An urn contains 5 red, 3 blue, and 2 green balls. You draw balls one at a time without replacement.
What is the expected number of draws until all balls of at least one color have been completely removed from the urn (i.e., only balls of at most 2 colors remain)?
Hints
Think of the draws as revealing positions in a random permutation of all 10 balls. When does each color's last ball appear?
The expected position of the last of $k$ specific items in a random permutation of $n$ is $\frac{k(n+1)}{k+1}$. You need the minimum across three such quantities.
Use the identity $\min(a,b,c) = a + b + c - \max(a,b) - \max(a,c) - \max(b,c) + \max(a,b,c)$ and note that $\max(T_A, T_B)$ is just the last position among all balls of colors $A$ and $B$ combined.
Worked Solution
How to Think About It: Imagine laying out all 10 balls in a random line -- a random permutation. Drawing without replacement is the same as revealing balls left to right in this permutation. You stop when you first exhaust some color, meaning you have drawn all balls of that color. So define $T_R$, $T_B$, $T_G$ as the position of the *last* red, blue, and green ball respectively in the permutation. You stop at $T = \min(T_R, T_B, T_G)$. The question reduces to: what is the expected value of the minimum of three order statistics?
Quick Estimate: Green has only 2 balls, so it is the most likely color to be exhausted first. The last green ball's expected position is roughly
\cdot 11/3 \approx 7.3$. But we take the *minimum* across three colors, which pulls the answer well below 7.3. A rough guess: the minimum of three correlated random variables with means around 7.3, 8.3, and 9.2 should land somewhere around 6. Let's see if the exact calculation confirms this.
Approach: Use the random permutation model and the min-max inclusion-exclusion identity to get an exact answer.
Formal Solution:
Place all 10 balls in a uniformly random permutation. For a color with $k$ balls among $n = 10$ total, the position of its last ball is the maximum of a random subset of size $k$ from $\{1, \ldots, 10\}$. By the order statistic formula for sampling without replacement:
$E[\text{last of } k \text{ items among } n] = \frac{k(n+1)}{k+1}$
$\min(a, b, c) = a + b + c - \max(a,b) - \max(a,c) - \max(b,c) + \max(a,b,c)$
Taking expectations, we need $E[\max(T_A, T_B)]$ for each pair. The maximum of the last positions of two color groups is simply the last position among all balls of both colors combined. If colors $A$ and $B$ have $k_A$ and $k_B$ balls, then:
On average, you need about 6.18 draws before one color is completely exhausted.
Intuition
The key insight is the permutation reframing: drawing without replacement is the same as fixing a random ordering of all the balls and reading left to right. Once you see it that way, "when does a color run out?" becomes "where is the last ball of that color in the permutation?" -- a clean order-statistic question. The min-max inclusion-exclusion identity then lets you decompose the minimum of correlated random variables into a sum of expected maxima, each of which has a simple closed-form formula.
This technique shows up frequently in coupon-collector variants and urn problems. The broader lesson: whenever you need the expected value of the first (or last) occurrence of some event in a random sequence, try rephrasing in terms of positions in a permutation. It often converts a messy conditional probability calculation into a clean combinatorial identity. In quantitative finance, similar reasoning applies to order-of-default problems in credit portfolios -- the "time to first default" among correlated names can sometimes be handled with analogous inclusion-exclusion arguments.