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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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    if $x$ is a variable, how is $x/c$ a probability?2012-09-27
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    @Avatar if c > maxx, then for any value x in [minx, maxx], 0 < x/c < 12012-09-27
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    @Avatar 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)*2/3 = 26/3, because the values 2/c and 5/c and 6/c are all not very far from 1/2. Thought, I don't know if there a more optimal value than (2+5+6)*2/3 for this example ...2012-09-27
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    Just because a number is between zero and one does not mean it is a probability ...2012-09-27
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    @kjetilbhalvorsen Well forget about the word "probability". Let's talk about 1/2. Ok?2012-09-27
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    You have to decide what you mean by "optimal" in order to have a ghost of a chance of getting an answer. Suppose your set is just 2 and 8. Is $c=10$ optimal because 2/10 and 8/10 are equally far from 1/2? Is $c=8$ optimal because 2/8 and 8/8 are equally far, multiplicatively, from 1/2? Maybe you want a least squares solution, the one that minimizes $((2/c)-(1/2))^2+((8/c)-(1/2))^2$? Think about what you really want to do.2012-09-27
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    @GerryMyerson Suppose I have a jar E containing 6 elements {e1,e2,e3,e4,e5,e6} mapped to a set X of 6 values: X = {x1=2, x2=4, x3=6, x4=1, x5=8, x6=3}. I take each element ei from E and put it in another jar E` with a probability of xi/c (if the probability event of xi/c occurs, I put ei 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 ?2012-09-27
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    @GerryMyerson so for your example of x={2,8}, an optimal value of c would make it possible to have only one element added to E`2012-09-27
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    Then I think you want joriki's answer.2012-09-27
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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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    Ok joriki, now I'll edit my post to make the question more difficult and more realistic for what I want to do. I'll suppose this time that each value xi is min distance between the corresponding item ei and the items that are already in E' !2012-09-27
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    @user995434: Please consider whether it's best to edit this question, or perhaps better to ask a new one. Changing the question too far away from the original question makes it difficult to follow the development of comments and answers and sometimes tends to confuse things.2012-09-27
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    Well ok, I've edited the post (see edit2), but ok, I'll make it a new question later if it is better.2012-09-27
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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

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