Worked solutionOne step at a time
Each flip multiplies your wealth
Bet a fraction f of what you have. A win multiplies your wealth by 1+f, a loss by 1−f. After many flips the order does not matter, only the count of wins and losses.
The typical player wins 60% of flips
Over many flips, wins ≈ 0.6n and losses ≈ 0.4n. Take the log and divide by n to get the growth rate per flip that the typical (median) player sees.
Maximise it
Differentiate and set to zero. The answer is the edge: win probability minus loss probability.
Answer: C, 20%
g(0.2) = 0.6·ln 1.2 + 0.4·ln 0.8 = +0.0201 per flip, the best possible. For comparison g(2%) = +0.0038, g(50%) = −0.034 and g(100%) = −∞. The average wealth always prefers bigger bets; the typical player does not, because a 50% loss needs a 100% gain to undo.
What the simulation showedSame flips, three bet sizes
The reel gives 10,000 players $100 each and one shared set of coin flips (seed 7), then replays them under different bet sizes.
- All-in: 9,940 of 10,000 were broke after 10 flips; 60 survived (0.6¹⁰ × 10,000 = 60.5 expected), and 1 was left after 20.
- Yet the average wealth of all 10,000 after 10 flips was $614, carried entirely by the 60 survivors holding $102,400 each. The theoretical average, $100 × 1.2¹⁰, is $619.
- Bet 50%: after 100 flips the typical (median) player had $3.34, despite the edge.
- Bet 2%: after 100 flips the typical player had $146.24, growth so slow it barely shows.
- A numerical scan of g(f) over 0 ≤ f < 1 in steps of 0.001 peaks at exactly f = 0.20. No 20% strategy was shown in the reel.
Wrong answersWhy each one is tempting
| Pick | What went wrong |
|---|---|
| A · 100% | Maximised the average wealth, which is carried by a handful of lucky survivors. |
| B · 60% | Confused the win probability with the bet size; growth is negative there. |
| D · 2% | Safe, but leaves most of the growth on the table: about a fifth of the best rate. |
Follow-ups an interviewer asks nextTry these before opening them
What if a win pays 2 to 1 instead of even money?
Kelly in general is f = p − q/b, where b is the net payout per dollar. With p = 0.6 and b = 2, f = 0.6 − 0.4/2 = 40%.
Above what fraction does the typical player shrink?
Solve g(f) = 0: 0.6·ln(1+f) + 0.4·ln(1−f) = 0 gives f ≈ 0.38. Betting more than about twice Kelly loses money for the typical player even with the edge.
Why do people bet half Kelly in practice?
Near the peak, g is flat: half Kelly (10%) keeps g = 0.015, three quarters of the best growth, with much smaller swings. And if your estimate of p is too high, overbetting hurts far more than underbetting.
How long until the Kelly bettor reliably beats the 2% bettor?
Their log-growth differs by about 0.016 per flip, while the Kelly bettor's log-wealth has a per-flip standard deviation near 0.2. The edge beats the noise once 0.016n exceeds about 0.2√n, roughly n ≈ 150 flips.
