I was recently asked this question in an interview, but was completely stumped as to how to even begin answering it – it's been bugging me ever since, and I thought it was quite a nice question, so hopefully someone on here can help me out. Any help would be appreciated! Here goes:
You start off with £100 and you toss a coin 100 times. Before each toss you choose a stake $S$ which cannot be more than your current balance $x$ (so your maximum stake for the first toss is £100). If the coin comes up heads, you win $2S$ and your new balance is $x+2S$. If it comes up tails, you lose your stake and have $x-S$. How do you choose your stake so as to maximise your expected winnings from the game, not including the initial balance?
Cheers,
Boris
Best Answer
It really is as simple as "the bet is in your favor-take it." $S=x$. You win $100(3^{100}-1)$ with probability $2^{-100}$ and lose $100$ with almost certainty. This presumes somebody can pay you that much. Then the expected win is $$\frac{1}{2^{100}} \cdot 100(3^{100}-1) -(1-\frac 1{2^{100}}) \cdot 100\approx 4\cdot 10^{19}.$$
To maybe make this less unbelievable, imagine a two round game. Clearly on the last throw, you want to bet all you have, increasing your expected fortune by $50\%$. On the first throw, your expected balance is $\frac {x-S}{2}+\frac {x+2S}{2}=x+\frac{S}{2}$ which (given the rules) is maximized when $S=x$. If you won on the first flip, you now have 3S, so you can now bet $S_{2}$ with the condition that $S_{2} \leq 3S$. Your expected balance after the second flip is then $\frac{3S-S_{2}}{2} + \frac{3S+2S_{2}}{2} = \frac{6S+S_{2}}{2} = 3S+\frac{S_{2}}{2}$, which is again maximized if $S_{2}=3S$, so that your expected balance will be $3S+\frac{3S}{2} = \frac{9S}{2}$, and your expected profit of the second round (assuming you won on the first flip) will now be $\frac{9S}{2}-3S=\frac{3}{2}S=\frac{S_{2}}{2}$.
Alternately, your result is the same if you interchange the two flips. Since you should be all on the last flip, you should on the first as well.