Condition probability distributions: Two people flipping fair coins Suppose that two people are playing a game where they each flip a fair coin 100 times. The winner of this game is the person who has flipped the most heads. 
What is the expected number of heads flipped by the winner?
I understand that in general the probability of a given number of heads flipped will be given by the binomial distribution and we can approximate it using a normal distribution. On average, we expect them to both flip around the same number of heads, but conditional on the fact that there will be a winner, we should expect the number of heads of the winner to be slightly above 50. How does one get the distribution of the winning player from the initial distribution?
 A: To approach the problem via normal approximations, let's first think about normal distributions in general
to understand the distribution of the maximum we assume that $X_1,X_2$ are independently distributed as normal variables with mean $\mu$ and standard deviation $\sigma$.  We see that $$P(\max (X_1,X_2)<\mu+t\sigma)=P(X_1<\mu+t\sigma)\times P(X_2<\mu+t\sigma)=\Phi(t)^2$$
Where $\Phi$ denotes the standardized normal cdf.  Differentiating yields $$E[max]=\mu +\sigma\int_{-\infty}^{\infty}t\frac d{dt}\Phi(t)^2dt=\mu+\frac {\sigma}{\sqrt {\pi}} $$
In your case, you want $\mu=50$ and $\sigma =5$.  This approximation gives $\fbox {52.8209479}$.
A few remarks:  
As expected, this answer is quite similar to the answer obtained using the more exact method employed by @heropup.
Using the normal lets us ignore the question of ties, which are pretty low probability for large numbers of tosses anyway.  
There is a simple closed formula for the expectation of the max of three such variables as well, but above three I am not aware of a pleasant computation for the relevant integral.
A: $\sum_\limits{x=0}^{99} P(x)\sum_\limits{y=x+1}^{100} P(y) = P(Y>X)$
$(\sum_\limits{x=0}^{99} P(x)\sum_\limits{y=x+1}^{100} P(y))y = E[y|y>x]P(y>x) = \frac 12 E[Y]$
I am having a hard time remembering exactly why this is....more later...
$\frac {E[Y]}{2P(Y>X)} = \frac {E[Y]}{1-P(X=Y)} = \frac{50}{1-0.0563} = 52.98$
If we relax the condition to $y\ge x$
$E[Y|y=x] = E[Y]\\
E[Y|y\ge x] = E[Y] (1 + P(Y=X)) = 50 (1.0563) = 52.82$ 
