What is the probability of ending in battle royale Top 5? In a battle royale, there are 100 players. I'm one of those players. For the exercise, we suppose that each player have same game skills.
What is my chance of ending in the Top 5 ?
This is what I tried:
Let $N$ be the number of possible top 5 : $N=nCr(100,5)$
Let $M$ be the number of top 5 I'm in : $M=99*98*97*96*1$
My chances of being in Top 5 are : $M/N = 1.2$
Obviously, this is wrong. What am I missing ?
 A: First off, your idea is great. Assume each outcome is equally likely, count the number of good outcomes and the number of possible outcomes, and divide them by one another.
But as you can tell, something went awry. The exact thing that went awry is that you count the number of top5's differently in the two cases. For the number of possible top5's, you are counting the number of 5-man-groups. That's fine.
However, when you count the number of top5's with you in them, what you're actually counting there is the number of distinct top5's, including internal ranking, with you at fifth place. There are 99 possible winners, and after that there are 98 possible runners-up, and so on. So you're counting an outcome where player A wins and player B comes second as different from an outcome where B wins and A comes second, and the ones where you win aren't counted at all.
The correct way to count the number of top5's with you in them is to do what you did for the total number of top5's, except you reserve one seat for you. Then there are 99 players left for the remaining four spots in the top5, so the answer is $nCr(99, 4)$.
However, there is an easier way. Consider the entire ranking list for the game. Each possible ranking (among the $100!$ possible ones) is equally likely. Particularily, that means that you are equally likely to place in any one of the $100$ spots. The probability that you happen to place in one of the top 5 spots is therefore $\frac5{100}$.
