Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
There was a Project Euler problem where your objective was to maximize the chance you finish with at least (as I recall) $10^9$. The thought process is different. –  Ross Millikan Nov 19 '12 at 15:41
add comment

1 Answer 1

up vote 4 down vote accepted

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. The expected win is then $\frac {3^{100}-1}{2^{100}}\cdot 100 -100(1-\frac 1{2^{100}})\approx 4\cdot 10^{19}$

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, then your expectation is $1.5(\frac {x-S}2+\frac {x+2S}2)=1.5(x+\frac S2)$ which (given the rules) is maximized when $S=x$. 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.

share|improve this answer
    
Thanks Ross - I really appreciate your help. If you're willing, I extended the question slightly here: math.stackexchange.com/questions/246565/… –  Boris Nov 28 '12 at 16:11
    
Can you explain how you get 3^100 − 1, and why do you subtract a lose of 1, i.e. why is this not £100? –  timboj May 22 '13 at 23:02
    
@timboj: yes, it should be winning $100(3^{100}-1)$ You end with $100\cdot 3^{100}$ and started with $100$, so that is the profit. –  Ross Millikan May 23 '13 at 3:31
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.