Q:

Probabilty Scaling Question

So i have been working on this problem but i can't seem to figure it out.
$$\frac{e - h/100}{e-h}$$
$e$ is a big number, but it doesn't really matter what it is (it's $2^{52}$), and
$h$ is a random number between $0$ and $e-1$ inclusive.
A.) What i want is if $h$ is halfway through its range ($h=e*0.5$) the outcome to be $2$ but instead it gives me $(e-e*0.5/100)/(e-0.5*e)=1.99$. Same goes for if $h$ is $1/4$ through its range for example. I get $1.33$ when i want it to be $1+1/3$.
Then I want to adjust the calculation from being $0$ off when $h$ is at its minimum $0$ to being 1% off when $h$ is at its maximum.
What would i have to change on the formula above to achieve this?
B.) What would i have to do to get a formula which is always 2% off and doesn't scale like the one 1 have? E.g. In this case i'd want the result to be $1.96$ for $e*0.5$ and $1.306667$ for $e*0.25$.
Feel free to ask me if something is unclear.
Holderbert

A:

Ignoring for now the adjustment by 1% or 2%, the formula
$$
\frac e{e-h} = \frac1{1-\frac he}\tag1
$$
seems to give the right pre-adjusted answer: when $h/e$ is zero, you get $1$; when $h/e=1/2$, you get $2$; when $h/e=1/4$ you get $4/3$. To adjust (1) so that it is off by zero when $h/e=0$ and is off by 1% when $h/e=1$, with linear scaling, you should use
$$
\left(1- 0.01\frac he\right)\frac1{1-\frac he}\tag2$$
To achieve a consistent reduction by 2%, you should use
$$
(0.98) \frac1{1-\frac he}\tag3
$$

