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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 $\$26^{11}+\$26^{10}+...+\$26^2 + \$26$.

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  • $\begingroup$ What makes you think the game is not fair? $\endgroup$ – user940 Feb 11 '14 at 16:00
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    $\begingroup$ @ByronSchmuland: Presumably because the answer is incorrect, and therefore something must be wrong somewhere. Though it may not be that the game is unfair; perhaps the issue is somewhere else. (To be honest I don't quite understand the standard solution yet...) $\endgroup$ – ShreevatsaR Feb 11 '14 at 17:42
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    $\begingroup$ @ShreevatsaR I'm trying to prod the OP into considering alternative explanations. The game is still fair; it is his calculations that are wrong. $\endgroup$ – user940 Feb 11 '14 at 18:04
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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 $\$26^{10}$.

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  • $\begingroup$ What about the specific case ABRACADABRA? Why is the OP getting a different answer for this case? $\endgroup$ – ShreevatsaR Feb 11 '14 at 17:34
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    $\begingroup$ @ShreevatsaR It is not necessary for the total payout to be the same for all possible strings. As long as the expected total payout is the same as when the gamblers use the standard strategy, there is no contradiction. $\endgroup$ – Julian Rosen Feb 11 '14 at 18:14
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    $\begingroup$ Ah I see, you're considering the case when the sequence of letters seen is ABRABRACADABRA; I thought at first that you were changing the question, and considering the problem for ABRABRACADABRA instead of for ABRACADABRA. I understand now, thanks. $\endgroup$ – ShreevatsaR Feb 11 '14 at 18:29

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