Expected number of cards drawn to get 4 consecutive spades Here's a question from my probability textbook:

A person draws cards one by one from a pack and replaces them till he has drawn four consecutive spades. How many cards may he expect to draw?

I ended up solving the equation$$e = {3\over4}(e + 1) + {3\over{4^2}}(e + 2) + {3\over{4^3}}(e + 3) + {1\over{4^4}}4,$$getting $e = 82$.
However, the answer in the back of my book is $340$. And the answer following the equations given here:
The expected number of draws until four successive cards of the same suit appear


*

*$\mu_{3}=1+\frac{3}{4}\mu_{1}$

*$\mu_{2}=1+\frac{3}{4}\mu_{1}+\frac{1}{4}\mu_{3}$

*$\mu_{1}=1+\frac{3}{4}\mu_{1}+\frac{1}{4}\mu_{2}$

*$\mu_{0}=1+\mu_{1}$

Gets us the answer $\mu_0 = 85$ according to Wolfram Alpha: https://www.wolframalpha.com/input/?i=d+%3D+1+%2B+%283%2F4%29b%2C+c+%3D+1+%2B+%283%2F4%29b+%2B+%281%2F4%29d%2C+b+%3D+1+%2B+%283%2F4%29b+%2B+%281%2F4%29c%2C+a+%3D+1+%2B+b
So there's three possible answers of $82$, $340$, $85$. Which one is correct?
EDIT: I see the error of my ways, my equation is missing a term. It should be$$e = {3\over4}(e + 1) + {3\over{4^2}}(e + 2) + {3\over{4^3}}(e + 3) + {3\over{4^4}}(e + 4) + {1\over{4^4}}4,$$getting $e = 340$ as desired.
 A: Your equation is missing a term for the case where you begin by drawing $3$ spades followed by a non-spade. If you change the equation to
$$
   e = {3\over4}(e + 1) + {3\over{4^2}}(e + 2) + {3\over{4^3}}(e + 3) + \frac3{4^4} (e+4) + {1\over{4^4}}4,
$$
then you get the correct solution $e=340$.

The logic here, for future readers, is that we split into $5$ cases for what the first draws from the deck are. If $\textsf S$ stands for "spade" and $\textsf N$ stands for "non-spade", then:

*

*With probability $\frac34$ we draw $\textsf N$ and we're back where we started with $1$ extra card drawn (in expectation, $e+1$ total).

*With probability $\frac14 \cdot \frac34$ we draw $\textsf {SN}$ and we're back where we started with $2$ extra cards drawn (in expectation, $e+2$ total).

*With probability $\frac14 \cdot \frac14 \cdot \frac34$ we draw $\textsf{SSN}$ and we're back where we started with $3$ extra cards drawn (in expectation, $e+3$ total).

*With probability $\frac14 \cdot \frac14 \cdot \frac14 \cdot \frac34$ we draw $\textsf{SSSN}$ and we're back where we started with $4$ extra cards drawn (in expectation, $e+4$ total).

*With probability $\frac14 \cdot \frac14 \cdot \frac14 \cdot \frac14$ we draw $\textsf{SSSS}$ and then, after $4$ draws, we're done.

We combine these with the law of total expectation to get the equation above.
