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Let $X_1,\ldots,X_n$ be indepedent Bin(n,1/2), and let M be their minimum. What is $E[M]$? (or an upper bound for it).

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    What have you got? Where are you stuck. People at this site will help if you show some effort into answering your own question. As of now, your question is not well presented (what is 'average lowest value', do you mean 'average of lowest value', what is 'random binomial experience'). Hint: You are looking for order statistics. – 2017-02-17
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    I know that it can be written as $\sum_{m=0}^n P(M>m)$ where $P(M>m)=P(X_i>m)^n$ but it does not have a closed form solution so I was wondering if there is an upper bound for that – 2017-02-17

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Its easy to write a summation for it:

$P (\min_i X_i >m ) = P(X_1,\ldots,X_n >m) = P(X_1 >m) P(X_2 >m) \ldots P(X_n >m)$ when $X_1,\ldots,X_n$ are independent.

In the i.i.d. case, $P(\min_i X_i >m) = (P(X_1 >m))^n$. Now, you can use the fact that $\min_i X_i$ is non-negative integer valued to note that $E[\min_i X_i] = \sum_{m >0} P[\min_i X_i > m] = \sum_{m=0}^n (P(X_1 > m))^n$.

If you want to upper bound it, substitute a binomial tail bound for $P(X_1 > m)$.

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    Thanks. The problem is that is there a tail that results in a closed form expectation? – 2017-02-17
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    And also I am more interested in having a general bound (not necessarily for the tail). – 2017-02-17