 Solution

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 $\mathbb{P}(T>4k)\le (1-6^{-4}{)}^k$ and $\mathbb{E}T<\infty$. 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: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.