Let $\{X_t\}$ be an iid sequence of fair coin tosses and $\tau_{HTH} = \inf\{t\geq 3: X_{t-2}X_{t-1}X_t=HTH\}$. I want to determine $E\tau_{HTH}$.
I don't understand a part of the explanation which is:
Gamblers place bets on each individual toss. On each bet, gambler pays an entrance fee of $k$ and is paid $2k$ in return if the outcome is $H$ or $0$ if the outcome is $T$. $k$ can be negative which corresponds to a bet on $T$.
Suppose that at each unit of time until $HTH$ first appears, a new gambler enters and employs the following strategy: on his first bet, he wagers $1$ on $H$. If he loses, he stops. If he wins and $HTH$ has not yet appeared, he wagers his payoff of $2$ on $T$. If he loses, he stops. If he wins and $HTH$ has not yet appeared, he wagers his payoff of $4$ on $H$. This is the last bet placed bu this particular gambler.
The gambler who started at $\tau_{HTH}$ is paid $2$ and the gambler who started at $\tau_{HTH}-2$ is paid $8$ and every gambler has paid an initial entry fee of $1$. At time $\tau_{HTH}$, the net profit to all gamblers $=10-\tau_{HTH}$ and since the game is fair $E\tau_{HTH}=10$
Q1) I can see that the net profit for all gamblers is $10-\tau_{HTH}$ if the outcome string is say $TTTHTH$ and this wouldn't be the net profit if atleast one $H$ appears before $HTH$, am I correct? (say the outcome string is $HHHHTH$).
Q2) The explanation also says $\tau_{HTH}/3$ is bounded by a Geometric(1/8) random variable. How can I see that?
Best Answer
If the outcome is $HHHHTH$, then the first gambler will win their first bet, then bet all of their winnings on the second toss being a $T$, and then lose everything, having a total loss of their initial payment of $1$.
The second gambler will win their first bet, then lose the second bet on exactly the same way as the first gambler, having a total loss of $1$.
In fact, of the six gamblers, each one of them has paid an initial $1$ to enter the game, and only the fourth and sixth have come away with winnings ($10$ in total). So the net win for all the gamblers is indeed $10-6=4$.
In general, it turns out that only the last and the third-from-last gambler can walk away with winnings: The moment any gambler wins their $8$, the game stops because we reached $HTH$, the last gambler walks away with their $2$, and every other gambler has lost,
As for question $2$, consider the following alternative game (forgetting the gamblers): Toss a coin three times. If you got $HTH$, stop, and if not, try again. Keep going until you have gotten a triple of $HTH$. Say it takes $\sigma_{HTH}$ triples before you stop. Can you see that this is a geometric random variable?
Now say both of these games were being played simultaneously with the same coin tosses. Can you see why $\tau_{HTH}\leq 3\sigma_{HTH}$?