2021 AMC 10B Problem 18

Repeated faces are irrelevant; only the order in which new faces first show up matters. Keep rolling (imagine continuing past the first odd) and record each face the first time it appears. The first recorded face is uniform over the six faces; whenever some faces remain unseen, the next new face is equally likely to be any of them, since repeats are just skipped. So the order of first appearances is a uniformly random permutation of $\{1, 2, 3, 4, 5, 6\}$.

The event "every even number appears before the first odd number" says exactly that the first three faces to appear are $\{2, 4, 6\}$ in some order. Among the $\binom{6}{3} = 20$ equally likely choices for which three faces appear first, only one is $\{2,4,6\}$.

The probability is $\frac{1}{20}$. The answer is $\boxed{\textbf{(C)}\ \frac{1}{20}}$.

When a random process only asks about "which happens first," collapse repeated outcomes and look at the induced ordering. For independent uniform trials, the first-appearance order of the outcomes is uniform, which converts a messy sequential probability into a single combinations count. This is the same trick as "the last chip drawn is red" or "which color runs out first."

Sequentially: the first roll must be even, probability $\frac12$. Then ignoring repeats of that face, the next new outcome is a new even (probability $\frac{2}{6}$ per roll) or an odd ($\frac{3}{6}$), so a new even comes first with probability $\frac{2}{5}$. Finally the last even beats an odd with probability $\frac{1}{4}$. Product: $\frac12 \cdot \frac25 \cdot \frac14 = \frac{1}{20}$.

• Ignoring that even numbers can repeat before the first odd, so the three evens need not arrive on three consecutive rolls.
• Computing $\frac12 \cdot \frac13 \cdot \frac16$ or similar, using $\frac16$ for a new even without conditioning on "not a repeat."