Monkey typing ABRACADABRA and gamblers

Problem: A monkey is sitting at a typewriter, typing a letter (A-Z) independently and with uniform distribution each minute. What is the expected amount of time that passes before ABRACADABRA is spelled?

Standard Solution: Suppose that, before every keystroke is made, a gambler enters and wagers $\$1$ on the next keystroke being an A fairly (so that if the keystroke is indeed an A, then the payoff is $\$26$). If the keystroke is an A, the gambler stays and wagers everything (in this case $\$26$) on the next letter being B, and so on. If the gambler ever loses a wager, then it leaves. Now let’s analyze what happens when ABRACADABRA is finally spelled out. The gambler who kept making correct wagers all the way through won $\$ 26^{11}$. But another gambler who got in on the second ABRA won $\$ 26^4$, and a third gambler who got in on the final A got $\$26$. Hence the total payoff is $\$ 26^{11}+\$26^4+\$26$. But all the wagers are fair, and the house gets $\$1$ on every turn from the new gambler, so the expected time before ABRACADABRA is spelled is $26^{11}+26^4+26$.

My question: What if the strategy of the gamblers changes? In the "standard solution" of the problem each gambler first bet on A, then on B, and so on... If the new gambler looks at the sequence of past letters and wagers $\$1$ on the NEXT USEFUL LETTER, why the game is not fair anymore? In this case, if the last letters are "..ABRACA" the new gambler bets on D instead of A. In this way the total payoff at the end of the game will be $\$26^{11}+\$26^{10}+...+\$26^2 + \$26$.

With the new strategy, the total payoff isn't always $\$26^{11}+\ldots+\$26$. Say the sequence of letters is ABRABRACADABRA. The gamblers who enter on the 4th and 5th letter will both bet on the 5th letter being `C.' This means nobody will win $\$26^{11}$ or $\$26^{10}$.