Dice Rolling To Infinity

Take a look at how you can achieve odd $X.$ $O$ represents an odd roll and $E$ represents an even roll.

By definition, the sequence of rolls will be $O,O,\ldots,O, E$ for some number of $O'$ s. $E$ and $O$ have equal probabilities of $\frac{1}{2}.$ Since there is only $1$ way to achieve a certain value of $X$ for any given $X$ (it MUST be consecutive $O'$ s ending with an $E$ ), then the probability of achieving a given value of $X$ is $2^X.$ If $X$ is odd, then $X\in\{1,3,5\ldots\}.$

The probability is equal to the following.

$\begin{aligned} \dfrac{1}{2^1}+\dfrac{1}{2^3}+\dfrac{1}{2^5}+\ldots&=\dfrac{1}{2}+\dfrac{1}{8}+\dfrac{1}{32}+\ldots\\ &=\sum_{i=0}^\infty\left(\dfrac{1}{2}\right)^{2i+1}\\ &=\sum_{j=0}^\infty\dfrac{1}{2}\times\left(\dfrac{1}{4}\right)^j\\ &=\dfrac{\frac{1}{2}}{1-\frac{1}{4}}\\ &=\dfrac{\frac{1}{2}}{\frac{3}{4}}=\boxed{\dfrac{2}{3}} \end{aligned}$