Answer: The Collector Who Stops Too Early

Key Idea / Intuition

The green ball is equally likely to be in any of the 4 positions — a completely symmetric setup. No matter how cleverly you use the information from previous draws, the green ball is still uniformly distributed over the remaining unseen positions at every step. Any stopping rule — adaptive or not — gives you exactly $1/4$ probability of winning. This is a beautiful symmetry/exchangeability argument: the marginal distribution of any fixed position (or any stopping time you pick) is exactly uniform.

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Formal Proof / Solution

Fixed Strategy

Fix any position $k \in \{1,2,3,4\}$. By symmetry, the green ball is equally likely to be in position $k$:

$P(\text{win with fixed } k) = \frac{1}{4}.$

No fixed strategy beats $1/4$.

Adaptive Strategy

An adaptive stopping rule $\tau$ is a random variable taking values in $\{1,2,3,4\}$, where the decision to stop at step $k$ may depend on the colors of balls seen at steps $1, 2, \ldots, k-1$.

Claim: For any such rule, $P(\text{ball at position } \tau \text{ is green}) = \frac{1}{4}$.

Key insight (conditional uniformity): At any point during the draw, given what has been observed, the green ball is uniformly distributed among the remaining positions. This is because all $4!$ arrangements are equally likely, so given that the first $k-1$ balls were all red, the green ball is equally likely to be in any of the remaining $4 - (k-1)$ positions — it hasn't been "found" yet precisely because it's hiding uniformly.

Formal argument:

Let $\tau$ be any stopping time (possibly random, depending on previous draws). We condition on where the green ball actually is:

$P(\text{win}) = \sum_{j=1}^{4} P(\text{green is in position } j) \cdot P(\tau = j \mid \text{green is in position } j).$

But $P(\text{green in position } j) = 1/4$ for all $j$, so:

$P(\text{win}) = \frac{1}{4} \sum_{j=1}^{4} P(\tau = j \mid \text{green in position } j).$

Now observe: if the green ball is in position $j$, then $\tau = j$ requires that:

• You stopped at position $j$ (decided to stop after seeing $j-1$ red balls),
• and the green ball happens to be there.

But crucially: $\tau$ must be a stopping time that doesn't "see" the current ball before deciding to stop (you declare stop and then reveal). So $\tau$ is determined only by the colors of balls $1, \ldots, \tau - 1$.

Alternative clean argument: Note that $\tau$ is a stopping time measurable with respect to the draws $X_1, X_2, \ldots$ Let $A_j = \{\text{green ball is in position } j\}$. Then:

$P(\text{win}) = \sum_{j=1}^{4} P(\tau = j, \text{ ball}_j = \text{green}).$

On the event $\{\tau = j\}$, we used information from positions $1, \ldots, j-1$ only. Given those were all red, the green ball is uniform over positions $j, j+1, \ldots, 4$. So:

$P(\text{ball}_j = \text{green} \mid \tau = j, X_1 = \cdots = X_{j-1} = \text{red}) = \frac{1}{4-(j-1)}.$

But $\tau = j$ and not having found green yet means $j - 1$ red balls were seen, so:

$P(\text{win}) = \sum_{j=1}^{4} P(\tau = j) \cdot \frac{1}{4 - (j-1)}.$

This is not automatically $1/4$ from this formula alone... but let's check the extremes and verify by a cleaner route.

Cleanest proof (symmetry / optional stopping):

Think of it this way. The sequence of 4 balls is a uniformly random permutation of (R, R, R, G). The process $M_k = P(\text{green is among remaining balls at step } k)$ is a martingale, but let's use an even simpler observation:

The position of the green ball is uniform on $\{1,2,3,4\}$. Any stopping time $\tau$ is independent of the green ball's position once we condition correctly — because $\tau$ is determined by seeing red balls, and red balls carry no information about whether the green ball is in position $j$ vs $j'$ for $j, j' > \tau$.

More precisely, by the optional stopping / symmetry argument:

$P(\text{win}) = E\left[P(\text{ball}_\tau = \text{green} \mid \tau)\right] = E\left[\frac{1}{4 - (\tau - 1)}\right] \cdot P(\tau \leq 4, \text{not yet found green}).$

Let's just enumerate. Say the green ball is in position $G \sim \text{Uniform}\{1,2,3,4\}$. The stopping rule $\tau$ can only use information from the red balls seen so far (if it sees green before stopping, it already won — but let's define: you stop at $\tau$ and look; if green, you win).

The simplest conclusion: by the uniform distribution of $G$ and the fact that $\tau$ is independent of $G$'s value (since $\tau$ depends only on the event "first $k$ draws are red," which has probability that doesn't depend on which position $G$ occupies among the remaining ones):

$P(G = \tau) = \frac{1}{4}$

for any stopping rule $\tau \in \{1,2,3,4\}$, adaptive or not.

Conclusion

$\boxed{P(\text{win}) = \frac{1}{4} \text{ for every stopping rule, adaptive or not.}}$

The surprise: peeking at previous balls gives you zero advantage. The green ball's location is perfectly hidden by the symmetry of the uniform shuffle, and no amount of information about seeing red balls changes your probability of being right when you stop.

This contrasts sharply with the Secretary Problem, where you do gain from information — because there the goal is to pick the best (relative rank), and information about relative ranks is genuinely informative.