Coin Betting Expectation Suppose I have a biased coin with probability of heads p, and tails q=(1-p).
It is then used in a game which lasts at most N tosses, and start with a stake of £1.  Each time the coin is tails my money is doubled.  The first time it comes down heads my money is reduced to £1, and the second time it comes down heads, I lose all my money.  The game ends after N tosses, or after the second head.  What is the expectation of my money at the end of the game?
 A: With probability $(1-p)^N$ you get no heads, and end up with $2^N$.
For $k=1,..,N$, with probability $p(1-p)^{N-1}$ you get a single "heads" on the $k$th toss.  Then you end up with a value of $2^{N-k}$.
Any other result you end up with zero.
So the expected value is:
$$(1-p)^N2^N + p(1-p)^{N-1}\left(1+2+2^2+...+2^{N-1}\right) = $$
$$(1-p)^{N-1}\left[(1-p)2^N+p(2^N-1)\right]=$$
$$(1-p)^{N-1}\left(2^N-p\right)$$
A: Let $m$ be the amount of money you have before your first toss. (The problem specified $m=1$, but I did not notice. So we might as well look at this slightly more general situation.) If you are very lucky, all tails, probability $q^N$, you end up with $m2^N$.
Or else there could be $1$ head and the rest tails. The probability this happens at any particular one of the $N$ tosses is $pq^{N-1}$.
Suppose it happens at the last toss. Then you end up with $1$. If it happens at the next to last toss, you end up with $2$.  If it happens on the toss before that, you end up with $4$, and so on. Finally, if head happens on the very first toss, you end up with $2^{N-1}$. (Doesn't seem reasonable if $m=1$: the toss was bad but the casino let you keep your $1$ pound!) So the expectation is
$$m2^Nq^N+ (1+2+\cdots +2^{N-1})pq^{N-1}$$
The expression can be simplified by noting that $1+2+\cdots+2^{N-1}=2^N-1$.
Now put $m=1$ for the actual problem as asked. 
