Coin Tossing Game Optimal Strategy 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   
 A: In my opinion this calls for the Kelly Criterion (http://en.wikipedia.org/wiki/Kelly_criterion). In this case, the fraction of your wealth that should be bet is $\frac{0.5 \times 3-1}{3-1}$.
A: 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.
A: Long comment on the risk :
While this's not an answer , it's still highly related to the question and I want to show how crazy the risk is growing using the idea of sharp ratio .
Assume the bet is a constant fraction of current wealth $f$ .
The expected return is (which's maximized at $f=1$)
$$
E[X] = \sum_{k=0}^{100}\frac{1}{2^{100}}\binom{100}{k} (1+2f)^k(1-f)^{100-k}100
$$
$$
= \frac{100}{2^{100}} (2+f)^{100}
$$
and
$$
E[X^2] = \sum_{k=0}^{100}\frac{1}{2^{100}}\binom{100}{k} (1+2f)^{2k}(1-f)^{2(100-k)}100^2
$$
$$
= \frac{100^2}{2^{100}} (5f^2 + 2f + 2)^{100}
$$
The sharp ratio is
$$
\frac{E[X] - 100}{\sqrt{E[X^2] -E[X]^2}}
$$
$$
= \frac{(\frac{1}{4}f^2 + f + 1)^{50} - 1}{\sqrt{(\frac{5}{
2}f^2 + f + 1)^{100} -  (\frac{1}{4}f^2 + f + 1)^{100}}}
$$
A plot shows that the sharp ratio is maximized very near zero and quickly converges to $0$

We could also check that variance of return
$E[X^2] -E[X]^2$ grows pretty crazily . So a risk averse person may not bet any money at all .
A: I'm not sure that you wrote out the explanation of the question correctly (please correct me if I'm wrong), but if you bet S, then right after betting and before the coin is tossed your wealth is x-S.. now if the game is such that you either lose everything or gain 2S, then your wealth after one coin toss is either x-S or x-S+2S=x+S with prob. 1/2 each. In this case, no matter how much you bet your expected wealth will be x and since we are risk averse you probably don't want to bet anything.
