X is number of (fair) coin flips needed to land m heads OR m tails. m is arbitrary natural number. Delfine CDF of X. (in It's simplest form)

Here is another way to get to the result that avoids using a computer.

Since $X$ is integer valued, it suffices to compute the CDF $F(y)$ for $y$ integer. If desired to extend to the reals, one may set $F(y) = F(\lfloor y \rfloor)$, where $\lfloor \cdot \rfloor$ denotes the floor function.

The CDF $F(y)$ is obviously $0$ for $y < m$, since the number of coin flips to land either $m$ tails or $m$ heads is at least $m$.

As such, $P(X \geq m) = 1$ and so $F(y) = P(X \leq y) = 1 - P(X > y) \geq 1 - P(X \geq m) = 0$, as claimed.

Next, for $y \geq 2m-1$, the CDF is $0$. Indeed, among $2m-1$ coin flips, by the Pidgeonhole principle there must be either $m$ heads or $m$ tails. Thus $P(X \geq 2m) = 0$ and so $P(X \leq y) = 1 - P(X \geq y) \leq 1 - P(X \geq 2m-1) = 0$.

For $m \leq y \leq 2m-2$, we shall first compute the probability $P(X > y)$. This can be done as follows - write $H$ for the number of heads in $y$ coin flips, and $T$ for the number of tails. Note that by definition $X > y$ if and only if $H < m$ and $T < m$.

But $T = y - H$, and so the second inequality becomes $H > y - m$. Thus we have $X > y$ if and only if

$$y - m < H < m.$$
Writing $F_H$ for the CDF of H, and noting that $H$ is a binomial random variable with success probability $0.5$ and number of trials $y$, we may compute

$$P( y - m < H < m)= P(H \leq m-1) - P(H \leq y - m)$$

$$= F_H (m-1) - F_H (y - m) $$

$$=\sum_{i = y - m + 1}^{m-1} \binom{y}{i} \left (\frac{1}{2} \right )^y$$ $$= \left (\frac{1}{2} \right )^y \sum_{i = y - m + 1}^{m-1} \binom{y}{i} $$
for which there is no further closed form.

Thus we finally have, for $m \leq y \leq 2m - 2$,

$$F(y) = P (X \leq y)$$

$$= 1 - P(X > y)$$

$$ = 1 - P(y - m < H < m)$$

$$ = 1 - \left (\frac{1}{2} \right )^y \sum_{i = y - m + 1}^{m-1} \binom{y}{i}.$$

Thus the full CDF is

$$F(y) =$$

$$0 \text{ if } y < m,$$ $$ 1 - \left (\frac{1}{2} \right )^y \sum_{i = y - m + 1}^{m-1} \binom{y}{i} \text{ if } \leq y \leq 2m-2,$$ $$ 1 \text{ if } y \geq 2m - 1.$$