Derivatives Foundation puzzles, solved step by step

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  1. 051You have the same 60% coin as before and even money on every flip, but now you bet half your capital every time. After 100 bets, what does your typical outcome look like, even though every single bet has positive expected value?Betting and sizingHardRisk managementHedge funds

    Try it first

    Each bet has a 10% edge on the money at risk. Where does betting half your stack 100 times leave the typical player?

    Show the worked solution

    The typical player ends with about 3% of the starting capital, while the average outcome is about 13,781 times the start. With 60 wins and 40 losses the stack is multiplied by 1.5 to the 60 and 0.5 to the 40, which is 0.033. Betting half the stack is 2.5 times the Kelly fraction of 20%, and anything beyond twice Kelly turns a positive-edge game into a losing one for the person actually playing it.

    Why can every bet be favourable and the typical result still be a loss?

    Think of a shopkeeper who doubles the stock every time a line sells and halves it every time it does not. A good line sells six times in ten, yet after a hundred cycles the shelf is nearly bare, because growth is a chain of multiplications and the order of operations does not rescue you. Expected value is an average across many imaginary players, but your capital follows one path, and a path is a product of factors, not a sum. Betting half the stack makes each win a factor of 1.5 and each loss a factor of 0.5. Sixty wins and forty losses, the most likely split, multiply to 0.033. The arithmetic mean of 13,781 is carried by the rare paths with 80 or 90 wins, which almost nobody lives on.

    The same 100 bets: the average flies, the typical player is nearly wiped outAverage path (mean)start = 1050100betsx 13,781Typical path (60 wins, 40 losses)start = 1050100betsx 0.033, about 3% left0.01110010,000x startEach bet: 0.6 x log 1.5 + 0.4 x log 0.5 = -0.034, so the typical path shrinks 3.4% a bet
    On a shared log scale, the average path compounds at 1.1 per bet to about 13,781 times the start after 100 bets, while the typical path of 60 wins and 40 losses shrinks by 3.4% per bet to about 3% of the start, because capital multiplies and the half-stack bet makes the loss factor too harsh.

    What is the right quantity to look at instead of the mean?

    Look at the expected growth rate of the logarithm of capital, because the log turns products into sums and the law of large numbers then works. Each bet changes log capital by 0.6 x log(1.5) + 0.4 x log(0.5), which is -0.034, a negative number, so the typical player loses about 3.4% of capital per bet on average. Over 100 bets that is a factor of exp(-3.4), the 0.033 you saw above. The chance of finishing below the start is 76%, so the mean is a number you will almost never meet.

    The relationship
    g(f)=p ln⁡(1+f)+(1−p) ln⁡(1−f)g(0.5)=0.6ln⁡1.5+0.4ln⁡0.5≈−0.034f∗=2p−1=0.2g(f) = p\,\ln(1+f) + (1-p)\,\ln(1-f) \qquad g(0.5) = 0.6\ln 1.5 + 0.4\ln 0.5 \approx -0.034 \qquad f^* = 2p-1 = 0.2
    g(f)the expected growth of log capital per bet when you stake fraction f
    pthe chance of winning each even-money bet, here 0.6
    f*the Kelly fraction that maximises g, here 20% of capital
    What it says in wordsStaking half the stack makes the expected log growth negative, while the Kelly stake of a fifth makes it positive.

    Where does the sizing go wrong, and what would fix it?

    The Kelly fraction for an even-money bet is 2p minus 1, here 20% of capital, and it maximises the growth rate at +0.0201 per bet, which turns 100 bets into a typical multiple of about 7.5. Growth is zero at exactly twice Kelly, 40% here, and negative beyond it, so a half-stack bettor is on the wrong side of the hill even though the coin is in their favour. The limitation is that the Kelly rule assumes you know p exactly and will play many times; with an uncertain edge, desks size below Kelly on purpose, often at half of it.

    Where candidates lose it

    Most candidates say the player ends up rich, because they multiply the 10% edge through 100 bets and quote the mean. The interviewer wants to hear the word median, or typical, and the observation that capital compounds multiplicatively, so wins and losses of equal size do not cancel.

    The second loss is stopping at the picture without the growth rate. Say the log growth per bet, show it is negative at a half stake and positive at a fifth, and name twice Kelly as the point where the game turns against you.

    What the interviewer asks next

    • What fraction maximises the typical growth, and what is the growth rate there?
    • At what stake does the typical player break even despite the edge?
    • If you only get to play 5 times rather than 100, does the answer change?
    • Your estimate of the 60% is itself noisy. Does that push the stake up or down?