comparison of first hitting times re: discrete, symmetric random walk on $\mathbb{Z}$

Let $$c, d \in \mathbb{Z}$$ be such that $$c < 0 < d$$. Let $$\tau_k = \inf_{n \geq 0} \left\{n: S_n = k\right\}$$ be the first hitting time of state $$k$$, where $$S_n$$ denotes the position of a symmetric random walk on $$\mathbb{Z}$$ after $$n$$ steps. The question is to show that $$\mathbb{P}(\tau_D < \tau_C) = \frac{-c}{d-c}$$. The random walk starts at position $$0$$ at time $$0$$.

I have been beating my head on this for a little bit, and I suspect it is a lot simpler than I am thinking. I played around with finding the probability of $$\tau_A = n$$ but wasn't able to use that to find the answer to the question. Any suggestions?

Hint: Let $$\tau=\min(\tau_C,\tau_D)$$. By Wald's equation, $$E[S_\tau]=0$$. On the other hand, $$S_\tau$$ is equal to either $$d$$ or $$c$$, according to whether $$\tau_C$$ is less than or more than $$\tau_D$$.
There is the hairy detail that in order to use Wald's equation, you need to show that $$E[\tau]<\infty$$. To prove this, note that if $$X_i=1$$ for $$d-c$$ steps in a row at any point, then the process will leave the interval $$[c,d]$$ if it has not already. This shows that $$P(\tau>(d-c)k)\le (1-\frac1{2^{d-c}})^k$$. This bound on the tail probabilities of $$\tau$$ can be used to show $$E[\tau]$$ is finite.