<s>[INST] 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?
 [/INST] Let $p$ be the required probability. We condition on the result of the first roll.
If we get a $1$ or a $5$, we are finished. If not, we need to roll something other than $1$ or $5$ the first and second time. Given that this happened, the conditional probability we then ultimately get a first $1$ or $5$ on an odd numbered roll is $p$. Thus
$$p=\frac{1}{3}+\frac{4}{9}p.$$
Solve for $p$.
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