break
-U$ at a uniform point: the sub-pieces are $V$ and-U-V$ where $V \sim \text{Unif}(0, 1-U)$. The three pieces are $U$, $V$, and-U-V$.Triangle inequality requires all three pieces
lt; 1/2$: - $U < 1/2$ -- already guaranteed by our conditioning. - $V < 1/2$ -Loading problems...- U - V < 1/2$, i.e., $V > 1/2 - U$So we need
/2 - U < V < 1/2$, with $V \in (0, 1-U)$.Since $U < 1/2$, we have
/2 - U > 0$ and/2 < 1-U$, so the constraint is valid. The probability given $U$ is:$P(\text{triangle} \mid U) = \frac{1/2 - (1/2 - U)}{1 - U} = \frac{U}{1 - U}$
Integrating over $U \in (0, 1/2)$ and doubling (by symmetry of the first break):
$P = 2 \int_0^{1/2} \frac{U}{1 - U} \, dU$
Substitute $w = 1 - U$, $dw = -dU$:
$P = 2 \int_{1/2}^{1} \frac{1 - w}{w} \, dw = 2 \int_{1/2}^{1} \left(\frac{1}{w} - 1\right) dw$
$= 2 \left[\ln w - w\right]_{1/2}^{1} = 2\left[(0 - 1) - (-\ln 2 - 1/2)\right]$
$= 2\left[-1 + \ln 2 + 1/2\right] = 2\left[\ln 2 - 1/2\right] = 2\ln 2 - 1$
Answer:
$P(\text{triangle}) = 2\ln 2 - 1 \approx 0.386$
Intuition
This problem is a nice variation on the classic stick-breaking triangle question. The classic version (break at two independent uniform points) gives probability
/4$. Here, the rule "break the longer piece" introduces a form of self-correction -- you're prevented from making one piece even more dominant. Despite this, the probability\ln 2 - 1 \approx 0.386$ is higher than/4$ but still well below/2$, because the first break can still produce a short piece that's close to/2$, making it hard for the other two to both stay below/2$.The integral $\int U/(1-U)\,dU$ producing a logarithm is a signature of these "conditional uniform" problems. The log arises because the density of the second break scales as
/(1-U)$ (it's uniform on a shrinking interval), and integrating that ratio against a linear numerator gives a log. This is a pattern worth recognizing -- it shows up any time you have a ratio of polynomial to linear in a uniform integration. - Condition on the first break point $U$ and use the symmetry around $U = 1/2$. For $U < 1/2$, the longer piece is