Once again, Kitkat has saved us from an oversight that I made concerning genetics. One could be 100% sure of having a het (AKA 100% het) by crossing a (homozygous) recessive individual (like an albino) with a 100% normal individual (homozygous dominant). Then all offspring would be 100% hets. One could also cross a (homozygous) recessive individual with a confirmed het. Then 50% of the offspring would be recessive (like albinos) or else they would be 100% het and have a normal appearance (phenotype).
If, however, you breed two hets, you get one aberrant homozygous recessive (e.g. albino), two hets, and one homozygous dominant, but the hets and the homozygous dominant all look alike. All one would know is that, statistically speaking, two of the three wild-type offspring would likely be heterozygotes, giving a 66% chance that they are hets.
A 100% het is certain to have one recessive allele (by definition). Thanks again, Kitkat
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In order to "design" a quad het, one would probably use two snakes that are double hets for different genes. If one crossed an anery x snow (double het) with an albino x (something else), they would produce a quad het, but this doesn't account for having a melanistic garter in the picture, unless it's unrelated. Also, as Kitkat mentioned, the wild-type is basically the opposite of anery, snow, albino, or whatever, and it is implied (but not counted) when you say you have hets.
Rick