What is the probability of getting equal numbers of heads and tails?

Question

I'm doing my hw, there is a question that I am not sure if I'm correct.
Here is the question

A fair coin is thrown repeatedly.
What is the probability that on the $n$ th throw, the numbers of heads and tails to date are equal?

My answer is as follows:
$ \mathrm{The\ required\ situation\ holds\ only\ when} \ n\ \mathrm{is\ even.}\\ \mathrm{Let}\ n=2k,\ \mathrm{then\ the\ required\ probability} = {2k \choose k} \left( \frac{1}{2} \right)^{k} \left( \frac{1}{2} \right)^{k} = {2k \choose k} \left( \frac{1}{2} \right)^{2k} \\ \mathrm{And\ I \ searched\ this\ question\ online\ and\ found\ that\ someone\ said\ that\ the \ probability\ is} \left( \frac{n}{2} \right)\left( \frac{1}{2} \right)^{n} $ Here is the link: https://www.algebra.com/algebra/homework/Probability-and-statistics/Probability-and-statistics.faq.question.779221.html

I also tried to expand ${2k \choose k}$ but failed to get the expression $\left( \frac{n}{2} \right)$

May I know whether I am correct or where did I do it wrongly, thanks.

Answer 1

The linked answer is wrong even for $n=2$. For $n=2$ we have $HT$ and $TH$ as the only possible winning paths, so the answer is $\frac 12$. But $\frac 22\times \left( \frac 12\right)^2=\frac 14$.

Worth noting that the answer you provide correctly gives $$\binom 21\times \left( \frac 12\right)^2=\frac 12$$

Indeed, your answer is correct for all (even) $n$.

Answer 2

Firstly, I think you should also give the (trivial) solution for the case when $n$ is odd. For the case where $n$ is even your result is correct and the linked answer is incorrect. Presumably what that person meant to say was ${2k \choose k} = {n \choose n/2}$, because the assertion that ${2k \choose k} = \tfrac{n}{2}$ is not generally true.