3
$\begingroup$

Consider the following game, lets call it $G$. You flip a fair coin $100$ times, but instead of having a fixed stake, you can freely choose the stake for each flip, just before the flip.

You start out with $£100$. After each flip, if it comes up heads you win twice your stake (and your stake is returned), and if it comes up tails you lose your stake, ie if you start with $x$ and select a stake of $s$, then after the flip you will either have $x − s$ or $x + 2s$. You can never make your stake larger than your current balance.

how should you select your stake in each round in order to maximize (Here $G$ denotes the profit from playing the game):

a) $E[G]$

b) $E[\log(G + 100)]$

Improve this question
$\endgroup$
4
  • $\begingroup$ Intuitively, in order to maximize the average gain, I thing the best way is to bet at each step all the available money (at least for the first question). $\endgroup$
    – digital-Ink
    Commented Sep 24, 2016 at 18:11
  • $\begingroup$ If this is homework, you should tag it or mention it as such. $\endgroup$
    – usul
    Commented Sep 24, 2016 at 20:39
  • $\begingroup$ Have you tried solving it for just a single round? $\endgroup$
    – usul
    Commented Sep 24, 2016 at 20:39
  • $\begingroup$ This is not homework. For single round it comes out as you have to wager all your money for part a, and 1/4 of your money for part b $\endgroup$
    – vishmay
    Commented Sep 24, 2016 at 20:39

1 Answer 1

Reset to default
0
$\begingroup$

Since the odds are even but the gain is twice in your favor, at each step it is "best" to bet the entire available sum (this is to be understood in the sense of maximizing the expected value; that means it becomes profitable in the long run, but not every time). That means that at the end of each game, there is a chance of $\dfrac{1}{2^{100}}$ of winning (practically, close to impossible), but the sum to be won is $100\cdot 3^{100}$, hence the expected value is $100\cdot \left( \dfrac{3}{2} \right) ^{100}$ which exceeds all the money in the world. Unfortunately, in order to "expect" this kind of money, you will need more time (to play the game) than the age of the Universe (or Multiverse :-) ).

So, in practice, depending on how much time and money is available, there should be a different strategy.

Improve this answer
$\endgroup$
3
  • 1
    $\begingroup$ How do I prove that mathematically this maximises the Expected value? Also what about part b? here log(0) is negative infinity so a different strategy may be needed? $\endgroup$
    – vishmay
    Commented Sep 24, 2016 at 20:08
  • $\begingroup$ For b), is $G+100$ which ensures that you won't have $\mathrm{log} 0$. $\endgroup$
    – digital-Ink
    Commented Sep 24, 2016 at 20:13
  • $\begingroup$ Here G denotes the profit from playing the game which can go negative.. $\endgroup$
    – vishmay
    Commented Sep 24, 2016 at 20:16

Not the answer you're looking for? Browse other questions tagged .