Sorry to zombie this thread, but I could use some help.

Hmm.. I'm going with 1/3 for 2 reasons. i might be wrong, but maybe I can explain how I am wrong well enough for someone to help me see it, because I'm stumped here.

reason one, from the bottom end: when the answers are yes-yes, there are only 3 possible types hands, ace-ace, and (2) ace-null, each as likely. Weighting probabilities based on how one may answer "yes-no" and still have ace-ace seems erroneous when the answers are yes-yes.

reason two, from the top end: It seems that a false set is being used in the explanations favoring argument 2. here's how I think this occurs. the set of possibilities does NOT equal the set(s) of probabilities. random pairings yield six possibilities. having an ace eliminates one of them, leaving ace-ace, and (4) ace-null pairings. this trimmed possibility set is being inherited to form a probability tree, erroneously in my opinion. the way i see it, once an ace is detected, the POSSIBILITIES equal 6-1=5 but the PROBABILITIES are within one of two distinct, exclusive sets each of ace-ace and (2) ace-nulls. the possible combinations of pairings do not represent the scope of the probability. in other words one may have [(ah-as or ah-2c or ah-2d) OR (as-ah or as-2c or as-2d)]. so if one knows that the hand contains at least 1 ace, the likelihood of having the other ace is 1 out of 3. notice that the ace-ace appears in each set. to call the probability 1/5 (by propagating the possibility set as a probability tree) is to mesh two exclusive sets and throw out one of the (2) ace-ace pairings before doing the math. detecting an ace is the key, knowing which type shouldn't change anything.

i think maybe the Bayesian concept could have been demonstrated by asking how many yes-no's may have ace-ace or something else.