<s>[INST] Die is rolled until 1 appears. What is the probability of rolling it odd number of times? Problem: Die is rolled until 1 appears. What is the probability of rolling it odd number of times?  
So, so far I have this:
$\frac{1}{6}$ - this is a probability of rolling "1" on first try
$\frac{5}{6} \cdot \frac{5}{6} \cdot \frac{1}{6}$ - three tries (two times something else than "1" and "1" on the 3rd try
$\frac{5}{6} \cdot \frac{5}{6} \cdot \frac{5}{6} \cdot \frac{5}{6} \cdot
\frac{1}{6}$ - five tries
and so on...
By the looks of it, it seems I could come up with following formula
$$ p = \left( \frac{5}{6}\right)^{n-1} \cdot \frac{1}{6}$$
where p would be a probability for nth try.
However, I cannot think of any way to get probability of all odd number of tries and not sure if I am even on right track here.
 [/INST] $p(1)=\frac 1 6 $ $P(not 1)=\frac 5 6$ 
1 on $1^{st}$ roll: $$\frac 1 6=\frac{5^0}{6^1}$$
1 on $3^{rd}$ roll: $$\frac 5 6 \cdot \frac 5 6  \cdot \frac 1 6 = \frac {5^2} {6^3} $$
1 on $5^{th}$ roll: $$\frac 5 6 \cdot \frac 5 6  \cdot \frac 5 6 \cdot \frac 5 6\cdot \frac 1 6=\frac {5^4}{6^5}$$
keep going upto $\infty$: $$\frac{5^0}{6^1}+\frac{5^2}{6^3}+\frac{5^4}{6^5}+...\infty$$
use infinite GP summation: $$\frac 1 6 \cdot \frac {1}{ (1- \frac {5^2}{6^2})} $$ 

$$=\frac 6 {11} $$

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