Tweaking formulas to increase scoring I am building a model for SVM classification. However, the confidence score that i have would be from negative to positive. 
This is the formula i am using to normalize the confidence score 
$\frac{1}{1 + e^{-score}}$
However, the results show very poor scoring such as of scores 0.3 and 0.4 
I would like the result of the formula to show 0.7 and 0.8. Any suggestions to tweaking the formula in such a way that i do not end up with a result of more than 1. 
I do have considered adding an arbitrary number but this might result in a value greater than 1 which is incorrect. 
Say for example: 
I would have a score which could be in the range $(-\infty, \infty)$
However, I would just need to compare the relative scores. 
In this formula, i am able to map them to $[0,1]$ to compare their relativity. 
However, now the scores that the formula is giving me is about 0.3 or 0.4 only, which does not seem to be a good score. Is there anyway i could increase the scores to have 0.7 without tweaking the formula. Maybe keeping the range from $[0.2,1]$ or something
 A: I guess your current raw scores are in the neighborhood of $-0.7$ or so, which yields a scaled score of
$$
\frac{1}{1+e^{0.7}} \doteq 0.33
$$
If you want, you can replace the $1$ in both the numerator and the denominator by $6$, and then you would get
$$
\frac{6}{6+e^{-score}} = \frac{6}{6+e^{0.7}} \doteq 0.75
$$
But this is fairly ad hoc.  I don't know if you wanted something like that.
A: You said formula give scores in $[0.3,0.4]$, which makes me guess your your raw data is in $[-0.6,-0.2]$. Try:


*

*$n\over{n+{(e^{-score})}}$, try varying the values of $n$ (n=5,5.5,6,7 etc will give you output in range $[.7,.8]$ if your raw data is in the domain mentioned above.)

*$n\over{n+n^{-score}}$, try varying the values of $n$. (n=6,7,8 etc will give you output in range $[.6,.85]$ if your raw data is in the domain mentioned above.)

*$n\over{n+{(e^{-score})}^n}$, try varying the values of $n$.(n=8,9,10 etc will give you output in range $[.02,.65]$, widening the range, if that's what you are looking for)
