Why does the reduction of the total cards (pool) make no difference to the probability that I will get a certain card in my hand Some friends and I had a heated debate about this and I am still quite confused. 
The problem
I will use an example of


*

*8 cards in a bag, Each card has a letter on it, e.g. A,B,C,D,E,F,G,H. 

*You draw 3 cards. 

*What are the chances of getting a specific card (e.g. B) in that final hand (i.e. what are the chances that one of the cards you now have is the one you wanted)?


My logic:
I thought surely its 1/8 then 1/7, then 1/6 as each time a card is taken there are less cards? This would be
1/8 + 1/7 + 1/6 = 43.5% chance.
Someone pointed out to me that on the second, and third times your card may already be chosen, but in my head this should not be a reason your chances get worse... but i'm sure that's the key thing i don't understand.
They said it actually remains a 1/8 each time so
1/8 + 1/8 + 1/8 = 3/8 = 37.5%
Which is correct (see simulation at bottom), and someone else pointed out that if you extrapolate my logic to picking 8 cards 
1/8 + 1/7 + 1/6 + 1/5 + 1/4 + 1/3 + 1/2 + 1/1
You get over 200% which is obviously wrong. 
The Question/Confusion
Why does the reduction not affect (appear in) the maths?
I feel the "reduction" in the pool should impact the maths or at least be cancelled out and that's what I struggle to grasp. Why does the reduction in the pool seemingly make no difference whatsoever, its seems at odds with what I feel I learnt from the Monty Hall problem.... 
I even felt the need to simulate it and sure enough *the answer was 37.5%. My gut tells me that the reduction might become statistically relevant in some cases, but my brain says there is missing (i.e. cancelled out maths) above that would explain that this is not true. Could someone help the penny drop and enlighten me? 
simulation: https://dotnetfiddle.net/ebTl5B
--- [Full Maths] --- (Thanks Ethan!)
Added here as it will render nicer then in the comments, but you will want to read Ethans answer below for why you do this. 
$$\frac{1}{8}  + (\frac{7}{8} * \frac{1}{7}) + (\frac{7}{8} * \frac{6}{7} * \frac{1}{6}) = $$
$$\frac{1}{8}  + \frac{1}{8} + \frac{1}{8} = $$
$$\frac{3}{8} =$$
$$ 0.375 $$
 A: @JakeMirra 's answer is correct and elegant. This one may help you see where you went wrong.
Try your reasoning on a smaller example. Suppose there are three cards in the deck, you like one of them, and you're dealt two. Clearly there's a $1/3$ chance that the good card is the one left, so a $2/3$ chance that you have it. 
Your logic would say your chances are $1/3 + 1/2 = 5/6$.
Here's how to correct that calculation. The chance that you get the good card first is indeed $1/3$. If you succeed  the first time you can go home happy, ignoring what happens next. If you fail  the first time the chance that you succeed the second time is $1/2$, but that only matters for the $2/3$ of the time you care about that second draw. So the probability that you succeed the second time (probability $1/2$) given that the first draw failed (probability $2/3$) is
$$
\frac{2}{3} \times \frac{1}{2} = \frac{1}{3} .
$$
Add that to the probability that you succeeded the first time and you get the correct $2/3$.
A: An elegant way to think about it, is to imagine that all the cards are black except for one card (the one you want) is white.  Let's imagine that you select three cards before the cards' colors are determined.  So you pick three cards.  And then one of the eight cards, at random, is painted white.  Then clearly, the probably of painting one of your three cards is 3/8, and the probability of painting one of the five cards you didn't pick is 5/8.
