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You're invited to a one-on-one coin-flip gambling game. Your opponent has 1 million USD on hand (the max you can bet is 1 million USD). The payouts for flipping heads and tails are as follows:

  • Tails: You win 2x your bet.
  • Heads: You lose your entire bet.

This game is a one-off gamble; no second chances. How much would you bet?

(This was from a trading interview so it likely involves choosing the right risk-reward profile. Is there a right answer? Or is this a subjective question?)

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    $\begingroup$ sounds like the heads and tails are analogy for limits and stops (with a risk reward ratio of 1:3). "Don't risk more than you can afford to lose" - a rule of thumb is 2% of equity. $\endgroup$
    – xiaomy
    Commented Sep 21, 2017 at 14:05
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    $\begingroup$ I think it's just meant to see how you think and there's no right answer. If I asked that and someone said "use the Kelly criterion cause it maximizes long term growth" I'd receive that much better than, say, "you have an edge; bet it all" but I'd ask followup questions to see if they understood the limitations of the approach and what circumstances would suit it best/worst. I'd be more impressed if the person asked me some questions that showed they understood what issues are pertinent to a risk management decision before answering. $\endgroup$
    – spaceisdarkgreen
    Commented Sep 21, 2017 at 16:26

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The Kelly criterion gives the fraction, $f$, of the current bankroll to bet in order to maximize the longterm growth. The criterion is given by $$ f = \frac{bp-q}{b}, $$ where $b$ is the winnings received on \$1 bet, $p$ is the probability of winning, and $q=1-p$ is the probability of losing the bet of \$1.

In your case $b=2$, $p=q=0.5$ so the optimal fraction to bet is $$ f = \frac{2\cdot0.5-0.5}{2} = 0.25. $$ That is 25% of your bankroll or \$250k.

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    $\begingroup$ I guess it's a good answer. The problem for me is that for most "trading" b and p are unknowns, or are interest rates / coupon rates where b = rate and p = 100% less credit risk. So other than credit defaults, trading isn't much to do with probability, it's all about rates, spreads, skews and competition with other traders. $\endgroup$
    – rupweb
    Commented Sep 25, 2017 at 10:46
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    $\begingroup$ @SMeaden Problem sez " you will get 2 + your bet" so $b=2$ $\endgroup$
    – nbbo2
    Commented Oct 18, 2017 at 15:26
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An alternative approach is to size your bet to maximize your expected utility, which is assumed to be given by a function $u(w)$ of your total wealth $w$. This could be a better approach than using the Kelly criterion, because the Kelly fraction gives the amount to bet if you want to maximize your long-term growth rate, assuming that you will bet a large number of times, but in this case you are told that you only get one chance to bet.

If you bet a fraction $x$ of your bankroll, you will have $1+2x$ if you win and $1-x$ if you lose, so your expected utility is

$$ \tfrac{1}{2}u(1 + 2x) + \tfrac{1}{2}u(1 - x) $$

Maximizing this is equivalent to maximizing $u(1+2x) + u(1-x)$. In the special case of log utility $u(w)=\log w$ you require

$$ \frac{d}{dx} \left( \log(1+2x) + \log(1-x) \right) = \frac{2}{1+2x} - \frac{1}{1-x} = 0 $$

which you can solve to give $x = 1/4$, the same answer as if you used Kelly betting to maximize your long-term growth. Other utility functions will give different results.

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    $\begingroup$ As your answers basically says, maximizing expected utility with $\log$ utility over terminal wealth is mathematically equivalent to maximizing the expected growth rate (as is done in Kelly criterion betting). Let $\frac{V_t}{V_0} = e^{Rt}$ hence $\log V_t - \log V_0 = R t$. Maximizing $E[\log V_t]$ or $E[R]$ are equivalent objectives. Log utility is also a special case of a broader set of constant relative risk aversion utility functions. $\endgroup$
    – Matthew Gunn
    Commented Sep 26, 2017 at 16:36
  • $\begingroup$ Nice solution. But I think if you want to maximise the expected winning, you should bet all you have. The bet is in your favour, take it! $\endgroup$
    – dynamic89
    Commented Nov 6, 2018 at 3:38

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