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.
Best Answer
Avoid summing of infinite series. Argue as follows instead: Call it a success, if you obtain the first $1$ after an odd number of throws. Let $p$ be the probability of a success. Since the probability that the game never ends is $0$, with probability $1-p$ it then ends after an even number of throws. Conditioning on whether the first throw is a $1$ or not we therefore obtain the following equation for $p$: $$p={1\over 6}\cdot1+{5\over6}\cdot(1-p)\ .$$ It follows that ${\displaystyle p={6\over11}}$.