Answering without looking at other comments, will check those after:
The assumption in the first argument is wrong, you in fact have different information than the info you have in scenario 1. In scenario 1, the information you have is “I answer yes when asked if I have a spade”
In this situation you have the information that I have an ace and if I pick one of the aces at random to reveal to you, I reveal that I have a spade.
The relevant likelihoods are DIFFERENT:
P(“I have at least one ace, and if I choose one at random (or just the one if I have only one), to reveal to you, I reveal to you I have a spade” | I have both aces) = 1⁄2
P(“You asked me if I have a spade? Well, yes, I do have a spade” | I have both aces) = 1
(modulo standard disclaimers on assigning P = 1 to any state of affairs, of course)
Argument 2 is correct. What changes is the probability that I hold just an ace of spades or just an ace of hearts.
In other words, the remaining possibilities are the same in both scenarios (AS/AH, AS/2C, AS/2D), but the probabilities attached to these possibilities are different. In scenario 2 the observation that one of the cards is the ace of spades does more than rule out possibilities, it also shifts the probability mass from the AS/AH possibility toward the AS/2C and AS/2D possibilities, in such a way that the probability attached to the AS/AH possibility goes ‘back’ to what it was when there were five possibilities.
It shows that drawing little boxes corresponding to each possibility and carefully crossing out those that were contradicted by the evidence is only a poor approximation of Bayes’ theorem.
Well, drawing boxes and crossing out would work here if you explicitly have boxes for “how does the card holder answer the questions” or in this case “how does the card holder answer the “if you have aces, pick one at random, is it a spade”?
Answering without looking at other comments, will check those after:
The assumption in the first argument is wrong, you in fact have different information than the info you have in scenario 1. In scenario 1, the information you have is “I answer yes when asked if I have a spade”
In this situation you have the information that I have an ace and if I pick one of the aces at random to reveal to you, I reveal that I have a spade.
The relevant likelihoods are DIFFERENT:
P(“I have at least one ace, and if I choose one at random (or just the one if I have only one), to reveal to you, I reveal to you I have a spade” | I have both aces) = 1⁄2
P(“You asked me if I have a spade? Well, yes, I do have a spade” | I have both aces) = 1
(modulo standard disclaimers on assigning P = 1 to any state of affairs, of course)
Argument 2 is correct. What changes is the probability that I hold just an ace of spades or just an ace of hearts.
Yup.
In other words, the remaining possibilities are the same in both scenarios (AS/AH, AS/2C, AS/2D), but the probabilities attached to these possibilities are different. In scenario 2 the observation that one of the cards is the ace of spades does more than rule out possibilities, it also shifts the probability mass from the AS/AH possibility toward the AS/2C and AS/2D possibilities, in such a way that the probability attached to the AS/AH possibility goes ‘back’ to what it was when there were five possibilities.
It shows that drawing little boxes corresponding to each possibility and carefully crossing out those that were contradicted by the evidence is only a poor approximation of Bayes’ theorem.
Well, drawing boxes and crossing out would work here if you explicitly have boxes for “how does the card holder answer the questions” or in this case “how does the card holder answer the “if you have aces, pick one at random, is it a spade”?
It just takes a bit more detail.
I made a diagram already. Note that we were using ‘prefer’ to refer to the ace you pick when you pick between the two if you have both.
Ditto. Psy-Kosh has correctly observed that argument 2 is correct, and has correctly pointed out the flaw in argument 1.