= Solution
{c}
Let $T$ be the first completion time of the word. A fresh gambler arrives before every throw with capital one, betting it all on the successive symbols $1,2,1,2$. Each correct bet multiplies capital by six and each wrong bet loses it, so every bet is fair. A gambler who completes the word keeps the winnings.
At time $T$, the gambler who bet on all four final throws has capital $6^4$. The gambler who bet on the final two throws has correctly matched the suffix $12$ and has capital $6^2$. There are no other survivors: the only nonempty suffixes of $1212$ that are also prefixes are $12$ and $1212$. Thus the total surviving capital is $6^4+6^2$.
To justify the expectation identity, independent blocks of four throws each have probability $6^{-4}$ of equalling the pattern, so $\mathbb P(T>4k)\leq(1-6^{-4})^k$ and $\mathbb ET<\infty$. Total gambling capital minus the number of arriving unit stakes is a martingale with bounded increments before $T$. <Optional stopping> therefore equates expected terminal capital with expected total stakes:
$$
\boxed{\mathbb ET=6^4+6^2=1332.}
$$
This is the <waiting time for a word in independent uniform symbols>. The extra $6^2$ reflects the overlapping occurrence of the prefix $12$ at the word's end.
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