This is the way that I see the problem (please correct me if I’m wrong).
ANSWER: 1⁄5
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
:)
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