I figure it a little different. Starts out the same, each egg from het pied X het pied has a 1/4 chance of being a homozygous pied so a 3/4 chance of not being visible pied. Then to find the odds of all three eggs not being pied you multiple the chance of each not being one together, so (3/4)^3 = 27/64 (so instead of multiplying by 3 you multiply by it's self three times, or how ever many eggs you have times). So if 27/64 is your odds of getting none then 37/64 is your chance of getting at least one. So I'm showing 57.8% chance of getting at least one pied from three eggs from het X het. So a good bit more likely to get at least one than to get none if you hatch all three eggs. However, as long as you are sure the parents are both hets all normal offspring are 66% chance hets.
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