How it unfolds, step by step
The setup
On this chart the bracket is the size of a win, and of a loss. It is varied from 1% of the position up to 10%, with the round-trip cost held fixed. The line is the hit rate needed for the sequence to come out level.
What happens next
At a wide bracket the required hit rate sits close to 50%, because the cost is small next to the move. As the bracket narrows, the line climbs steeply. The same fixed cost is now a large share of what is being won.
What it means — and does not
The requirement is not a matter of skill or of view. It is division. A process aiming at small moves is being asked to be right far more often than one aiming at large ones. That is true before anything about markets is considered at all. What is more, 'be right more often' is exactly the thing that is hardest to improve.
- Hit rate needed to break even (%)
- 50%, the free-of-cost floor · 50
A fixed round-trip cost against a widening bracket. The narrower the target, the more often the process has to be right.
Show the underlying data
| Bracket size, % of position | Hit rate needed to break even (%) |
|---|---|
| 1 | 65 |
| 2 | 57.5 |
| 3 | 55 |
| 4 | 53.8 |
| 5 | 53 |
| 6 | 52.5 |
| 7 | 52.1 |
| 8 | 51.9 |
| 9 | 51.7 |
| 10 | 51.5 |
What this does not establish
The arithmetic assumes wins and losses are the same size. That is the assumption to check, and the honest version of the calculation asks for both numbers. A process whose wins are larger than its losses needs a lower . One whose losses run further than its wins needs a higher one than this line shows. The level at which the reading fails is not on the chart. It is whether the win and loss sizes actually are symmetric in practice.
Things to check
4What is the round-trip cost as a fraction of the bracket being aimed at?
Are wins and losses actually the same size, or is one systematically larger?
What hit rate does the arithmetic require, and is there evidence of achieving it?
The breakeven hit rate follows from the assumed win, loss and cost. It does not predict your hit rate.
Common mistakes
2Reasoning from the win rate alone, without the size of a win against the size of a loss.
Treating a required hit rate above 50% as a small adjustment. Moving a real process by ten percentage points is very hard.
Aiming at smaller moves, with the cost per trade unchanged. What happens to the hit rate needed to stand still?
- It rises, because the same cost is a bigger share of a smaller win
- It stays at 50%, since wins and losses are still equal
- It falls, because small moves happen more often
Show the answer
A. It rises, because the same cost is a bigger share of a smaller win
The requirement is division: the cost sits in the numerator and the bracket in the denominator. Narrowing the target while holding the cost fixed pushes the required rate up, and it climbs steeply once the cost is a large fraction of the bracket.
Frequently asked
3What is the formula?
With equal win and loss sizes, the break-even hit rate is (loss size + cost) divided by (win size + loss size). With no cost that is one half. The cost is what lifts it.
Does a bigger position help?
It changes the rupees, not the arithmetic, because both the bracket and most of the charges scale with the position. What changes the ratio is the size of the move being aimed at, set against the cost.
Can I work this out for my own numbers?
Yes. The break-even frequency calculator on this site takes the segment, the position and the bracket, and returns the required rate, with every charge listed.