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.
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 chance the person chose spades over hearts rather than just being forced in the other two scenarios to be forced to choose spades.
We know that the person has an Ace of spades in their hand which means the following are the only combinations the person could have is AH and AS, 2C and AS, 2D and AS. The probability of the person answering the question with “I have an Ace of spades” when the person has the combination of AH and AS is 1⁄2. The probability with 2C and AS is 1 and the probability of 2D and AS is 1 as there is only one ace in these combinations meaning the person would be forced to choose these. There for the probability is (1/2)/(1/2+1+1) which simplifies to 1⁄5. Mathematically this can be shown as:
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 chance the person chose spades over hearts rather than just being forced in the other two scenarios to be forced to choose spades.
We know that the person has an Ace of spades in their hand which means the following are the only combinations the person could have is AH and AS, 2C and AS, 2D and AS. The probability of the person answering the question with “I have an Ace of spades” when the person has the combination of AH and AS is 1⁄2. The probability with 2C and AS is 1 and the probability of 2D and AS is 1 as there is only one ace in these combinations meaning the person would be forced to choose these. There for the probability is (1/2)/(1/2+1+1) which simplifies to 1⁄5. Mathematically this can be shown as:
AH AS P(choose AS)=1/2 P(choose AH)=1/2
2C AS P(choose AS)=1
2D AS P(choose AS)=1
therefore P(both) = (1/2)/(1/2+1+1) = 1⁄5