x percent chance of winning In a game there is a 3% chance of winning "a prize" on every try.  The game doesn't  list any more information than that.  How many tries will it take before the chance to win the prize is nearly 100%? 
Logically, it feels like the answer is never, since it's always 3%?  Can anyone clarify?
 A: Of course, you have a greater chance of winning with 2 tries than with 1 and greater again with 3.  Naively, you might expect that the chances of winning with 2 tries is 6% but actually it is a bit less.  While the probability is small and the number of tries is also small, this naive logic will be close.  As the numbers get bigger, they will be less close.  With 33 tries, you will not have a 99% chance and with 34 tries, it will not be more than certain. 
The easiest way to calculate the probability of winning at least once in $n$ trials is to first calculate the probability of not winning at all in $n$ trials.  The probability not winning once in your game is $0.97$, the probability of not winning twice is $0.97^2 = 0.9409$ so the probability of winning at least one in two goes is $1 - 0.9409 = 0.0591$.  As I said, slightly less than the naive 6%.  The probability of winning at least once in 33 tries is about 87% and not the naive 99%.  
Why?  Well, the simple answer is that the difference is explained by the probability of winning more than once.  
How many trials to get to nearly 100%?  Well that depends on what you regard as nearly.  It will never be certain, not even after a million tries.  
To convince yourself of this, play with some dice.  Assuming that they are fair, throw one a few times, you will have a $\frac{1}{6}$ probability of getting a $6$.  Now throw two, you will be more likely to get a $6$ but it will be $\frac{11}{36}$ which is slightly less than $\frac{1}{3}$ you might expect.  However, there will be a $\frac{1}{36}$ chance of getting two $6$s.  Throw 6 times and you won't always get a 6.  
A: You can use a geometric distribution to answer this question, since the tries are independent. The number of tries needed to win the prize (denoted $X$) has the following PMF:
$P(X=k)=(1-p)^{k-1}p, \; k \geq 1$, where $p=0.03$. 
Now consider $q_n = \sum_{k=1}^n P(X=k)$, which is the probability of winning in $n$ tries or fewer. The expression simplifies to $q_n = 1-(1-p)^n$. 
You can see that when $n=1$, $q_n=p$, implying you have a probability $p$ of winning (only 3%) if you try just once. As $n \rightarrow \infty$, $q_n \rightarrow 1$. This means that your probability of winning approaches 1 as you keep trying the game. 
Your question is not clearly worded though. You want the probability of winning to be nearly 100%, but "nearly" needs to be defined clearly. For example, if "nearly" means 99.9%, you can find the number of tries required by solving $1-(1-p)^n \geq .999$, and so on. In the example I gave, $n=227$.
