Consider \(n\) is even and coin is flipped \(n\) times.
Let us assume \(X\) is the number of heads that appear, then \(X\) be a binomial random variable with parameters \(\left( {n\;,\;\frac{n}{2}} \right)\).
The probability of an equal number of heads and tails is given as:
\(\begin{aligned}{}P\left( {X = \frac{n}{2}} \right) &= C\left( {n,\frac{n}{2}} \right){\left( {\frac{1}{2}} \right)^{\frac{n}{2}}}{\left( {\frac{1}{2}} \right)^{n - \frac{n}{2}}}\\P\left( {X = \frac{n}{2}} \right) &= C\left( {n,\frac{n}{2}} \right){\left( {\frac{1}{2}} \right)^{\frac{n}{2}}}{\left( {\frac{1}{2}} \right)^{\frac{n}{2}}}\\P\left( {X = \frac{n}{2}} \right) &= C\left( {n,\frac{n}{2}} \right){\left( {\frac{1}{2}} \right)^n}\end{aligned}\)
Therefore, the probability that when a fair coin is flipped \(n\) times an equal number of heads and tails appear for even \(n\) is \(C\left( {n,\frac{n}{2}} \right){\left( {\frac{1}{2}} \right)^n}\).
Consider \(n\) is odd and coin is flipped \(n\) times.
There is no chance to get equal number of heads and tails.
Therefore, the probability for odd \(n\) of an equal number of heads and tails appear is zero.
