What would be the value of this joint probability? 
We roll a die until we get 1 or 5. What is the probability that we will make odd number of rolls?

Let, A = 1 or 5, and, B = otherwise.
Therefore, $$\Omega = \{A, BA, BBA, BBBA, ...\}$$
So, $$P(A) = P(1) + P(5) = 2 \times \frac{1}{6} = \frac{1}{3}$$
$$P(BA) = P(B \cap A) = P(B) * P(A) = \frac{4}{6} \times \frac{1}{3} = \frac{2}{3} \times \frac{1}{3} $$
Likewise, $$P(B........A) = { (\frac{2}{3} ) }^{n-1} \times \frac{1}{3}$$
Therefore, $P($Odd number of rolls$) = \frac{1}{3} + \frac{1}{3} . (\frac{2}{3})^{2} + \frac{1}{3} . (\frac{2}{3})^{4}+... ... = \frac {\frac{1}{3}}{1 - (\frac{2}{3})^2} = \frac{3}{5}$

My question is, is there any alternative way to calculate this?

Feel free to correct me if I'm wrong, but I think your final sum should be $ \frac{1/3}{1-(2/3)^2} = \frac{3}{5} $
Other than that, I think you have the simplest method.
Edit: I realise this is probably just a typo
