What is the variance of the expected time until one can construct an ABRACADABRA sequence? Suppose that I am interested in the expected time until one can type out a sequence ABRACADABRA, assuming that each letter has $\frac{1}{26}$ probability of occurring. If we use a martingale $$X_n = n -26^{11} - 26^4-26 $$ where $X_n$ is the expected waiting time until stop is $26^{11} + 26^4+26$ by Doob's optional stopping theorem. However, I am not quite sure how to obtain the variance. Would it require constructing another martingale? Thanks.

Edit:
The definition of $X_n$ is wrong and needs correction. But the question regarding the variance is an interesting one.
 A: Pretend the letters are generated one at a time and a new gambler arrives before each letter is generated. Denote $G_n$ as the gambler who arrives before the $n$th letter is generated. Suppose that the gamblers initially bet \$n that the next letter will be an $A$. If they are successful, $G_n$ will receive \$26 and then bet that \$26$n$ that the next letter will be a $B$. If unsuccessful, $G_n$ leaves. Suppose the gamblers continue this way by betting everything at each step and that the payouts are such that each bet is fair. This would mean that if successful, $G_n$ should receive \$ $26^{i+1} n$ after the $n+i$th letter is generated.
Let $X_n$ represent the amount of money gamblers earn after the $n$th letter is generated. Define $M_n = X_1 + \cdots + X_n$ to be the total amount of money gamblers earn after $n$ rounds. Consider a variation of the game described above where gambler $G_n$ bets $\min(n,m)$ where $m$ is some positive constant. Let $X_{m,n}$ represent the amount of money gamblers earn after the $n$th letter is generated and define
$$M_{m,n} = X_{m,1} + \cdots + X_{m,n}$$
analogously. We claim that the $M_{m,n}$ are a martingale with respect to the $X_{m,n}$.  Since each bet is fair, we know that $E[X_{m,n}]$ = 0. It is also clear that the $X_{m,n}$ are independent since the monkey is typing randomly. It follows that
$$E[M_{m,n+1}|X_{m,1},\cdots, X_{m,n}] = M_{m,n} + E[X_{m,n+1}|X_{m,1},\cdots, X_{m,n}] = M_{m,n}$$
It obvious that the $M_{m,n}$ are functions of $X_1,\cdots,X_n$ and so we have a martingale. It is not difficult to show that $T$ is a stopping time. We can also easily bound $E[T] < 26^{12}$. In addition, we can see that
$$|M_{m,n} - M_{m,n-1}| = |X_{m,n}| < m(26^{11} + 26^{4} + 26)$$
Thus, we can apply Doob's Optional Stopping Theorem to see that
$$E[M_{m,T}] = E[M_0] = 0$$
It is clear that
$$\lim_{m\to\infty} M_{m,T} = M_T$$
We can use the Dominated Convergence Theorem to see that
$$E[M_T] = E[\lim_{m\to\infty} M_{m,T}] = \lim_{m\to\infty} E[M_{m,T}] = 0$$
Finally, we observe that
$$E[M_T] = E[26^{11}(T-10) + 26^4(T-3) + 26T - \frac{T(T+1)}{2}]$$
Setting this equal to zero, we can rearrange a few terms to see that
$$\mathrm{Var}(T) = E[T^2] - E[T]^2 = \boxed{(26^{11} + 26^4 + 26)(26^{11} + 26^4 + 25) - 20\cdot 26^{11} - 6 \cdot 26^4}$$
