You have two 16-ounce cups — cup A and cup B. Both cups initially have 8 ounces of water in them.
You take half of the water in cup A and pour it into cup B. Then, you take half of the water in cup B and pour it back into cup A. You do this again. And again. And again. And then many, many, many more times — always pouring half the contents of A into B, and then half of B back into A.
When you finally pause for a breather, what fraction of the total water is in cup A?
Extra credit: Now suppose both cups initially have somewhere between 0 and 8 ounces of water in them. You don’t know the precise amount in each cup, but you know that both cups are not empty. Again, you pour half the water from cup A into cup B, and then half from cup B back to A. You do this many, many times. When you again finally pause for a breather, what fraction of the total water is in cup A?
Let $A_n$ and $B_n$ denote the volume of water in each cup after $n$ pours, for even $n$.
$A_{n+1} = \dfrac{A_n}{2}$
$B_{n+1} = B_{n} + \dfrac{A_n}{2}$
$A_{n+2} = A_{n+1} + \dfrac{B_{n+1}}{2}$
$B_{n+2} = \dfrac{B_{n+1}}{2}$
After two pours, the amount in each cup should revert to what it was before the two pours.
$A_{n+2} = A_n$
$B_{n+2} = B_n$
Using the first equation from the above two parts and substituting:
$2 \cdot A_{n+1} = {A_n} = A_{n+2}$
$2 \cdot A_{n+1} = A_{n+1} + \dfrac{B_{n+1}}{2}$
$A_{n+1} = \dfrac{B_{n+1}}{2}$
Similarly, using the second equations
$2 \cdot B_{n+2} = B_{n+1} = 2 \cdot B_{n}$
$2 \cdot B_{n} = B_{n} + \dfrac{A_n}{2}$
$B_{n} = \dfrac{A_{n}}{2}$
After some time, after an odd pour, A contains $\dfrac{1}{3}$ the water volume.
After some time, after an even pour, A contains $\dfrac{2}{3}$ the water volume.