Card Trick of the 'Prime Obsession' book I saw this problem at the first chapter of the book called Prime Obsession by John Derbyshire:

Take an ordinary deck of 52 cards, lying on a table, all four sides of
the deck squared away. Now, with a finger slide the topmost card
forward without moving any of the others. How far can you slide it
before it tips and falls? Or, to put it another way, how far can you
make it overhang the rest of the deck?
The answer is half a card length. If you push it so that
more than half the card overhangs, it falls. The tipping point is at
the center of gravity of the card, which is halfway along it. Now
let’s go a little further. With that top card pushed out half its
length—that is, to maximum overhang—over the second one, push that
second card with your finger. How much combined overhang can you get
from these top two cards? The trick is to think of these top two cards
as a single unit. Where is the center of gravity of this unit? Well,
it’s halfway along the unit, which is altogether one and a half cards
long; so it’s three-quarters of a card length from the leading edge of
the top card. The combined overhang is, therefore, three-quarters of a
card length.
If you now start pushing the third card to see how much you can
increase the overhang, you find you can push it just one-sixth of a
card length. Again, the trick is to see the top three cards as a
single unit. The center of gravity is one-sixth of a card length back
from the leading edge of the third card.

Now, in case of the first two cases I don't have any trouble to understand. I understand how I get $1/2$ for the first card and $(1/2 + 1/4) = 3/4$ overhang for the two cards. But in case of overhanging three cards together I don't get how we have extra $(1/6)$ amount of overhanging. Instead of getting the overhanging of $(1/2 + 1/4 + 1/6)$ for the three cards I get $(1/2 + 1/4 +1/8)$ amount of overhanging. Here is my reasoning:
As the author said, to consider the three cards as a single one, we have a combined length of $1 + 3/4 = 7/4$ for these three cards[$3/4$ is the overhang for the first two cards and $1$ is the length of the third card] and their center of gravity will be at the halfway of their length, that is at $(7/4)/2 = 7/8$ unit from the left side of the rest of the cards. So, the extra overhang is $(1 - 7/8 = 1/8)$. Can anyone please explain what is wrong in my reasoning? I would appreciate that.
Update: I think I already found out where I went wrong. As the system with three cards combined is not symmetrical, unlike the first two cases, it is wrong to assume that the center of gravity is at the midpoint. In this case we have to use the formula for finding the CoG of compound object. Have a look at this answer for more insight. Still, if any of you can have any thoughts just deliver it, I would appreciate your comments.
 A: You have to divide by three to get the center of mass of the three card system. When we start out, all cards have their centers of mass at $x=\frac{1}{2}$ where $x$ is the distance from the left side of the deck which we will call $x=0$.
As you and the author have pointed out the center of mass of the first two cards is the average of the two centers of mass. Before moving the second card we get the average of $\frac{1}{2}$ and $1$ which is $\frac{3}{4}$. Since we can move the center of mass to $1$ this allows us to push the second card to $x=\frac{1}{4}$ giving us a center of mass  that is the average of $\frac{3}{4}$ and $\frac{5}{4}$ which again gives $1$. So far so good.
The three card system prior to moving the third card has a center of mass that is the average of $\frac{1}{2}$, $\frac{3}{4}$ and $\frac{5}{4}$ which is $\frac{5}{6}$. Why? (we must divide by $3$ to get the average)
This allows us to move the third card by $\frac{1}{6}$ which is the author's claim. We can verify this by noting that after the move we are taking the average of $\frac{2}{3}$, $\frac{11}{12}$ and $\frac{17}{12}$ which gives $1$.
