I'm open to suggestions for better erasure coding storage ratio :)
Based on https://en.wikipedia.org/wiki/Reed%E2%80%93Solomon_error_cor... which gives (n=1 - k=3)/2 is 1 as the answer to the question of the number of missing symbols that RS(1,3) can recover which means that RS(1,3) is not a worse way of storing four copies but a worse way of storing 2 copies. It takes the space of four copies to store what only has the redundancy of two copies.