Conditional Probability of Uniforms on a Simplex
Let $X$, $Y$, $Z$ be i.i.d. $\text{Uniform}(0,1)$ random variables. Compute the conditional probability
$P(\max\{X, Y, Z\} < 0.5 \mid X + Y + Z = 1).$
Hints
- Conditioning on $X+Y+Z=1$ makes the joint distribution uniform on the standard 2-simplex -- think of it as a probability measure on a triangle.
- The fraction of a simplex where coordinate $x_i \ge c$ is $(1-c)^{n-1}$; use this with inclusion-exclusion over the three bad events.
- Check that the pairwise intersections (e.g., $x \ge 0.5$ and $y \ge 0.5$ simultaneously) are impossible given the constraint $x+y+z=1$, so inclusion-exclusion collapses cleanly.
Worked Solution
How to Think About It: Conditioning on $X + Y + Z = 1$ pins the joint distribution to the standard 2-simplex -- the triangle $\{(x,y,z) : x,y,z \ge 0,\ x+y+z=1\}$. By symmetry and uniformity, the conditional distribution is uniform on this simplex. So the question becomes: what fraction of the simplex's area lies in the region where all three coordinates are strictly less than $0.5$?
Quick Estimate: The full simplex has area $\frac{\sqrt{3}}{2} \cdot 1^2$, but since we are taking ratios, just work with the 2D projected measure. The region $\{x < 0.5, y < 0.5, z < 0.5, x+y+z=1\}$ is a smaller triangle inside the big one. By inclusion-exclusion: start with the full simplex (measure 1), subtract the three "bad" corners where some coordinate $\ge 0.5$. Each bad region (say $x \ge 0.5$) is a simplex of side length $0.5$, contributing area $(0.5)^2 = 0.25$ of the total. No two bad regions overlap (if $x \ge 0.5$ and $y \ge 0.5$ then $x+y \ge 1$ but $z \ge 0$, so $x+y+z > 1$ -- impossible). So the good region has measure
Approach: Uniform measure on the simplex; inclusion-exclusion on the three boundary constraints.
Formal Solution:
Given $X + Y + Z = 1$ with $X, Y, Z \ge 0$, the conditional distribution is uniform on the standard 2-simplex $\Delta_2 = \{(x,y,z) : x,y,z \ge 0,\ x+y+z=1\}$.
The area of a sub-simplex where one coordinate $\ge c$ (say $x \ge c$) and $x+y+z=1$, $y,z \ge 0$, is proportional to $(1-c)^{n-1}$ relative to the full simplex, where $n = 3$ here. More precisely, the fraction of the simplex where $x \ge c$ is $(1-c)^2$ for $c \in [0,1]$. With $c = 0.5$, each single bad event has probability $(0.5)^2 = 0.25$.
For intersections: $P(x \ge 0.5 \text{ and } y \ge 0.5) = 0$ since $x + y \ge 1$ forces $z \le 0$, which is outside the simplex.
By inclusion-exclusion: $P(\max\{X,Y,Z\} \ge 0.5) = 3 \cdot (0.5)^2 - 0 + 0 = \frac{3}{4}.$
Therefore: $P(\max\{X,Y,Z\} < 0.5 \mid X+Y+Z=1) = 1 - \frac{3}{4} = \frac{1}{4}.$
Answer: $\boxed{1/4}$
Intuition
The key move here is recognizing that conditioning on a linear constraint like $X+Y+Z=1$ reduces the distribution to the uniform measure on a simplex -- a lower-dimensional object with a known geometry. This is a specific instance of a general fact: for i.i.d. uniform or exponential random variables, conditioning on the sum gives a Dirichlet distribution, which in the uniform case is just uniform on the simplex.
The inclusion-exclusion argument is clean because the bad regions (any coordinate $\ge 0.5$) are non-overlapping corners of the triangle, each similar to the full triangle with linear scale factor $0.5$, so area factor $0.25$. Geometric symmetry does all the heavy lifting. This type of simplex-volume calculation appears in multivariate order statistics, Dirichlet pricing models, and any problem where you need to count constrained allocations -- like how many ways to split a budget of 1 among three assets such that no single asset takes more than half.