Suppose you set down three cards face down: two hearts, and a spade. You ask someone to pick one of the cards; if they pick the spade, they win. They pick card 1. You then reveal card 3: a heart. You offer to switch their card with 2. Is it favorable to switch?
Apparently, it's supposed to be, but I'm a little confused: there are two lines of reasoning that both seem to make sense to me, but they contradict each other:
I. There was a 2 in 3 chance you were wrong, and since you know 3 is wrong, you are basically asking yourself a question: were you right or wrong on your first pick? If you pick right you keep your card, which is a 1 in 3 chance; if you pick 2 you are betting on wrong, 2 in 3.
Seems to make sense, but...
II. We know the third card is a heart. That means the face-up setup is either spade-heart-heart, or heart-spade-heart; what is it really that favors one over the other? When we're deciding what completes the threefold heart-spade complex (i.e., the first two cards, which we know contains a spade and heart), can we not, for all intents and purposes, instead focus on the first two cards, the ones we don't know? We know that one has to be a spade, and one has to be a heart. Either combo seems plausible, doesn't it? Like in algebra, it seems you could cancel out heart, since we know that one; it really boiles down to heart-spade versus spade-heart; the fact that there is an equivalent third card added to both sides doesn't seem to mean much. It's like saying x = y, but x+1 does not equal y+1.
Couldn't we just treat this as a new game, one where you put two cards face down, one a spade, other a heart, and ask what they choose? When the heart is revealed, isn't that essentially what you're doing? Why, really, should it matter what you picked before? You're still choosing the completion of the heart-spade complex, aren't you? Isn't that neutral?
"I" seems very logical, yet I can't quite refute "II."
You know what I think it is? I think it's that when 3 is revealed, your #1 is not eliminated right off the bat, but #3 is. So for example, if you were told that #1, your first pick, was wrong, then the chances of being correct would indeed be 1 in 2 since you know no distinct differences between 2 and 3. But since you know 3 is wrong, you can compare 1 in 2, which, with 3 out, is the choice between right and wrong, reverting back to I. Because 3 is wrong, the value of 1 and 2 I guess do have different values, unlike if 1 was wrong when there is no conflict between 2 and 3.
The thing about I is, it is taking into account slightly more information (what you picked) than the more objective postulating in II; that extra information I guess is surprisingly useful!