Why does a 50% loss need a 100% gain to recover?
Because the loss is taken on a large base and the recovery has to be earned on a small one. That asymmetry is not a psychological quirk — it is division, and it is the mathematical reason risk control exists at all.
The formula
After losing a fraction D, you hold (1 − D) of what you had. To get back to 1 you need to multiply by 1 ÷ (1 − D), which is a gain of:
Gain needed = D ÷ (1 − D)
| Drawdown | Gain needed to recover |
|---|---|
| 10% | 11.1% |
| 20% | 25.0% |
| 33.3% | 50.0% |
| 50% | 100% |
| 60% | 150% |
| 75% | 300% |
| 90% | 900% |
Concretely: $100,000 falls 50% to $50,000. Recovering the same $50,000 you lost is now a 100% gain, because the denominator halved. The dollars are symmetric; the percentages never are.
The time cost
Gains compound at a finite rate, so a drawdown is also a bill payable in years. At a hypothetical 8% annual compounding:
- Recovering a 20% drawdown (+25% needed) takes ln(1.25) ÷ ln(1.08) ≈ 2.9 years.
- Recovering a 50% drawdown (+100% needed) takes ln(2) ÷ ln(1.08) ≈ 9.0 years.
Nine years is not a setback; it's a decade of a working life spent returning to a starting point.
Connecting it back to sizing
Unit 1 said the fraction risked per trade matters more than it looks. Here is why, with n consecutive losses leaving (1 − f)ⁿ of the account:
| Risk per trade | Equity after 6 losses | Drawdown | Gain needed |
|---|---|---|---|
| 1% | 94.1% | 5.9% | 6.2% |
| 5% | 73.5% | 26.5% | 36.0% |
| 10% | 53.1% | 46.9% | 88.2% |
Five times the risk per trade produces roughly six times the recovery burden, because the recovery function is convex — it punishes deep holes far more than proportionally.
Are six losses in a row unusual?
No. For a system that wins 40% of the time, any specific block of six trades comes up all losers with probability 0.6⁶ ≈ 4.7%. Across a hundred trades, a run like that is ordinary rather than remarkable. The planning implication is blunt: a streak that hurts should be something the arithmetic already survived on paper, because it will arrive eventually and you don't get to choose when.
Try it now
- Compute the gain needed to recover from −15%, −40% and −65% using D ÷ (1 − D).
- Assume a hypothetical process compounds at 10% a year. Roughly how many years does each of those recoveries take? Use ln(1 + gain) ÷ ln(1.10).
- Now find the same shape in a record nobody authored. The full history below contains a fall of more than 50%: locate the peak, locate the trough, then find the date price first closed back at that old high. Count the years. The arithmetic on this page is not theoretical.
Unit done. Next: the unit about the number everyone believes and shouldn't.