猫狗排队：求狗的比例

设猫与狗的比例分别为 $c,d$，$c+d=1$。

由“猫后面跟狗”的结构，可得狗的比例满足

$$d=\frac{6}{7}c\left(1+\frac{3}{4}+\left(\frac{3}{4}\right)^2+\cdots\right)=\frac{6}{7}c\cdot\frac{1}{1-3/4}=\frac{24}{7}c.$$

代入 $c+d=1$ 得 $c=\frac{7}{31}$，所以

$$d=\frac{24}{31}.$$

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Original Explanation

Let $c$ and $d$, respectively, be the proportion of cats and dogs on the sidewalk. We know $c + d = 1$. Let's find $d$ in terms of $c$. Note that of all the cats, $\frac{6}{7}$ are followed by a dog. Then, of those dogs, $\frac{3}{4}$ of them are followed by another dog. Of those dogs, $\frac{3}{4}$ are followed by another dog, etc. Therefore, we can write

$$d = \frac{6}{7}c + \frac{6}{7} \cdot \frac{3}{4}c + \frac{6}{7} \cdot \left(\frac{3}{4}\right)^2 c + \dots = \frac{6c}{7} \sum_{k=0}^{\infty} \left(\frac{3}{4}\right)^k = \frac{24c}{7}$$

Substituting this into our initial equation,

$$\frac{31c}{7} = 1$$

so

$$c = \frac{7}{31}$$

This means

$$d = \frac{24}{31}$$

Alternatively, you can create a Markov chain representing this scenario. Namely, if state $1$ is dog and state $2$ is cat, the Markov chain is

$$\begin{bmatrix} 3/4 & 1/4 \\ 6/7 & 1/7 \end{bmatrix}$$

The steady state of this Markov chain yields the same answer as above.