As a follow up for the "it's complicated" comment, I'll attempt to do the math for a simple example to show you why.

You're over at a friend's house and playing 4-handed hold'em. You're on the turn with a spade flush draw and your opponent is holding 2 red aces. The other two players folded before the flop. Example:

VS

The board is

Now, you can see 8 cards. So there are 44 unknown cards. 9 of those unknown cards give you a winning hand. Your 'naive' odds are 9/44 or about 20.45%.

You don't know what your opponents folded, but you can figure out the probabilities that they folded outs of yours.

Odds they folded 0 spades: 52,360/135,751 ~= 38.57%

Odds they folded 1 spades: 58,905/135,751 ~= 43.39%

Odds they folded 2 spades: 21,420/135,751 ~= 15.78%

Odds they folded 3 spades: 2,940/135,751 ~= 2.17%

Odds they folded 4 spades: 126/135,751 ~= 0.09%

Since we 'know' the 4 folded cards now, there are only 40 unknown cards left.

Your adjusted odds for them folding 0 spades: 9/40 = 22.5%

Your adjusted odds for them folding 1 spades: 8/40 = 20.0%

Your adjusted odds for them folding 2 spades: 7/40 ~= 17.5%

Your adjusted odds for them folding 3 spades: 6/40 ~= 15.0%

Your adjusted odds for them folding 4 spades: 5/40 ~= 12.5%

But, we need to adjust these odds with the probability that we're facing them.

Weighted total odds 0 spades folded: 52,360/135,751 * 9/40 = 471,240/5,430,040 ~= 8.68%

Weighted total odds 1 spades folded: 58,905/135,751 * 8/40 = 471,240/5,430,040 ~= 8.68%

Weighted total odds 2 spades folded: 21,420/135,751 * 7/40 = 149,940/5,430,040 ~= 2.76%

Weighted total odds 3 spades folded: 2,940/135,751 * 6/40 = 17,640/5,430,040 ~= 0.32%

Weighted total odds 4 spades folded: 126/135,751 * 5/40 = 630/5,430,040 ~= 0.00%

The total weighted odds are 1,110,690/5,430,040 ~= 20.45%

You will note that 1,110,690/5,430,040 is equivalent to 9/44. (Multiply the top and bottom by 123,410 to get the larger fraction.)

So... ALL that work to try and account for the fact that some of our outs might have been folded, and the end result is the exact same value we got by ignoring those cards to begin with. And those calculations are not easy to get. I didn't show all my work for how to get the number of different ways they could hold your out.

And this was a SIMPLE example with only 2 people who folded. Imagine trying to do all this math for a 9 handed table where 7 people folded. Fortunately, for everyone, we don't need to because it will work out the same in the end. Burn cards work the same way. In the end, cards we don't know the value of do not change our odds in any way.