Birthday Prize Analysis
This front matter follows the pattern of other files in the blog directory, includes math support for the calculations, and reflects the content about birthday prize money analysis with mathematical calculations.
| Years | Prize | Recipient |
|---|---|---|
| 21 | 50,20 | Aarav, Jisu |
| 22 | 50,100 | Erick, Sarro |
| 23 | 100,150 | Unclaimed, Aarav |
| 24 | 300 | Unclaimed |
rule: take the most expensive offering and then double it: 21: 50,20 | 50 22: 50, 100 | 100 23: 100, 150 |
200
started at 50 for top prize then 100 for top prize then 150 for top prize now should have been 200 for top prize, but I’ve collapsed it with a second place prize money of 100 which you can somewhat consider to be the unclaimed money from last year.
I wander how I structure the remaining years to be most interesting. and poetic as possible.
if this years money goes unclaimed, then I probably should offer just a single prize pool in 2026?
I might go with 250 as per usual
300, 350
Tn = 300 + 50*n
# use list comprehensions! that is the python way :)
years = [300 + 50*i for i in range(100)]
print(years)
birthdays = [n+24 for n in range(100)]
print(list(zip(birthdays, years)))
print(f"sum of first place prizes: {sum(years)}")
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