Math riddle about 4 cups

Question

There are 4 cups filled till half. In 2 cups there are one drink and in other 2 another drink. They are placed randomly on a table. How to drink half cup of each drink?

Answer 1

Take two glasses and empty them into the other two glasses. You now have two completely full glasses. Drink half of each.

Answer 2

Disclaimer: I don't know how to solve the problem in finitely many steps.

Label drinks as $A$ and $B$, and the cups as $1,\, 2,\, 3,\, 4.$ Then we have six possible starting positions, but we can just look at three cases due to the symmetry of the problem (just exchange $A$ and $B$).

I will represent this in a table writing a fraction of a cup each drink occupies at a given step. Steps will always consist of pouring from a full cup to a half full cup until the half full one becomes full, the only exception being the first step. This is because I assume that we do not have the ability to measure a half of rhe content of a cup, so if at any point we get two full and two empty cups, we are stuck.

\begin{array}{ c | c | c c c c | c c c c | c c c c } \text{Drink} & \text{Step} & \text{Case}\, 1 & & & & \text{Case}\, 2 & & & & \text{Case}\, 3 & & & &\\ & & 1 & 2 & 3 & 4 & 1 & 2 & 3 & 4 & 1 & 2 & 3 & 4 \\ \hline A & 0 & 1/2 & 1/2 & 0 & 0 & 1/2 & 0 & 1/2 & 0 & 1/2 & 0 & 0 & 1/2 \\ B & & 0 & 0 & 1/2 & 1/2 & 0 & 1/2 & 0 & 1/2 & 0 & 1/2 & 1/2 & 0 \\ \hline \end{array}

The steps will be the following: pour from $1$ to $2$, pour from $2$ to $3$, pour from $3$ to $2$, pour from $2$ to $3$, and so on. We will make some conclusions after observing what happens after couple of steps:

\begin{array}{ c | c | c c c c | c c c c | c c c c } \text{Drink} & \text{Step} & \text{Case}\, 1 & & & & \text{Case}\, 2 & & & & \text{Case}\, 3 & & & &\\ & & 1 & 2 & 3 & 4 & 1 & 2 & 3 & 4 & 1 & 2 & 3 & 4 \\ \hline A & 0 & 1/2 & 1/2 & 0 & 0 & 1/2 & 0 & 1/2 & 0 & 1/2 & 0 & 0 & 1/2 \\ B & & 0 & 0 & 1/2 & 1/2 & 0 & 1/2 & 0 & 1/2 & 0 & 1/2 & 1/2 & 0 \\ \hline A & 1 & 0 & 1 & 0 & 0 & 0 & 1/2 & 1/2 & 0 & 0 & 1/2 & 0 & 1/2 \\ B & & 0 & 0 & 1/2 & 1/2 & 0 & 1/2 & 0 & 1/2 & 0 & 1/2 & 1/2 & 0 \\ \hline A & 2 & 0 & 1/2 & 1/2 & 0 & 0 & 1/4 & 3/4 & 0 & 0 & 1/4 & 1/4 & 1/2 \\ B & & 0 & 0 & 1/2 & 1/2 & 0 & 1/4 & 1/4 & 1/2 & 0 & 1/4 & 3/4 & 0 \\ \hline A & 3 & 0 & 3/4 & 1/4 & 0 & 0 & 5/8 & 3/8 & 0 & 0 & 3/8 & 1/8 & 1/2 \\ B & & 0 & 1/4 & 1/4 & 1/2 & 0 & 3/8 & 1/8 & 1/2 & 0 & 5/8 & 3/8 & 0 \\ \hline A & 4 & 0 & 3/8 & 5/8 & 0 & \\ B & & 0 & 1/8 & 3/8 & 1/2 & \\ \hline \end{array}

First notice that Cases $2$ and $3$ became essentially the same after first step (just exchange $A$ and $B$). That's why we won't concern ourselves with Case $3$ anymore.

Also, I purposely didn't include step $4$ for Cases $2$ and $3$ because I want you to notice that Case $1$ is essentially the same as Cases $2$ and $3$, but one step behind. This tells us something important: if we can solve the problem in $n$ steps for Cases $2$ and $3$, it will be solved in $(n+1)$-st step for Case $1$. Since we do not know in which case we are, we can't solve the problem in finitely many steps using this process! That's why we will introduce a sequence $(a_n)$ which will tell us the fraction of Drink $A$ in the full cup at particular step. This sequence obeys recurrence relation $a_{n+2} = \frac 12 a_{n+1} + \frac 12 a_n$, which will hopefully be explained by the table:

\begin{array}{ c | c | c c c c | c c c c} \text{Drink} & \text{Step} & \text{Case}\, 1 & & & & \text{Case}\, 2 & & & & \\ & & 1 & 2 & 3 & 4 & 1 & 2 & 3 & 4 \\ \hline A & 0 & 1/2 & 1/2 & 0 & 0 & 1/2 & 0 & 1/2 & 0 \\ B & & 0 & 0 & 1/2 & 1/2 & 0 & 1/2 & 0 & 1/2 \\ \hline A & 1 & 0 & 1 & 0 & 0 & 0 & a_0 & 1/2 & 0 \\ B & & 0 & 0 & 1/2 & 1/2 & 0 & 1-a_0 & 0 & 1/2 \\ \hline A & 2 & 0 & 1/2 & a_0 & 0 & 0 & a_0/2 & a_1 & 0 \\ B & & 0 & 0 & 1-a_0 & 1/2 & 0 & (1-a_0)/2 & 1-a_1 & 1/2 \\ \hline A & 3 & 0 & a_1 & a_0/2 & 0 & 0 & a_2 & a_1/2 & 0 \\ B & & 0 & 1-a_1 & (1-a_0)/2 & 1/2 & 0 & 1-a_2 & (1-a_1)/2 & 1/2 \\ \hline A & 4 & 0 & a_1/2 & \small{a_2 = a_0/2 + a_1/2} & 0 & \\ B & & 0 & (1-a_1)/2 & 1-a_2 & 1/2 & \\ \hline \end{array}

From the table you can also notice that we chose $a_0 = 1/2$, $a_1 = 3/4$. Solving the recurrence will give us $$a_n = \frac{1}{6} \left(\frac{(-1)^{n+1}}{2^{n}}+4\right)$$ which converges to $a = 2/3$.

So, if we repeat pouring from cup $2$ to cup $3$ and from cup $3$ to cup $2$ countably many times, we will end up with situation like this:

\begin{array}{ c | c | c c c c } \text{Drink} & \text{Step} & 1 & 2 & 3 & 4\\ \hline A & \omega & 0 & 2/3 & 1/3 & 0 \\ B & & 0 & 1/3 & 1/6 & 1/2 \\ \hline \end{array}

A quick explanation. It doesn't matter if the full cup is $2$ or $3$. Passing to infinitely many steps, we lose track of this. So just say that the full cup is $2$. So, what's in the cup $3$? Well, we do know that cup $1$ is empty, and the state of cup $4$ the whole time. So just calculate what's in the cup $3$ from the total amount of drinks. Also note that if we repeated pouring from cup $2$ to cup $3$, they would just switch places (which should be obvious since we took a limit and $\lim_n a_n = \lim_n a_{n+1}$.)

Finally, pour from cup $2$ to cup $4$ and then from cup $4$ to cup $3$:

\begin{array}{ c | c | c c c c } \text{Drink} & \text{Step} & 1 & 2 & 3 & 4\\ \hline A & \omega & 0 & 4/6 & 2/6 & 0 \\ B & & 0 & 2/6 & 1/6 & 3/6 \\ \hline A & \omega+1 & 0 & 2/6 & 2/6 & 2/6 \\ B & & 0 & 1/6 & 1/6 & 4/6 \\ \hline A & \omega+2 & 0 & 2/6 & 3/6 & 2/6 \\ B & & 0 & 1/6 & 3/6 & 2/6 \\ \hline \end{array}

and have your celebratory drink from cup $3$.