Question
A sponsor runs a repeatable even-money wager. Before each round, a player chooses a fraction f of their current bankroll to stake. The player wins with probability 70%, independently across rounds; on a win the stake is returned and the net bonus would normally equal the stake. However, the sponsor caps the net bonus at 20% of the bankroll immediately before that round. On a loss the entire stake is gone. There are no fees, and the player uses the same staking fraction each round.
What fraction maximises expected logarithmic bankroll growth per round? Show where the cap changes the objective, and derive the optimal fraction if the win probability is instead an arbitrary fixed p strictly between 0 and 1. Include the no-bet and bankruptcy boundaries.
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Hint 1
Write the bankroll multipliers for a win and a loss. The winning multiplier stops improving once the stake reaches 20%.
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Solution
Let W be the bankroll immediately before a round. A loss gives \(W(1-f)\). A win gives \(W[1+\operatorname{min}(f,0.20)]\), because the cash bonus is capped at \(0.20W\) and the stake itself is returned. For \(0\le f<1\), expected log growth relative to log W is
\[g_{p}(f)=p\operatorname{log}[1+\operatorname{min}(f,0.20)]+(1-p)\operatorname{log}(1-f).\]
For \(0\le f\le0.20\) this is the ordinary even-money Kelly objective, with derivative \(g^{\prime}_{p}(f)=\frac{p}{1+f}-\frac{1-p}{1-f}=\frac{(2p-1)-f}{1-f^{2}}\), and strictly negative second derivative wherever \(0<p<1\). Its unconstrained maximiser is \(2p-1\). For \(f>0.20\), the win term is constant but the loss penalty continues to increase; indeed \(g^{\prime}_{p}(f)=-\frac{1-p}{1-f}<0\). Therefore staking more than 20% can never be optimal. On \([0,0.20]\) the concave objective is maximised by clipping the unconstrained value to that interval: \(f^{*}=\operatorname{max}(0,\operatorname{min}(2p-1,0.20))\). This is a global, not merely local, optimum because the function decreases strictly above the cap.
With \(p=0.70\), the uncapped Kelly fraction would be 0.40, so the cap binds: \(f^{*}=0.20\), at the kink where the left derivative is positive and the right derivative negative. If \(p\le\frac{1}{2}\), \(f^{*}=0\); if \(\frac{1}{2}<p<\frac{3}{5}\), \(f^{*}=2p-1\); if \(p\ge\frac{3}{5}\) (and \(p<1\)), \(f^{*}=0.20\). A fraction \(f=1\) has a strictly positive chance \(1-p\) of zero final wealth, hence expected log wealth \(-\infty\) and cannot be optimal. No borrowing or shorting is allowed, so negative f and \(f>1\) are excluded. This repeated-round solution depends on the cap scaling with current bankroll; a fixed dollar cap independent of wealth would produce a different dynamic problem.
Assumptions
Each wager has independent win probability p, with \(0<p<1\); odds are even money before the cap.
The cap is 20% of pre-round current wealth, not a fixed dollar amount, so the fractional problem repeats unchanged.
Allowed stakes are \(0\le f\le1\) with no leverage or short bets; zero wealth has log utility \(-\infty\).
A win returns staked principal in addition to the capped net bonus.
Final Answer
With \(p=0.70\) the optimal stake is \(f^{*}=0.20\) of the current bankroll. More generally, for \(0<p<1\), \(f^{*}=\operatorname{max}(0,\operatorname{min}(2p-1,0.20))\). A full-bankroll stake \(f=1\) has log growth \(-\infty\) whenever losses have positive probability.