What is the probability that the first time that the front and back of a fair coin appear an equal number of times is the $2n$th trial?

What is the probability that the first time that the front and back of a fair coin appear an equal number of times is the $2n$th trial?

The probability that the first time that the number of appearances front and back sides of the coin are equal ($n$ times each) after continuously throwing a fair coin $2n$ times is $${\small \frac{1}{2n-1}\binom{2n}{n}\cdot\frac{1}{2^{2n}}=\frac{1}{2n-1}\cdot\frac{2n\cdot(2n-1)\cdots(n+1)}{n\cdot(n-1)\cdots 2\cdot 1}\cdot\frac{1}{4^n}.}$$ For example, take $n = 3$, that is, throw a fair coin $6$ times continuously. After two trials, it can't be "$1$ front side and $1$ back side" (only $2$ front sides or $2$ back sides); after the $4$th trial, it can't be $2$ front sides and $2$ back sides (only $3$ front sides and $1$ back side, or $1$ front side and $3$ back sides); after the $6$th trial, there are $3$ front sides and $3$ back sides (the number of appearances of the two sides are equal). The probability of this event is $$\frac{1}{6-1}\cdot\frac{6\cdot5\cdot4}{3\cdot2\cdot 1}\cdot\frac{1}{4^3}=0.0625.$$ How to prove the above results?

Solution 1:

Let's consider a sequence of Bernouilli trials, with probability $p$ of cuccess and $q= 1-p$ of failure, and let's represent it as a path with steps $E-N$ on a square lattice.

Now you are looking for the probability of taking a path from $(0,0)$ which means also without crossing it.
So the possible paths are those which, apart from the initial and the final leg, remain within the lower triangle $(1,0),(n,0),(n,n-1)$ or within the symmetrical upper triangle.

It is known that the number of paths within each triangle is given by the Catalan Number $C_{n-1}$.

Therefore the probability$ P(n)$ you are looking for is $$ P(n) = 2p\left( {C_{n - 1} p^{n - 1} q^{n - 1} } \right)q = {2 \over n}\left( \matrix{ 2\left( {n - 1} \right) \cr n - 1 \cr} \right)p^n q^n $$

which for $p=q=1/2$ and for $n=1, 2, \cdots,6$ gives $$ 1/2, \; 1/8, \; 1/16, \; 5/128, \; 7/256, \; 21/1024 $$ which checks with computations.