My question is regarding the problem in this video: https://www.youtube.com/watch?v=BcIDeZ2nLf0&list=PLADLdjl79_mhdmhdxA1Gk8sklGEWJkEKq
You are tossing 300 pieces of crumpled paper into a bin. You miss the first, and make the second. Subsequently, your chance of making each shot is your cumulative win-rate. What is the probability that you will make exactly 298 shots?
Before looking at the solution, here is how I set up this problem. I thought, we need to calculate the probability that exactly one of the remaining 298 shots is a miss: so, we can add the probabilities for the following events: third shot is missed and the rest are hits, fourth shot is missed and the rest are hits, ..., 300th shot is missed and the rest are hits.
Interestingly, I find that each of the above events has equal probability, $\frac{1}{(299)(300)}$ (please correct me if I am wrong). Note, that there are 298 such events (300-3+1). So, I conclude that the answer should be:
$$ \sum_{i=1}^{298} \frac{1}{(299)(300)}=\frac{298}{(299)(300)} $$
However, the answer is $\frac{1}{299}$.
Please help me find where I am going wrong. Also, if you could explain how to solve this problem from a Polya's urn approach (thinking about green balls as hits and red balls as misses, and starting with one of each ball in an urn), that would be great! I have an understanding of Markov chains.