# 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

• 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

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 $$\frac{3S-S_{2}}{2} + \frac{3S+2S_{2}}{2} = \frac{3S+S_{2}}{2}$$, which is again maximized if $$S_{2}=3S$$, so that your expected balance will be $$\frac{6S}{2} = 3S$$, and your expected profit (assuming you won on the first flip) will now be $$3S-S=2S=2x$$.

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.

• 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
• @RossMillikan I'm not sure why you say that the expectation is $1.5(x+s/2)$ and not just $(x+s/2)$. – bobsacameno Mar 20 '18 at 18:14

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 $\frac{0.5 \times 3-1}{3-1}$.

• To maximize the expected winnings you should bet your whole bankroll (as @Ross suggests in his answer). The problem with doing this, is that at some point (with probability 1) you will lose everything you have. This is the idea of Kelly's bet, it maximizes the growth of your wealth, and keeps you safe from losing it all. Note that the expectation of betting it all will be larger than betting Kelly, but with Kelly you will not lose everything. – bobsacameno Mar 23 '18 at 8:08

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.

• You didn't read the question carefully. You are getting 2:1 odds, so if you win your balance is $x+2S$, not $x+S$ – Ross Millikan Jan 21 '15 at 19:38
• I did, I was just asking him if it is possible that he made a mistake because on one hand he says you win 2S on the other your new balance is x+2S (which means you win 3S).. – andros Jan 21 '15 at 22:20
• No. Your stake is not forfeited unless you lose. Until then it's still yours, so it's not counted as part of your winnings just because you retain it after you win. If you win $\ 2S\$, your both retain your stake of $\ S\$ and also win $\ 2S\$. – lonza leggiera Sep 13 at 12:14