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.
Hi,
This is why I'm pretty sure its 1/5 (although I don't understand Eliezer's reasoning in Argument 2, this is my reasoning - which by all means could be the same, I'm just not very good at dissecting flowery language).
If you ignore the fact that the person has the ability to choose the Ace of Hearts (if they have both aces) in the second question, yes it would be 1/3, however if the person does have both spades and they choose to say spade instead of hearts then that changes the probability as it means that there's less of a chanc...
Suppose I have a deck of four cards: The ace of spades, the ace of hearts, and two others (say, 2C and 2D).
You draw two cards at random.
Scenario 1: I ask you "Do you have the ace of spades?" You say "Yes." Then the probability that you are holding both aces is 1/3: There are three equiprobable arrangements of cards you could be holding that contain AS, and one of these is AS+AH.
Scenario 2: I ask you "Do you have an ace?" You respond "Yes." The probability you hold both aces is 1/5: There are five arrangements of cards you could be holding (all except 2C+2D) and only one of those arrangements is AS+AH.
Now suppose I ask you "Do you have an ace?"
You say "Yes."
I then say to you: "Choose one of the aces you're holding at random (so if you have only one, pick that one). Is it the ace of spades?"
You reply "Yes."
What is the probability that you hold two aces?
Argument 1: I now know that you are holding at least one ace and that one of the aces you hold is the ace of spades, which is just the same state of knowledge that I obtained in Scenario 1. Therefore the answer must be 1/3.
Argument 2: In Scenario 2, I know that I can hypothetically ask you to choose an ace you hold, and you must hypothetically answer that you chose either the ace of spades or the ace of hearts. My posterior probability that you hold two aces should be the same either way. The expectation of my future probability must equal my current probability: If I expect to change my mind later, I should just give in and change my mind now. Therefore the answer must be 1/5.
Naturally I know which argument is correct. Do you?