Number of lottery tickets required to win with at least one with probability of $x$. I was solving some probability and combination problems and I came across this one,
that I couldn't solve. Any tips appreciated.

Chance to win with one lottery ticket is $\frac13$. How many tickets must be purchased for the probability of winning at least with one ticket to be greater than $0.9$?
Given:
Chance to win: $\frac13$
Number of tickets that have to win: $1$
Probability of that happening: $> 0.9$
Need to find number of tickets to buy

 A: So this question deals with a binomial distribution, as we want to know how many tickets must be purchased for $Pr(X \ge 1) > 0.9$, where $X$ is the number of wins. 
So as you already noted $Pr(win) = 1/3$, and since this is  binomial distribution then $X \sim Bin(1/3,n)$ where $n$ is what we are looking for, number of tickets purchased. 
Observe: $Pr(X \ge 1) = 1 - Pr( X = 0) = 1 - (2/3)^n $
So we want $ 1- (2/3)^n > 0.9 $, which will give you
$ n > \frac {ln(0.1)}{ln(2/3)}$, so $ n \ge 6 $.
This is a binomial distribution as we are counting two possible outcomes, and the winning or losing of one ticket does not affect the other tickets. And we may organize the tickets in any way we choose to. 
A: The question is poorly-phrased.
If $\frac13$ of lottery tickets win, then just buy $3$ tickets and the probability to win is $1>0.9$.
The correct phrasing should be that the probability of a lottery ticket to win is $\frac13$.

That being said, simply solve the inequality $[1-(1-\frac13)^n>0.9]$ for $n\in\mathbb{N}$:
$1-(1-\frac13)^n>0.9 \implies (\frac23)^n>0.1 \implies n>\log_{2/3}(0.1) \implies n>5.678 \implies n=6$
