April 11, 2021
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?
Answer: \(\frac{2}{3}\) for both original question and extra credit.
Explanation:
Let \(f\) be the fraction of total water in cup A after many iterations of transfering half the water from cup A to B and then vice-versa. Obviously \(0 \leq f \leq 1\). Now, if we are interested in the "equilibrium" value of \(f\), as defined by the fact that \(f\) remains the same after an iteration of the transfer process, we simply need to apply the definition.
After first transfering water from cup A to B, we have only \(\frac{f}{2}\) of the total water in cup A and therefore \((1 - f) + \frac{f}{2} = \frac{2-f}{2}\) of the total water in cup B. Moving half of what's now in cup B back to A yields cup A having \(\frac{f}{2} + \frac{2-f}{4} = \frac{f + 2}{4}\) of total water. For "equilibrium", this value should equal to "f", and hence we have:
\begin{align} \frac{f + 2}{4} &= f \\ f + 2 &= 4f \\ 3f &= 2 \\ f &= \boxed{\frac{2}{3}} \end{align}
We thereby confirm that this equilibrium \(f\) is between 0 and 1 and is the only possible value. This also makes intuitive sense as when A contains \(\frac{2}{3}\), a third of it will simply shift back and forth between the two cups, thereby keeping the fraction in cup A constant after every pair of iterations.
Now the remaining question is, will every initial amount of water in cups A and B converge to this equilibrium? We can easily "find out" the answer by running computer simulations and making speculations after many runs, but let's try to see the answer mathematically here. Since the equilibrium point is unique, this means that either the process will converge to this value, or not converge at all. Therefore, the only case where our answer of \(\frac{2}{3}\) might not work would be if the cup transfer process does not converge to the equilibrium value. This can only happen if each transfer moves more and more amounts of water to the other cup. This is not possible as we will always have transfer amounts see-sawing between being less than \(\frac{1}{3}\) of the total water and greater, and thereby eventually converging to this value.