On a trip to Africa the researcher Alison notices that zebras with an even amount of stripes have double the probability to be seen than zebras with an odd amount of stripes. Let $E_n$ denote the event that a specific zebra has exactly $n$ stripes. Create a probability space that assigns a strict positive probability to $E_n$ and that suffices Alisons observation.
To keep one solution short (using abbreviated notation at some points), let $\Omega=\mathbb N,E_n=\{n\}$ and $F$ be the event that a zebra has any even amount of stripes and $\Pr[F]+\Pr[\overline F]=1$ yields
$$\frac{1}{2}\cdot \Pr[F] = \frac{1}{2}\sum_{n=1}^\infty\Pr[2n] = \sum_{n=1}^\infty\Pr[2n-1] = \Pr[\overline F]\implies \Pr[F]=\frac{2}{3},\Pr[\overline F]=\frac{1}{3}.$$
One possible pdf would be
$$\Pr[n]=\begin{cases}\frac{2}{3}\cdot\left(\frac{1}{2}\right)^{n/2},&n\in F,\\\frac{1}{3}\cdot\left(\frac{1}{2}\right)^{(n+1)/2},&n\not\in F.\end{cases}$$
This is true since
$$\Pr[F] = \sum_{n=1}^\infty\Pr[2n] = \sum_{n=1}^\infty\frac{2}{3}\cdot\left(\frac{1}{2}\right)^{(2n)/2} = \frac{2}{3}\cdot\sum_{n=1}^\infty\left(\frac{1}{2}\right)^{n} = \frac{2}{3}$$
as well as
$$\Pr[\overline F] = \sum_{n=1}^\infty\Pr[2n-1] = \sum_{n=1}^\infty\frac{1}{3}\cdot\left(\frac{1}{2}\right)^{((2n-1)+1)/2} = \frac{1}{3}\cdot\sum_{n=1}^\infty\left(\frac{1}{2}\right)^{n} = \frac{1}{3}.$$
Now I stumbled upon another attempt which suggest
$$\Pr[n]=\begin{cases}\left(\frac{1}{2}\right)^{n-1},&n\text{ even},\\\left(\frac{1}{2}\right)^{n+1},&n\text{ odd}.\end{cases}$$
being a correct pdf. Is this reasonable to any extent?
Yes, this is also reasonable. The generalisation of these two solutions is
$$ \textsf{Pr}[n]=\begin{cases}\frac23(1-q)q^{(n-2)/2}&n\text{ even}\;,\\\frac13(1-q)q^{(n-1)/2}&n\text{ odd}\;,\end{cases} $$
with $q=\frac12$ for your solution and $q=\frac14$ for the other solution.