The chances for a double recessive from a double het breeding is 1:16...not 1:15. Not a big difference, but a difference all the same
On that note the chances of a double homozygous dominant (not showing either trait) is the same, the chance of being just albino or just patternless is 3:16 (each...of either is 6:16). 1:3 of each of the single recessive animals will be homozygous dominant for the other gene. The chance of being normal homozygous dominant at the patternless gene and het at the albino gene or homozygous dominant at the albino gene and het at the patternless gene would be 2:16 (each...of either is 4:16), and finally the chance of being double het is 4:16. This makes the total ratio of genotypes 1:2:2:4:1:2:1:2:1 (dbl hom dom:hom dom/het:het/hom dom:dbl het:hom rec/dom:hom rec/het: dom/hom rec:het/hom rec:dbl rec). Phenotypically the ratio is 9:3:3:1 (normal:rec 1:rec2:dbl rec).
I have no idea if that makes any sense whatsoever.....lol