An imperfect method by which to acquire hex colors 'manually' instead of relying on some automatic process. Requires three blank d20s, each representing the red, green, or blue of a hex color code. Two faces are reserved for 0 (00) and 255 (FF), nine faces are increments of 13 up to 130 (excluding 13 itself), and the last nine faces are whatever value happens to equal 255 when summed with their opposing face. It's possible to make the distribution a little more equal, but in practice the difference would be virtually imperceptible anyway. It's 8000 colors, more than enough! The actual perfect solution is to have 6 d16s labelled with 0-F, two per channel, allowing all 16,777,216 colors to be made. However: blank d20s are cheap, nobody seems to make blank d16s, and I'm not paying 10 dollars to buy 6 of those pre-hexadecimalized d16s. I mean, technically you could just roll a single die 6 times whenever you need a color, but i guarantee that isn't nearly as fun as getting one color per roll. Might be interesting to make d4-12s with hex values as well to encourage even less precision. Distribution of numbers stolen from this. If it seems to you that green is overrepresented, consider these as 20x20 slices of a 20x20x20 cube where the topmost face is entirely greenless, the leftmost is entirely blueless, and the farthest is entirely redless.