(Feel free to suggest the correct Stackexchange community - or otherwise - if this is not the correct one)

When trading financial markets, a gain of x%, won't recover a loss of x% (same applies to any gain-loss scenario, in any market, I assume)

e.g. I start with 100, I lose 2% (98% or original capital), I gain 2%. I'm still in loss (99.96% of original capital)

Is there a formula to calculate the necessary % gain of a single transaction over k loss of x% gain after y losses of x%? (if the formula is generalized to z gains of x% and k losses of x% is preferable - x% is fixed)

  • $\begingroup$ I don't quite understand the last paragraph; is your loss/gain $x$ fixed? So, you suppose that we occur $y$ losses and then $k$ gains of $x$. So the losses/gains are in exactly that order and not mixed und each of them is $x%$? Last, you ask on the necessary percentage gain of a single transaction needed, to recover all losses, right? $\endgroup$ – skoestlmeier Jun 15 '19 at 8:35
  • $\begingroup$ loss/gain % fixed and = to x% (of the current balance). My goal is how to find an expectancy model where by winning and losing x% z and k times respectively, I will still be having a gain higher than the initial capital. The formula would allow me to calculate how many wins I need to cover one loss. Another alternative is indeed what you say, the necessary % to recover the losses in 1 transaction, after k losses $\endgroup$ – dragonmnl Jun 15 '19 at 8:41

Let x represent the percent change-e.g. 2%, let k represent the number of decreases, and z the number of increases. Something like this? We want to find z such that:




And take log:

$z \ln \left(1+x\right)=-k \ln \left(1-x\right)$

and solve for z:

$z =-k \frac{\ln \left(1-x\right)}{\ln \left(1+x\right)}$

Or you want z to represent the percent increase such that:



  • $\begingroup$ could you elaborate the answer by specifying what k, z and x represent? for the sake of clarity (even though I assume you've just used the same variables as in the question) $\endgroup$ – dragonmnl Jun 21 '19 at 20:57
  • $\begingroup$ Added more explanation, hope it is clearer now. $\endgroup$ – Magic is in the chain Jun 21 '19 at 21:58
  • $\begingroup$ much clearer. thanks. but where is the k in the second formula, if based on what I read the second formula is the % increase necessary after k decreases of x% (which is in fact what'd fully give a complete answer) $\endgroup$ – dragonmnl Jun 28 '19 at 21:44
  • $\begingroup$ When you take the k-th root of both sides , the k disappear, thanks to the right hand side being equal to 1. $\endgroup$ – Magic is in the chain Jun 28 '19 at 22:19

This is a common question type on the GMAT.

Suppose you lose x% of your initial funds, $V_0$, so you now have $V_0(1-x/100)$. What y% return on the new funds do you need to return to your initial funds?

Stated this way, we express our problem as:

$V_0(1-x/100)(1+y/100)=V_0 $, or $(1-x/100)(1+y/100)=1$

Solving for y, we have:

$y= \frac{100}{1-x/100}-100$, or


  • 1
    $\begingroup$ this would cover the answer for "how much gain for one loss". is that correct? $\endgroup$ – dragonmnl Jun 21 '19 at 20:58
  • $\begingroup$ Yes. Or vice versa $\endgroup$ – David Addison Jun 22 '19 at 6:34

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