This is my first question so apologies if its unclear/vague.
There exists a contest with me in it, and $5$ others, thus $6$ people in total, along with $5$ prizes.
A person can only win one prize, and once they do, they're out of the contest. Winners are chosen at random.
So, what is the chance of me winning at a prize?
My initial thought was: $\frac16$ chance initially, then $\frac15$ if I don't win the first time, $\frac14$ if I don't win the second, ect. as people are removed once they win, resulting in $\frac16+\frac15+\frac14+\frac13+\frac12 = 1.45$ which is $145$%.
This is greater than $100$%, and obviously I am not guaranteed to win as I can be the $1$ loser, so how do you find the correct probability?
Thanks!
Edit: All of you have been extremely helpful. Thank you!
Best Answer
Well, if we are choosing $5$ winners at random out of $6$ people, then you have a $\dfrac{5}{6}$ probability of winning.
However, your approach is correct - we can sum the individual probabilities to get the same result, but we have to be cautious. Consider the first two prizes given. As you say, we have a $\dfrac{1}{6}$ probability of winning the first prize, and then a $\dfrac{1}{5}$ chance of winning the second prize if we don't win the first prize. We need to take into account this part - you only have a $\dfrac{5}{6}$ chance to get to the second round in the first place.
Using your method, this would give us a total probability of
$$\frac{1}{6}+\frac{5}{6}\left(\frac{1}{5}\right)+\frac{5}{6}\left(\frac{4}{5}\right)\left(\frac{1}{4}\right)+\frac{5}{6}\left(\frac{4}{5}\right)\left(\frac{3}{4}\right)\left(\frac{1}{3}\right)+\frac{5}{6}\left(\frac{4}{5}\right)\left(\frac{3}{4}\right)\left(\frac{2}{3}\right)\left(\frac{1}{2}\right) \\=\frac{1}{6}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}=\frac{5}{6}$$