<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] Yet another wording for the same argument.  See if this thinking/wording is convincing.
$P_m = $ prob of the first $1$ occurring on the $m$-th roll $= (\frac 56)^{m-1}\cdot \frac 16$.
Rolling the first one on the $m$th roll and rolling the first one on the $k$th roll are mutually exclusive events.  So
$P_{m\lor k} =$ the probability of rolling the first $1$ occurring on the $m$th roll OR on the $k$th roll $= P_m + P_k$.
So the probability of rolling the first $1$ occurring on one of the odd rolls is:
$P_{odd} = \sum\limits_{m\text{ is odd}} P_m = $
$\sum\limits_{k = 0}^{\infty} P_{2k + 1} = $
$\sum\limits_{k = 0}^{\infty} (\frac 56)^{2k}\cdot \frac 16 =$
$\frac 16(\sum\limits_{k=1}^{\infty}[(\frac 56)^2]^k) = $
$\frac 16(\sum\limits_{k=1}^{\infty}(\frac {25}{36})^k) = $
(.... and as $|\frac {25}{36}|<1$ and as for all $|a|< 1$ we know $1 + a + a^2 + a^3 + .... = \frac 1{1-a}$ we have.....)
$=\frac 16( \frac 1{1- \frac {25}{36}})=\frac 16(\frac 1{\frac {11}{36}})=\frac 16\frac {36}{11}= \frac 6{11}$
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