I am in a contest with 200 total participants. There will be 5 total scores to determine the top 20 finishers. Each is initially ranked according to their ranking from a previous contest. That previous ranking is your first score. Mine is 193. The best starting rank is 1. The worst starting rank is 200. There will be 4 additional ranked contests that contribute to my total score. My total score will be the sum of my ranked performances from each of these 4 contests and my previous score of 193. So, if I got first place in each of the 4 additional contests, my total score would be 197. The 20 lowest scores overall move on to the next round. What is the probability that I can make the top 20 after the next 4 contests with my starting score of 193 relative to the other contestants ranging from 1-200 in their initial scores?
2026-03-31 16:26:53.1774974413
Odds of placing top 20 in a contest of 200 participants out of 5 contests with the first score being 193
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Let's assume the contest rankings are ransom and independent - a very bold assumption! Then again, this boldness allows us to make very crude approximations in what follows.
For the sum of five uniform random vars, it is a bit of a stretch, but let us just do as if we can approximate the sum as a normally distributed variable. We use the mean $\frac{1+200}2\approx 100$ and variance $\frac1{12}(200-1)^2\approx 3333$ for a single contest, hence mean $\approx 500$ and standard deviation of $\sqrt{5\cdot 3333}\approx 129$ for the sum of five contests. The top ten percent ought to exceed the mean by at least $1.28$ standard deviations, i.e., the winners will have rank sums up to $500-1.28\cdot 129\approx 335$.
To be in that range, your rank sum of four contests must be $<142$. Similar to above, the rank sum of four contests has mean $\approx 400$ and std.dev. $\approx 115$. Thus you need to be better by $\frac{400-142}{115}\approx 2.2$ sigmas, which has probability $\approx 1.4\%$.
As mentioned, this is a very rough estimate, but shows how hard it is to raise to the top be chance.