Waiting time for a pattern in coin tossing

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 $\tau_{HTH}\leq 3\sigma_{HTH}$?

Tags:

Probability Theory

Stochastic Processes

Martingales

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