If you know, that one card is an ace, the probability that the second card is also an ace is 1⁄3, because there are two non aces and one ace remaining.
The tricky thing is, that it´s completely irrelevant, if the ace is of spades or of hearts. Remember, the question is if you hold both aces! Just distinguish between aces and non-aces..
After answering “Yes” to “Do you have an ace,” the possible worlds are
(1) AS, AH
(2) AS, 2C
(3) AS, 2D
(4) AH, 2C
(5) AH, 2D
That is, in world (6) you would not answer “Yes,” so it is eliminated.
After picking one of your aces randomly, the possible worlds are:
(1a) AS, AH → AS in one possible sub-world
(1b) AS, AH → AH in another possible sub-world
(2) AS, 2C → AS
(3) AS, 2D → AS
(4) AH, 2C → AH
(5) AH, 2D → AH
You’re counting (1a) and (1b) as 1 each, then dividing by the six worlds to get 1⁄3, but the trick is that those two worlds are not as likely as the others - (edited for clarity) half the time world (1) evolves into (1a); half the time it evolves into (1b).
So if you count (1a) and (1b) as 0.5 each and the rest as 1 each, then the probability of having both aces is (0.5 + 0.5) / (0.5 + 0.5 + 1 + 1 + 1 + 1) = 1⁄5.
I don’t understand what your tree is trying to show.
However, I think my explanation above shows why world’s (1a) and (1b) are weighted less than (2) through (5) - they both come from parent world (1). If you gave (1a) and (1b) weight=1, you would be giving parent world 1 weight=2.
I think Scenario 2 is wrong.
If you know, that one card is an ace, the probability that the second card is also an ace is 1⁄3, because there are two non aces and one ace remaining.
The tricky thing is, that it´s completely irrelevant, if the ace is of spades or of hearts. Remember, the question is if you hold both aces! Just distinguish between aces and non-aces..
It helps to enumerate the possible worlds:
In the beginning you can have:
(1) AS, AH
(2) AS, 2C
(3) AS, 2D
(4) AH, 2C
(5) AH, 2D
(6) 2C, 2D
After answering “Yes” to “Do you have an ace,” the possible worlds are
(1) AS, AH
(2) AS, 2C
(3) AS, 2D
(4) AH, 2C
(5) AH, 2D
That is, in world (6) you would not answer “Yes,” so it is eliminated.
After picking one of your aces randomly, the possible worlds are:
(1a) AS, AH → AS in one possible sub-world
(1b) AS, AH → AH in another possible sub-world
(2) AS, 2C → AS
(3) AS, 2D → AS
(4) AH, 2C → AH
(5) AH, 2D → AH
You’re counting (1a) and (1b) as 1 each, then dividing by the six worlds to get 1⁄3, but the trick is that those two worlds are not as likely as the others - (edited for clarity) half the time world (1) evolves into (1a); half the time it evolves into (1b).
So if you count (1a) and (1b) as 0.5 each and the rest as 1 each, then the probability of having both aces is (0.5 + 0.5) / (0.5 + 0.5 + 1 + 1 + 1 + 1) = 1⁄5.
I don’t get what your saying… see my probability tree diagram and tell me why you are weighting them differently?
I don’t understand what your tree is trying to show.
However, I think my explanation above shows why world’s (1a) and (1b) are weighted less than (2) through (5) - they both come from parent world (1). If you gave (1a) and (1b) weight=1, you would be giving parent world 1 weight=2.