Challenged in math, question about figuring odds

[really annoyed]

3/10 I don't understand and I'm really confused about the explanation of figuring odds. You're in a Hold em game. You hold two 7's. Two are left in the deck and the flop comes and so five cards are out of the deck and your odds of getting any one or two of the remaining 7's are based on calculating based on 47 remaining cards. But how can that be? And here is where I don't understand. Pre flop, if you are playing with 8 other participants, they each got two cards as well, taking 16 cards out of the deck and within those 16 cards any one or two of your 7's can be in there as well, right? So, how can you accurately figure your chances or odds of getting another one or two 7's based on 47 cards, when, if you are playing with 8 other players, each holding 2 cards, possibly 16 additional cards have already been removed from the 47 and any one or two of the 7's can be held within the 16. Am I making sense? That occurance has to be taken into account as well. Or am I wrong? All of those who have superior skills in math have to forgive my inablility in understanidng the expression of calculating odds but I just don't think it is right to base the odds on 47 remaining cards. I would appreciate any feedback Thanks in advance, from REALLY ANNOYED.

[FatJoe]

you're wrong. any card you havent seen counts toward making the odds, including cards your opponents have.

think about this, is there any difference between the cards that your opponents hold which will never be on the board and the cards at the bottom of the deck? if you're confused by the fact that somebody might have your 7, what about the fact that your 7 might be on the bottom of the deck? either way it cant come off.

dont see whats so hard about that.

[really annoyed]

3/10 Thanks Joe, I appreciate your explanation and your help. As I said, I'm really challenged. Guess I'll stick to the ponies, seems like the odds are a lot simpler there.