Yep. This idea is beautiful.
- First line: $.== 1 => 1/1 chance to keep this line1.
- Second line: $. == 2 => 1/2 chance to keep line2 (so 1-(1/2)=1/2 to keep line1) (Equiprobability)
- Third line: $. == 3 => 1/3 chance to keep line3 ( so (1-(1/3))=2/3 chance to keep a previous line (line1 or line2) : (2/3)*(1/2)=1/3 for line1 and same thing 1/3 for line2 (Equiprobability)
- (N+1)th line: $. == N+1 => 1/(N+1) chance to keep line(N+1) so (N+1 -1)/(N+1) to keep a previous line, one of all previous N lines (each one has 1/N): for each previous line probability is ((N+1 -1)/(N+1))*(1/N) = 1/(N+1) (Equiprobability again)
