Expected number of coins that come up tails on second round of flipping 
I flip $20$ coins. I then discard the coins that come up heads, and re-flip those that come up tails. What is the expected number of coins that again come up tails?

Here's what I got:
What's the probability of getting $1$ tails on the first pass? It's ${{\binom{20}{1}}\over{2^{20}}}$. What's the expected number of tails I get from flipping $1$ coin? It's ${{1 \binom{1}{1}}\over{2^1}}$.
What's the probability of getting $2$ tails on the first pass? It's ${{\binom{20}{2}}\over{2^{20}}}$. What's the expected number of tails I get from flipping $2$ coins? It's ${{1 \binom{2}{1} + 2\binom{2}{2}}\over{2^2}}$.
What's the probability of getting $3$ tails on the first pass? It's ${{\binom{20}{3}}\over{2^{20}}}$. What's the expected number of tails I get from flipping $2$ coins? It's ${{1 \binom{3}{1} + 2\binom{3}{2} + 3\binom{3}{3}}\over{2^3}}$.
Continuing in this fashion, the expected number of coins that come up tails with the second round of flipping should be$$\sum_{n = 1}^{20} \left( {{\binom{20}{n}}\over{2^{20}}} \left(\sum_{i = 1}^n {{i\binom{n}{i}}\over{2^n}}\right)\right) = \sum_{n = 1}^{20}\left({{\binom{20}{n}}\over{2^{n + 20}}} \sum_{i = 1}^n i\binom{n}{i}\right).$$However, I'm not sure how to further simplify this expression. Can anybody help?
Update: Okay, I just realized as I submitted my question to Math Stack Exchange a much easier to solve my problem. Each coin has a ${1\over2}\left({1\over2}\right) = {1\over4}$ probability of coming up tails twice in a row, and since we have $20$ coins, the expected number of coins that again come up tails is $20\left({1\over4}\right) = 5$.
But I am still wondering if anyone can help me evaluate that sum!
 A: Honestly, I read only the first few lines of your work. It might even be correct, but it appears needlessly complicated.
Suppose any coin is tossed twice. The probability of its falling tails on both tosses is $(\frac {1}{2})(\frac {1}{2}) = \frac {1}{4}$.
Now suppose each of the 20 coins is tossed twice. How many of them fall tails on both tosses? That is a binomial probability, so the expectation is simply the number of trials times probability for each trial.
$(20)(\frac {1}{4}) = 5$
This is not exactly what you asked, but it is equivalent. For the coins that you would have discarded, the outcome of the second toss is neither here nor there. They are already failures; it changes nothing to keep tossing them.
A: Evaluation of the sum: note that $i\binom{n}{i}=n\binom{n-1}{i-1}$ and $\binom{20}{n}n=20\binom{19}{n-1}$, therefore
$$\sum_{n = 1}^{20}\left({{\binom{20}{n}}\over{2^{n + 20}}} \sum_{i = 1}^n i\binom{n}{i}\right)=\sum_{n = 1}^{20}\left({{\binom{20}{n}}\over{2^{n + 20}}}\, n\sum_{i = 1}^n \binom{n-1}{i-1}\right)=20\sum_{n = 1}^{20}{{\binom{19}{n-1}}\over{2^{21}}}=20\frac{2^{19}}{2^{21}}=5.$$
