Why does light($color,50%) parse to white?

Question

This sass snippet:

$red: #f00

$lightred: lighten($red, 50%)

body
  background: $lightred

Is parsed to:

body {
  background: white; }

If you instead do 45% instead of 50%, it seems to work as expected:

$lightred2: lighten($red, 45%)
...
background-color: #ffe5e5

Why is this turning into white instead of the expected 50% of red?

You can see it in action here http://tinkerbin.com/OefelPoi

(Note - the save function seems to have a bug, on the css area, select Sass Old Syntax again, and rerun)

Update -

Ends up what I wanted was to mix in white with the original color to get a tinted shade:

mix($color,white, 10%)

Answer 1

Because lighten function is described on HSL colors and your red color #f90 is translated as hsl(0, 100%, 50%)

so lighten($lightred, 50%) is equal to hsl(0, 0, 100%), or white