What is more probable for a number? being the smallest element in the set or being the median. let's say there is a finite set of random integers with odd cardinality, pick any number from the set. What is more probable for a number ?
The answers I have received so far say that it is equiprobable but here is my way of solving 
it please correct me if I am wrong.
lets say this is my set
            { a1, a2, a3, a4, a5, ......, an } 
        and take two other empty set call one of it 'smaller' and other one 'larger' 
        I randomly chose an element 'ak' from the set and started comparing it with every element from the beginning.
        if the first element is smaller put it in the 'smaller' set,if larger, then in 'larger' set(it can fall in either set with equal probability), then compare the next one and do it for all elements
        now the probability that both set will contain the same elements is larger than where one is empty.
 A: I can think of two ways to construe the question, and either way, being the median and being the minimum are equally probable.
One way is that you have a list of numbers, e.g. $1,2,3,4,5$, and you choose one at random, each having an equal probability of being chosen.  The minimum is $1$ and the median is $3$, and each has probability $1/5$ of being chosen, so they're equally probable.
Another way is this:  You're choosing from some probabilty distribution, and let's take it to be a continuous distribution so that you won't get the same number twice.  Let $X_1$ the first number you get.  Then independently pick another and call it $X_2$.  And so on up to $X_n$, where $n$ is some odd number. What is the probability that the first number chosen ends up being the smallest, and what it the probability that the first number chosen is the median?  And similarly the second number.  And the third, etc.
But any of the $n$ indices, $1,2,3,\ldots,n$, is equally likely to be the smallest, and any is equally likely to be the median, and so again the two probabilities are equal, each $1/n$.
A: If the distribution is uniform then the probabilities are the same.  Otherwise, it depends on the distribution.  A sample from a distribution such as a Bernoulli whose values are concentrated at the median will tend to have more duplicate values there than at the extremes, so an element chosen randomly from the sample is more likely to be equal to the median.  Alternatively, we could define a distribution where $0$ and $1$ each occur $\frac{2}{5}$ of the time, and $\frac{1}{2}$ occurs the other $\frac{1}{5}$ of the time.  In this case, the median of a large sample is almost certainly $\frac{1}{2}$, but there will be twice as many occurrences as $0$.
In the above cases we are talking about a multiset of random variables chosen independently from the same (discrete) distribution.  But if by $set$ you mean the values are distinct, then there is exactly one median and one minimum so the probabilities are equal.
