What if there's a voting system where one needs 43% or more to win? (explained)

by vitarnixofntrnton 2/22/2024, 2:24 PMwith 7 comments

In my previous post, I made a pseudo random number generator (licensed under the Public Domain) where every bit is 42.5% random in 10 lines of code and it outputs a byte every clock cycle. Thoughts on this?

Basically, it's a filter against my simple pseudorandom number generator so it keeps narrowly losing and winning for it is an extreme rarity.

To win, one idea gets 43% or more of the votes. If two ideas get 43% or more of the votes, restart. Or it could not restart but something else, am open to changing my mind about the restarting idea.

by pavel_lishinon 2/22/2024, 2:33 PM

> In my previous post, I made a pseudo random number generator

Someone pointed out that what you made is just a function that outputs triangle numbers: https://news.ycombinator.com/item?id=39462194

So, it's nowhere near a pseudo-random number generator; for any given output, it's trivial to guess the next (and previous) numbers.

I also don't understand where your 42.5%/43% numbers are coming from. What does it mean for "every bit is 42.5% random"? How does your code output a byte every clock cycle?

by necovekon 2/22/2024, 2:31 PM

In most electoral systems, winning 43% of the vote is usually sufficient to gather a representative majority.

Even in USA which is highly bipartisan, people vote is not directly linked to electoral votes. In Serbia (and possibly many other countries), there is a census and parties under it have all the votes for them basically ignored (I remember a vote from ~10 years ago with a census of 5% when less than 20% of the voting population resulted in over 67% of parliament seats).