Decreasing an amount of number that person receives by calculating a percentage for each I need to calculate a percentage for each person in my list (100 participants) with certain conditions:


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*We know how many percents to put for first 3 people



1st person - 20%
2nd person - 15%
3rd person - 5%



*We also have a percentage that receives the last 100th participant



100th place - 0.25%

Please, can you help me to calculate the decreasing percentage of other participants? I also have a photo which describes this process. Hope it can help you. Thanks!
https://monosnap.com/file/pykMhyerN4J6s91Cs9AGtOhSAJ1Jqo
 A: On average the remaining people get $\frac{1-0.4025}{96}$. You can give all the same fraction or let their prizes decrease linearly, for example. If you want strictly decreasing prizes, for a fraction $p\in(\frac{96}{239},1)$ you can give all $96$ people a fixed amount of $p\frac{1-0.4025}{96}$ and divide the rest linearly. This implies that you look for extra pieces of magnitude $x$. You give place $99$ $0x$, place $98$ $1x$, etc and place $4$ $95x$. So you need $\sum_{i=0}^{95}i=\frac12\cdot95(95+1)=4560$ extra pieces $x$. Then $x:=(1-p)\frac{1-0.4025}{4560}$. The exact prices for place $4$ to $99$ are then given by, for place $n\in\{4,5,\ldots99\}$
$$p\frac{1-0.4025}{96\cdot2}+(1-p)\frac{1-0.4025}{4560}(99-n).$$
However, different prize schemes might make much more sense since you don't have a linear prize scheme for places $1$ to $3$. The point of this post is that you have to be more specific in what you expect from the prize scheme, except from being decreasing. For every $p\in(\frac{96}{239},1)$, you find a different linear prize scheme that satisfies your conditions.
