Discussed
here, more or less....
But your way of looking at it is far better than saying, "I can't calculate the pot odds unless I know EXACTLY how many flush cards are out." You're taking into account the EV for the number of flush cards in folded or live hands. It's not wrong, just adding needless complexity that doesn't affect your calculation at all.
You have two spades, two spades are on the board, one card is to come. Just for kicks, let's try accounting for the spades in the nine unseen hands (whether mucked or live). There are 18 cards in those hands, which is 18/46 of the remaining deck. So the EV for number of spades in those hands is (18/46) * 9, or 3.52.
So we expect that 9 - 3.52 => 5.48 spades are left in the remaining 28 cards (stub plus burn cards). Hence there are 22.52 non spades left on average. Our odds against catching a spade on the river are 22.52 : 5.48 => 4.11 : 1. This happens to be precisely the same as 37 : 9.
Do you see why this is? The proportion of spades in the pool of unseen cards isn't changing just because those cards were or weren't dealt out. None of those cards in someone else's hand is more or less likely to be a spade. (NOTE: That's slightly untrue. One or more of your live opponents might also have a spade draw. But this possibility is negligible.)
You could do the same math I just did for the three burn cards, too, taking them out as candidates for the river card. That wouldn't affect your pot odds either, since you don't know their identity. But if the dealer accidentally exposed the 5 [img]/images/graemlins/diamond.gif[/img] and burned it, then it would affect your calculation.
Since it's easier to consider all 46 cards to be one pool of cards, and doesn't change the result one iota, that's how most people choose to calculate it.
Cross posted to the wiki. Everyone's invited to help clean up that article.