I'm doing some prep work for a job interview and one of their difficult sample questions is as follows:
You flip a coin 100 times, but instead of having a fixed stake, you can freely choose the stake for each flip. You start with £100. After each flip, if it comes up heads you win twice your stake (and your stake is returned), if it comes up tails you lose your stake. So if you start with $x$ and select a stake of $S$, then after the flip you will either have $x+2S$ or $x-S$. You can never make your stake larger than your current balance.
If your profit is $P$ how should you select your stake, $S$, to maximise:
a) $E[P]$
b) $E[\log (P+100)]$
Edit: to clarify $0<S<100$ is required so you are betting an amount but you are also protecting yourself from losing all of your money (of course successive losses would whittle your total money to zero eventually)
For the second, you are trying to maximize the expectation of the log of how much you have. For one flip, you start with $1$ and bet $x$. Your expectation is then $\frac 12(\log (1-x)+\log (1+2x))$ We take the derivative, set to zero, solve for $x$. That gives $\frac {-1}{1-x}+ \frac 2{1+2x}=0$ or $x=\frac 14$ so you should bet one quarter of your bankroll at each step. A win and a loss multiply your bankroll by $\frac 98$, so fifty wins and fifty losses will have you at about $36110$ at the end.
If you are required to bet at least $1$ every time there is no strategy in b that has positive expectation. You could lose every time, winding up with $0$ and the log is negative infinity. When you do questions like this it is essential to get the rules specified precisely.