Chocolates and jelly beans

Idea. In the long run, the flow from chocolates to jelly beans must balance the flow back. That one equation gives the stationary distribution of a two-state chain.

Step 1: balance. Let $c$ and $j$ be the long-run fractions of chocolates and jelly beans. Switches from chocolate to jelly bean happen at rate $c\cdot \frac{4}{5}$, and back at rate $j\cdot \frac{2}{3}$. These must be equal: $$\frac{4}{5}c=\frac{2}{3}j.$$

Step 2: solve. With $c+j=1$: $$j=\frac{4/5}{4/5+2/3}=\frac{12/15}{22/15}=\frac{6}{11}.$$

So $\frac{6}{11}\approx 55\%$ of the candies are jelly beans.

In general. For a two-state chain that leaves state A with probability $a$ and state B with probability $b$, the long-run fraction of time in B is $\frac{a}{a+b}$.