Rolling a die and 99% chance of getting a 6 I am trying a question from Probability Theory by Rota and Baclawski. The question asks "How many times must one roll a die in order to have a 99% chance of rolling a 6".
Now my first attempt was to find this probability like this:
Let $R_n$ be the event that $n^{th}$ roll is a 6. 
\begin{equation}
R_1 = S\\
R_2 = \overline{S}S \\
R_3 = \overline{S}\overline{S}S, etc.
\end{equation}
Then the probability of rolling a 6 will be the sum of these probabilities, upto $n$. I did not take this to be an infinite sum.
\begin{equation}
P(\text{Roll a six}) = R_1 + R_2  + ... + R_n \\
\implies 0.99 =R_1 + R_2  + ... + R_n \\
\implies 0.99 = S(1 + \overline{S} + \overline{S^2} + ... + \overline{S^n}) \\
\end{equation}
I thought for a while and my second approach would be the concept of $independence$, because I could follow the path in the decision tree to get to $n$, whatever it might be.
In this approach, 
\begin{equation}
P(\overline{R_1}\land\overline{R_2}\land\overline{R_3}\land...\land R_n) = P(\overline{R_1})P(\overline{R_2})...P(R_n) \\
\implies 0.99 = (\frac{5}{6}^{n-1}) . \frac{1}{6}
\end{equation}
So my question is which of these (if any) would give me the correct answer?
 A: Presumably, the question is asking about a 99% chance of rolling a 6 at some point in the $n$ rolls.
André's comment is the most straightforward approach. The complement of the event "rolling a 6 [at some point]" is "never rolling a 6," and it is much easier to compute this quantity.

You can complete your first approach to recover the same answer.
Let $p=1/6$ be the probability of rolling a $6$.
With your first method, the geometric series can be rewritten as
$$p(1+(1-p) + (1-p)^2 + \cdots + (1-p)^{n-1}) = p \frac{1-(1-p)^n}{1-(1-p)} = 1-(1-p)^n.$$
For this quantity to be $ \ge 0.99$, you need $(1-p)^n < 0.01$.

Your second approach is flawed: the probability you have computed is the probability that you roll a 6 exactly once and on a particular roll (e.g. the last roll).
A: A dice will have a specific side with exactly $1/6$ probability. Therefore, the probability of not rolling a specific side is $1-1/6=5/6$. You are solving the equation $99/100=1-(5/6)^x$ for $x$.
$e^{1/100}=x \ln(5/6)$, $x=\frac{e^{1/100}}{\ln{5/6}}$ which is simplified to $ \frac{2(log(2)+log(5))}{log(2)+log(3)-log(5)}$, which rounded up is 26 rolls.
A: No solution.  $25$ times comes close.  $26$ times goes above $99$% so no solution to the question as stated.
