# Fitness based ranking

If 3 variables $x[0]$, $x[1]$, $x[2]$ and their sum give the probability of the variable being selected as follows

\begin{align} p[0] &= \frac{\text{sum}-x[0]}{2\times\text{sum}}\\ p[1] &= \frac{\text{sum}-x[1]}{2\times\text{sum}}\\ p[2] &= \frac{\text{sum}-x[2]}{2\times\text{sum}} \end{align}

then, it is not a very good fitness based probability. because for the numbers $0.426, 0.036, 0.325$ the corresponding probabilities are $0.23, 0.48, 0.29$, although the second number is 10 times smaller than the other two, it only receives half the share.

Can someone suggest a better approach?

• Mar 6, 2015 at 15:07

## 1 Answer

[assuming I understood what you meant] use Exponential ranking: $p_j = \frac{e^{- \alpha x_j}}{\sum_j e^{- \alpha x_j}}$

Where $\alpha$ is a coefficient $>1$ if you need to increase the 'weight' of the individual with the lower value

• Well, I tried your approach but the answers are pretty much the same. 0.279, 0.412, 0.308 for the probability which is actually the opposite effect. Instead of increasing the share of the second number it decreased it.
– user221505
Mar 6, 2015 at 15:50
• Add coefficients as per edit
– Alex
Mar 6, 2015 at 15:56
• Thanks, that works. Now I just have to decide on a suitable coefficient
– user221505
Mar 6, 2015 at 16:23