What justification is there for a 50% probability of Hh and 50% probability of HH. It's one of the most common logical fallacies in probability; because you have two possibility, people somehow assume that the probability of each situation must be 50%.

You can, however, justify saying that nearly all people with Huntington's disease are Hh. We know that Huntington's disease is a rare genetic disorder. Let's assume that Huntington's affects 1 person per 100 people (the real figure is about 1 person per 10,000 people of European descent). Therefore, 99% of people have genotype hh. Assuming that the population is in Hardy-Weinberg equilibrium (which is fairly valid since the HD phenotype doesn't show up until after one reaches reproductive age). This means that:

p^{2} = 0.99, and

p = 0.99499

where p is the prevalence of the h allele in the population. Therefore:

q = 1-p = 0.00501

Therefore, we can find the fraction of the population that are heterozygous, f(Hh), and the fraction of the population that are homozygotes, f(HH):

f(Hh) = 2pq = 2(0.99499)(0.00501) = 0.00997

f(HH) = q^{2} = (0.00501)^{2} = 2.51x10^{-5}

So, among the people with Huntington's disease, the probability that an afflicted individual is heterozygous, f(Hh|Huntingtions), and the probability that an afflicted individual is homozgous, f(HH|Huntingtons), are:

f(Hh|Huntingtons) = 0.00997/.01 = 0.997

f(HH|Huntingtons) = 2.51x10^{-5}/0.01 = 0.00251

In our example, 99.7% of the people with our "rare" disorder are heterozygotes and we can pretty safely ignore the population of homozygotes.