Milk and Coffee will they ever finish Let two glasses, numbered 1 and 2, contain an equal quantity
of liquid, milk in glass 1 and coffee in glass 2. One does the following: 


*

*Take one spoon of mixture from glass 1 and pour it into glass 2, and 

*then take the same spoon of the new mixture from glass 2 and pour it back into the
first glass. 


What happens to the concentrations of milk and coffee in the glasses after this operation is repeated $n$ times, and what as $n$ tends to infinity?
 A: Suppose the first glass contains a fraction $x$ of milk, and a fraction $1-x$ of coffee. Initially, before you did the transfer, the total quantity of milk was the volume of one glass. Now, as $x$ of that volume is in the first glass, the remaining $1-x$ of the milk must be in the second glass! So the second glass contains a fraction $1-x$ of milk, and therefore $x$ of coffee. After you do the operation one time, the amount of coffee in the first glass and the amount of milk in the second glass are the same.
Clearly, as you do this $n$ times, the property is maintained. At any point of time, after $n$ steps, if $x_n$ is the fraction of milk in the first glass, then the first glass has $1-x_n$ of coffee, and the second glass has $x_n$ of coffee and $1-x_n$ of milk. As $n \to \infty$, because of all the mixing, both glasses will tend to contain the liquids in the same composition, i.e., $x_n \to \frac{1}{2}$.

This much we did without any calculation beyond using the fact that $x + (1-x) = 1$. If we want to calculate the actual value of $x_n$, we need to do some work. But this also depends on the volume of the spoon (how big the spoon is relative to the glass). Say the volume of the spoon is some fraction $s$ of the glass's volume. Then, after the $n$th operation,


*

*We start with $x_n$ milk and $1 - x_n$ coffee in the first glass.

*We transfer one spoon to the second glass: this spoon-in-transfer accordingly contains $sx_n$ of milk and $s(1-x_n)$ of coffee.

*We put this spoon in the second glass: before the spoon came in, the glass contained $1 - x_n$ milk and $x_n$ coffee; now it contains $1-x_n + sx_n= 1 + (s-1)x_n$ milk and $x_n + s(1-x_n) = s + (1-s)x_n$ coffee. The sum of the two volumes is $(1 + s)$ (as expected), so the fraction of milk is $\dfrac{1 + (s-1)x_n}{1+s}$.

*When we transfer a spoon of this back to the first glass, it carries $s\dfrac{1 + (s-1)x_n}{1+s}$ of milk, and the rest coffee.

*So the total quantity of milk in the first glass after the transfer is 
$$x_{n+1} = (1-s)x_n + s\frac{1 + (s-1)x_n}{1+s} = \frac{s + (1-s)x_n}{1+s}$$
and that of coffee is
$$1 - x_{n+1} = \frac{1 - (1-s)x_n}{1+s}.$$


Just as a sanity check, note that if $s = 0$ (empty spoon) nothing changes, and if $s = 1$ (you transfer the whole thing in one gulp), $x_{n+1} = \frac12$ immediately.
To show the convergence of $x_n$, we look at 
$$\frac{1}{2}-x_{n+1} = \frac{1-s}{1+s} \Big(\frac{1}{2}-x_n\Big).$$
$0<\frac{1-s}{1+s}<1$, so $\frac{1}{2}-x_n$ converges to $0$ monotonically and exponentially fast.
A: Let one glass of liquid be 1. Let $X_n$ be the percentage of milk in glass 1 and $Y_n$ be the percentage of milk in glass 2 after the $n$th operation. Let a spoon be $\epsilon$ times the size of one glass. We have
$$ Y_{n+1} = \frac{Y_n + \epsilon X_n}{1 + \epsilon} $$
and
$$ X_{n+1} = (1 - \epsilon)X_n + \epsilon Y_{n+1} = (1-\epsilon)X_n + \frac{\epsilon}{1+\epsilon}(Y_n + \epsilon X_n) = \frac{X_n + \epsilon Y_n}{1+\epsilon}$$
We can write this as a dynamical system:
$$ \begin{pmatrix} X_{n+1} \\ Y_{n+1} \end{pmatrix} = \begin{pmatrix}
\frac{1}{1+\epsilon} & \frac{\epsilon}{1+\epsilon} \\
\frac{\epsilon}{1+\epsilon} & \frac{1}{1+\epsilon}
\end{pmatrix}\begin{pmatrix} X_n \\ Y_n\end{pmatrix} $$
Its eigenvalues are $1$ and $\frac{1-\epsilon}{1+\epsilon} < 1$. Since the second eigenvalue is $<1$ we have that as $n \to \infty$ the configuration will converge to $1$ eigenspace, which is given by $X = Y$. 
Therefore: independently of the initial amounts of milk and coffee in either cup, the limiting distribution is for the amount of milk to be equal in the two cups and $n\to \infty$. 
