Last Marble Standing
An urn contains $r$ red marbles, $b$ blue marbles, and $g$ green marbles. You draw marbles one at a time, uniformly at random, without replacement.
What is the probability that immediately after the last red marble is drawn, there is still at least one blue marble AND at least one green marble remaining in the urn?
Derive a general formula, then evaluate it for $r = 10$, $b = 20$, $g = 30$.
Hints
- Think about the problem as a random permutation of all marbles. The position of the last marble of each color is what matters -- not the full draw sequence.
- Use inclusion-exclusion on the two bad events: event $A$ (all greens drawn before the last red) and event $B$ (all blues drawn before the last red). Each bad event has a clean symmetry argument.
- For $P(A)$: among the $r + g$ red and green marbles, each is equally likely to be the last one drawn. So $P(A) = r/(r+g)$. Apply the same logic for $P(B)$ and $P(A \cap B)$.
Worked Solution
How to Think About It: The key move is to stop thinking about the draw sequence and instead think about the position of the last marble of each color in a random permutation. Drawing $r + b + g$ marbles without replacement is exactly the same as choosing a uniformly random ordering of all of them. You need the last red to come before the last blue and before the last green -- equivalently, you want red NOT to be the last color still standing among the relevant pairs. Inclusion-exclusion on the two bad events handles this cleanly.
Quick Estimate: For $r = 10$, $b = 20$, $g = 30$: the last red is the last among all $r + b + g = 60$ marbles with probability $10/60 = 1/6$ -- that is the probability both other colors run out first, which is a lower bound on the bad events. The answer should be comfortably above $1/2$ because red is the smallest group. Rough bracket: probably between $0.55$ and $0.65$.
Approach: Reformulate as a permutation problem, identify the two bad events, and apply inclusion-exclusion.
Formal Solution:
Model the process as a uniformly random permutation of all $r + b + g$ marbles. Define:
- $\text{last}(C)$ = the position of the last marble of color $C$ in the permutation.
We want the event: $\text{last}(R) < \text{last}(B)$ and $\text{last}(R) < \text{last}(G)$ -- i.e., after the last red is drawn, at least one blue and one green still remain.
Define the two bad events: $$A: \text{last}(G) < \text{last}(R) \quad \text{(all greens drawn before the last red)}$$ $$B: \text{last}(B) < \text{last}(R) \quad \text{(all blues drawn before the last red)}$$
By inclusion-exclusion: $$P(\text{good}) = 1 - P(A \cup B) = 1 - P(A) - P(B) + P(A \cap B)$$
Computing $P(A)$: Among the $r + g$ marbles that are red or green, the last one is equally likely to be any of them by symmetry. So: $$P(A) = P(\text{last of } \{R, G\} \text{ is red}) = \frac{r}{r + g}$$
Computing $P(B)$: By the same symmetry argument on red and blue marbles: $$P(B) = \frac{r}{r + b}$$
Computing $P(A \cap B)$: Both bad events occur if and only if the overall last marble among all three colors is red -- i.e., red's last position exceeds both the last green and last blue. Among all $r + b + g$ marbles, the very last one is equally likely to be any marble, so: $$P(A \cap B) = \frac{r}{r + b + g}$$
General formula: $$\boxed{P(\text{good}) = 1 - \frac{r}{r+g} - \frac{r}{r+b} + \frac{r}{r+b+g}}$$
Numerical case $r = 10$, $b = 20$, $g = 30$: $$1 - \frac{10}{40} - \frac{10}{30} + \frac{10}{60} = 1 - \frac{1}{4} - \frac{1}{3} + \frac{1}{6} = \frac{12 - 3 - 4 + 2}{12} = \frac{7}{12}$$
Answer: $\dfrac{7}{12} \approx 0.583$
Intuition
The elegant trick here is the "last marble of each color" symmetry. In a uniformly random permutation, the last element among any fixed subset is equally likely to be any member of that subset. This is because you only need to know the relative order of the subset, and every ordering is equally likely. So asking "is the last red before the last green?" reduces to "is the last marble in the red-green subset red?" -- which is just $r/(r+g)$. No combinatorial counting needed.
This symmetry argument -- focusing on the last (or first) element of a color group -- is a recurring device in urn and permutation problems. It shows up in runs problems, in the "last success" family of optimal stopping results, and in race-to-the-bottom models in market microstructure. The inclusion-exclusion structure here is also worth internalizing: "at least one of each surviving" is most naturally attacked by complementing to "some color runs out first," then deducing the overlap.