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$x$ is a variable which take its real values in the interval $[\min x, \max x]$ and $c$ is a real constant value that I want to determine. I want to determine a fixed value for $c$ such that $x/c$ approaches $1/2$ for any value of $x$ in $[\min x,\max x]$.

That is, I want that for any value of $x$ (in the interval $[\min x, \max x]$), $x/c$ approximates $1/2$. How can I determine a reasonable value of $c$ for this purpose (i.e. the most convenient value of $c$ such that $x/c$ is usually near $1/2$)?

Maybe by resolving an equations which looks like

$\int_{\min x}^{\max x} \frac{x}{c} dx = \frac{1}{2},$

or something similar? Or by making use of the average value of all possible values of $x$?

This is an example with a finite set of values that $x$ can take (not an interval):

Suppose that $x$ take its values from the set $X = \{2, 5, 6\}$, thus maybe a good value to choose for $c$ is $c = (2+5+6)\cdot2/3 = 26/3$, because the values $2/c$ and $5/c$ and $6/c$ are all not very far from $1/2$. This is the same as saying that $(2/c+5/c+6/c)/|X| = 1/2$ which gives $c = 26/3$. Thought, I don't know if there a more optimal value than $(2+5+6)\cdot2/3$ for this example ...

EDIT1: clarification of what I want to do:

Suppose I have a jar $E$ containing $6$ elements $\{e_1,e_2,e_3,e_4,e_5,e_6\}$ mapped to a set $X$ of $6$ values: $X = \{x_1=2, x_2=4, x_3=6, x_4=1, x_5=8, x_6=3\}$. I take each element $e_i$ from $E$ and put it in another jar $E'$ with a probability of $x_i/c$ (if the probability event of $x_i/c$ occurs, I put $e_i$ in $E'$). At the end, I want that the number of elements in the jar $E'$ equals approximately the half of the number of elements of $E$, that is, for this example ideally the number of elements in $E'$ will be $6/2=3$. How to determine a fixed value of $c$ in order to do that? I guess that the problems turns out to finding the value of $c$, such that for any value $x_i$, the probability $x_i/c$ approximates $1/2$, right?

Ok the answer is probably to choose $c$ as $2$ times the average value of $x_i$.

EDIT2: the second problem (more difficult) after joriki's answer:

Suppose this time that each value $x_i$ is actually the $\min$ distance between the corresponding item $e_i$ and the items that are already in $E'$. That is with previous example:

We have $E = \{e_1,e_2,e_3,e_4,e_5,e_6\}, X = \{\max x, ?, ?, ?, ?, ? \}$

We put $e_1$ in $E'$ according to probability $x_1/c$ which is $1$, thus: $E = \{e_2,e_3,e_4,e_5,e_6\}, E' = \{e_1\}$, and $X = \{x_1=\max x, x_2=\min_{e_j \in {E'}}(e_2,e_j), ?, ?, ?, ?\}$

We put $e_2$ in $E'$ according to probability $x_2/c$, thus: $E = \{e_3,e_4,e_5,e_6\}, E' = \{e_1,e_2\}$, and $X = \{\max x, x_2, x_3=\min_{e_j \in {E'}}(e_3,e_j), ?, ?, ?\}$

And so on ...

So how to determine $c$ in this case ?

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    @GerryMyerson Right, now I'll edit my post to make the question more difficult. I'll suppose this time that each value xi depends on the min distance between ei and the elements already in E' !2012-09-27

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The expected number of items in $E'$ is $\sum_ix_i/c=n\langle x\rangle/c$, where $n$ is the number of items and $\langle x\rangle$ is the average of the $x_i$. You want this to be $n/2$, so you want $\langle x\rangle/c=1/2$, or $c=2\langle x\rangle$. However, note that there is no guarantee that the average is at least half the maximal value, so with this choice the values $x_i/c$ are not guaranteed to be probabilities. To remedy this, you can either use $\min (x_i/c,1)$ as probabilities instead, or choose $c=\max (2\langle x\rangle,\max_ix_i)$.

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    I've created a new similar question at http://math.stackexchange.com/questions/203452/determining-an-optimal-value-of-c-such-that-the-probability-x-c-for-any-value-of2012-09-27
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Let's suppose $C_i$ is a continuous uniform random variable on the interval $[0,c)$ and you include element $e_i$ if $C_i \le x_i$. Then the expected number of elements included is

$\sum_{i: x_i \lt c} \frac{x_i}{c} + \sum_{i: x_i \ge c} 1 = \frac{\sum_{i}\min\left({x_i},{c}\right)}{c}.$

You want this to be equal to $\frac{n}{2}$. A reasonable first estimate $\hat{c}_1$ for $c$ is $2 \times \frac{\sum_{i} x_i}{n}$, twice the mean of the $x_i$, and if this is at least as large as the maximum $x_i$ then you have an answer.

If not then the expected number of elements included will be less than $\frac{n}{2}$, and you need to adjust your estimate (possibly more than once), perhaps using something like

$\hat{c}_{k+1} = \sum_{i: x_i \lt \hat{c}_k} \frac{x_i}{\hat{c}_k} \left/ \left(\frac{n}{2}-\sum_{i: x_i \ge \hat{c}_k} 1\right) \right.$