(Quant job interviews - questions and answers - Question 3.8)
For a fair coin, what is the expected number of tosses to get 3 heads in a row
The answer is stated as :
We gamble in such a way that we make money on heads but such that if we get a T on toss $n$, our position is $-n$. We therefore gamble one unit on the first toss and on each toss after a $T$. After one head, we gamble three. This guarantees that if we get a $T$ next then we go to $-n$. After two heads we are therefore up 4, and so we gamble 7 to get us to $-n$ again. our gambling winnings is a martingale since we are making finite trades in a martingale (any bounded trading strategy in a martingale is a martingale). After three heads our position is $11 - (n-3)=14-n$. the time taken to get three heads is a stopping time with finite expectation so if we stop at it we still have a martingale (Optional sampling theorem) thus $\mathbb E (14 -n) = 0 $ and we are done.
I realise there are a few answers to this already, however i am not sure the explanation above, which uses martingale theory, is entirely clear.
The author states that any bounded trading strategy in a martingale is a martingale but what is the underlying martingale here ?
Also I don't understand the underlying motivation for gambling the way described, please could someone put it in more mathematical terms so i can understand the reason for the sizes of the bets ?