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?)

  • 1
    $\begingroup$ 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
    Sep 21, 2017 at 14:05
  • 2
    $\begingroup$ 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. $\endgroup$ Sep 21, 2017 at 16:26

2 Answers 2


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.

  • 1
    $\begingroup$ 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
    Sep 25, 2017 at 10:46
  • 1
    $\begingroup$ @SMeaden Problem sez " you will get 2 + your bet" so $b=2$ $\endgroup$
    – nbbo2
    Oct 18, 2017 at 15:26

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.

  • 5
    $\begingroup$ 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. $\endgroup$ 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
    Nov 6, 2018 at 3:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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