If you toss an even number of coins, what is the probability of 50% head and 50% tail? If I toss an even number of coins, how can I calculate the probability to obtain head or tail?
This question is different from the other because I can fling the coin a different number of times but the number is always even.
What is the formula to compute the probability?
 A: Assume you toss the coin $2n$ times, then the total number of possible results is $2^{2n}$. How many of those have the same number of heads and tails? You need $n$ heads and $n$ tails, so to count this you choose $n$ spots for heads and you put tails in all the other spots. This can be done in $\binom{2n}{n}$ ways.
In this way you get $$P(\textrm{"same number of heads and tails with $2n$ flips"})=\frac{\binom{2n}{n}}{2^{2n}}$$
EDIT:
If you use Stirling's Approximation, you can get the asymptotic behavior of the probability, as $\binom{2n}{n}\sim \frac{4^n}{\sqrt{\pi n}}$ you get
$$P(\textrm{"same number of heads and tails with $2n$ flips"})\sim \frac{1}{\sqrt{\pi n}}$$
A: We throw 2n coins, what is the probability we get n heads and n tails?
In how many ways could this happen? $\binom{2n}{n}=\frac{2n!}{n!^2}$
What is the total number of throws possible? $2^{2n}$
Since all ordered scenarios are equally likely then the probability?
$\dfrac{\binom{2n}{n}}{2^{2n}}$
A: For $2n$ coins it is 
$${2n \choose n}{1\over 4^n}.$$
If you use Stirling's formula, it is asymptotic to $1\over \sqrt{\pi n}$.
A: Probability of getting exactly $n$ heads fron $2n$ throws of fair coin = $\frac{(2n)!}{n!n!} 0.5^{2n}$
http://en.wikipedia.org/wiki/Binomial_distribution#Probability_mass_function
A: If the number of coins is $2n$, then the probability of flipping exactly half heads is $$2^{-2n}\binom{2n}{n},$$ where   $\binom{2n}{n} = \frac{(2n)!}{(n!)^2}$.
