I have an interest in gaming management and practical probabilities. I have a great interest in economics as well. I stumbled onto this site and tyhe "Drawing 2 aces" post. I struggled with it for about a week, and then wrote a few things. The thread is old, but I look forward to any helpful responses.
Even further: the probability that I can guess your hand is 1/5. However, the probability that you have 2 aces is 1/3. No?
Further: write out your 12 possible trial pulls. Raw odds of ace-ace =2/12. Once an ace is pulled, do two things: cross out the two null-null pulls. The odds of ace-ace appear to be 2/10, but... We didn't draw out Schrodinger's ace; it is either ah or as. pick one (it doesnt matter which, they are symmetrical in distribution) and cross out the combinations that do not have this ace. As the waveform collapses the true odds of ace-ace appear- 2/6. Do you see that? We didnt draw an equally hearty or spadey ace, it had to be one of them or the other, which made combinations without it no longer a factor in our investigation.
Yes you can initially reduce the twelve trials to the six unique combinations, but ONLY because each of the six appears the same number of times as each of the other six (they are equally probable). In a set which favors some combinations, reducing the probability set to the possibility set will lose any meaningfulness.
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.
If we have an ace in the hand, 2 of those "equally likely cases" are no longer possible. (2 of those cases involve the other ace and a non-ace card.)