Half-Kelly Betting With Equal Wins and Losses
You start with $\$2$ and make 6 consecutive bets on independent fair coin flips. Each time, you bet exactly half of your current bankroll: if you win, you gain your stake (your bankroll is multiplied by $3/2$); if you lose, you lose your stake (your bankroll is multiplied by $1/2$).
Suppose that across the 6 flips you win exactly 3 times and lose exactly 3 times. What is your final bankroll, and what is your net profit or loss?
Hints
- Each win multiplies your bankroll by $3/2$ and each loss multiplies it by $1/2$. Does the order of wins and losses affect the final result?
- The final bankroll is $B_0 \cdot (3/2)^3 \cdot (1/2)^3$. Compute $(3/2)^3 \cdot (1/2)^3 = 27/64$.
- The growth factor $27/64 < 1$ even though you won half your bets. This illustrates why the geometric mean of the per-bet multipliers ($\sqrt{3/2 \cdot 1/2} = \sqrt{3/4} < 1$) determines long-run wealth, not the arithmetic mean.
Worked Solution
How to Think About It: The crucial insight is that the order of wins and losses does not matter. Each win multiplies your bankroll by $3/2$ and each loss multiplies it by $1/2$. Multiplication is commutative, so whether you win the first three and lose the last three, or alternate, or any other pattern -- the final bankroll is the same. This is a key property of multiplicative betting strategies.
Quick Estimate: After 3 wins and 3 losses, your bankroll is multiplied by $(3/2)^3 \times (1/2)^3$. Let's compute: $(3/2)^3 = 27/8 = 3.375$ and $(1/2)^3 = 1/8 = 0.125$. The product is $3.375 \times 0.125 = 0.421875$. Starting from $\$2$: $2 \times 0.421875 = 0.84375$. So you end up with less than a dollar despite winning half your bets. That's a loss of about $\$1.16$.
Approach: Compute the multiplicative growth factor directly.
Formal Solution:
Let $B_0 = 2$ be the initial bankroll. At each step, you bet half: - Win: bankroll goes from $B$ to $B + B/2 = (3/2)B$ - Loss: bankroll goes from $B$ to $B - B/2 = (1/2)B$
After $w$ wins and $\ell$ losses, the final bankroll is: $$B_{\text{final}} = B_0 \cdot \left(\frac{3}{2}\right)^w \cdot \left(\frac{1}{2}\right)^\ell$$
With $w = 3$, $\ell = 3$: $$B_{\text{final}} = 2 \cdot \left(\frac{3}{2}\right)^3 \cdot \left(\frac{1}{2}\right)^3 = 2 \cdot \frac{27}{8} \cdot \frac{1}{8} = 2 \cdot \frac{27}{64} = \frac{27}{32}$$
Net profit/loss: $$\text{Net} = \frac{27}{32} - 2 = \frac{27 - 64}{32} = -\frac{37}{32} \approx -\$1.15625$$
Answer:
- Final bankroll: $\frac{27}{32} = \$0.84375$
- Net loss: $\frac{37}{32} \approx \$1.16$
Despite winning exactly half the bets, you lose more than half your initial stake.
Intuition
This is the classic illustration of why the geometric mean matters more than the arithmetic mean in multiplicative settings. The arithmetic mean of the per-bet multipliers is $(3/2 + 1/2)/2 = 1.0$ -- suggesting you should break even. But in a multiplicative process, it's the geometric mean $\sqrt{(3/2)(1/2)} = \sqrt{3/4} \approx 0.866$ that determines your growth rate per bet. Since $0.866 < 1$, you lose money on average despite having a "fair" bet in additive terms.
This is directly relevant to portfolio management and the Kelly criterion. Betting half your bankroll on a fair coin is actually overbetting -- the Kelly-optimal fraction for a fair coin is zero (don't bet at all, since the edge is zero). More generally, even with a positive edge, betting too large a fraction destroys wealth because the geometric drag from losses outweighs the arithmetic gain from wins. The lesson: in any repeated multiplicative game, volatility is a drag on growth, and the drag increases with bet size. This is sometimes called "volatility drag" or the "variance tax."