Question
You found a way to play blackjack where you have the following payoffs:
\[
P(\text{win}) = 0.55, \qquad P(\text{lose}) = 0.45.
\]
Where if you win, you stake doubles, and if you lose, you lose the stake.
Each round you bet a fraction \(f\) of your current wealth.
Derive the value of \(f\) that maximises the long-run expected logarithmic growth rate of wealth.
Check your answer
Solution
Answer: 0.1
Let your current wealth be \(W\), and suppose you bet a fraction \(f\) of that wealth each round.
The amount staked each round is therefore
\[
fW.
\]
Because the game is even-money in the sense that a win gains the amount staked and a loss loses the amount staked, your wealth after one round is
\[
W(1+f) \quad \text{with probability } 0.55,
\]
and
\[
W(1-f) \quad \text{with probability } 0.45.
\]
To maximise long-run growth, we maximise the expected logarithmic growth rate:
\[
g(f)=0.55\ln(1+f)+0.45\ln(1-f).
\]
This is the right objective because wealth evolves multiplicatively over repeated rounds, and logarithms convert multiplication into addition. So maximising expected log-return maximises the asymptotic growth rate.
We can now differentiate and set to 0:
\[
g'(f)=\frac{0.55}{1+f}-\frac{0.45}{1-f}.
\]
\[
\frac{0.55}{1+f}=\frac{0.45}{1-f}.
\]
\[
0.55(1-f)=0.45(1+f).
\]
\[
0.10=f.
\]
So the optimal betting fraction is
\[
f^*=0.1.
\]
To confirm this is a maximum, take the second derivative:
\[
g''(f)=-\frac{0.55}{(1+f)^2}-\frac{0.45}{(1-f)^2}.
\]
This is negative for all feasible \(f\in(-1,1)\), so \(g(f)\) is strictly concave and the critical point is the unique maximiser.