There are several problems with your solution to 1. You are essentially picking $k$ throws to ask for a $6$ and ignoring the rest of the $n-k$ throws, which might give you a $6$. Then your figure $(5/6)^k$ is the chance that all those throws are non-sixes, so your figure $1-(5/6)^k$ is the chance of at least one six among the specific $k$ throws you picked.
What you have to do is pick a number of sixes to get, compute the probability of that many sixes, and sum over the allowable numbers of sixes. To get exactly $j$ sixes, you choose the $j$ rolls out of the $n$ (how many ways?), require that those are all sixes, and require that all the rest are non-sixes. Then sum these from $j=k$ to $j=n$
When you roll a die until $6$ appears, you can represent the sample space as all possible finite sequences from the set $\{1, 2, 3, 4, 5, 6\}$ ending in $6$, with probability of any sequence of length $k$ being $(1/6)^k$. The original question is asking for
$(1)$ the expected length of a sequence conditional on all throws being even.
You've correctly enumerated all sequences from $\{2,4,6\}$ that end in $6$, and calculated the sum
$$\sum_{n=1}^\infty n\left( {\frac{1}{6}} \right)^n 2^{n-1} = 0.375$$
properly, but you forgot to divide this by the probability of the event you are conditioning on, which is
$$
\sum_{n=1}^\infty\left(\frac16\right)^n2^{n-1}=1/4.
$$
So your approach does yield the correct answer, namely $4\times 0.375=1.5$.
The act of conditioning on all throws being even is tantamount to restricting the sample space to all possible finite sequences from the set $\{2, 4, 6\}$ that end in $6$, and rescaling the probability function (by a factor $4$) so that this new sample space has total mass $1$.
As for the Jin paper, he claims that the original question $(1)$ is equivalent to
$(2)$ the expected number of times you can roll only $2$'s or $4$'s until you roll a $6$.
I disagree with $(2)$; it is incorrect to compute an unconditional expectation, as he just explained in his previous paragraph. He still needs an expectation conditional on some event, and I would argue the original question $(1)$ is equivalent to computing
$(2')$ the expected number of times you can roll only $2$'s or $4$'s until you roll any other number, given that the other number is $6$.
The reason is that conditioning on the event "the other number is $6$" results in the same restricted sample space as before. In fact his subsequent argument that it suffices to compute the unconditional expectation
$(3)$ the expected number of times you can roll only $2$'s or $4$'s until you roll any other number.
(i.e., that $(2') = (3)$, which is what he actually proves) is relevant only if we intend $(2')$ instead of $(2)$.
EDIT: Here's a Python simulation of the experiment, based on code provided by @thecoder:
import random
times = 0 #number of times a successful (all-even) sequence was rolled
rolls = 0 #total of all number of rolls it took to get a 6, on successful sequences
curr = 0
alleven = True
for x in range(0, 100000):
num = random.randint(1,6)
if num % 2 != 0:
alleven = False
else:
if num == 6:
if alleven:
times += 1
rolls += curr + 1
curr = 0
alleven = True
else:
curr += 1
print(rolls * 1.0 / times)
#1.51506456241
Best Answer
If the question you found is this question, the expectation $E(X)=\frac{91}{36}$ is referring to part (b). You are correct that in part (a), $E(X)=\frac72$.
To get the value $\frac{91}{36}$, construct a $6\times 6$ table of all outcomes of the experiment. The $36$ possible values for (first roll, second roll) are $$ \begin{matrix}(1,1)&(1,2)&(1,3)&(1,4)&(1,5)&(1,6)\\ (2,1)&(2,2)&(2,3)&(2,4)&(2,5)&(2,6)\\ \cdots&\cdots&\cdots&\cdots&\cdots&\cdots&\\ (5,1)&(5,2)&(5,3)&(5,4)&(5,5)&(5,6)\\ (6,1)&(6,2)&(6,3)&(6,4)&(6,5)&(6,6) \end{matrix} $$ Here are the corresponding values for $X:=$ smaller of the two rolls: $$ \begin{matrix}1&1&1&1&1&1\\ 1&2&2&2&2&2\\ \cdots&\cdots&\cdots&\cdots&\cdots&\cdots&\\ 1&2&3&4&5&5\\ 1&2&3&4&5&6 \end{matrix} $$ Since all $36$ outcomes are equally likely, count up the occurrences of $1,2,\ldots,6$ to obtain $P(X=1)=\frac{11}{36}$, $P(X=2)=\frac{9}{36}$, $P(X=3)=\frac7{36}$, $P(X=4)=\frac5{36}$, $P(X=5)=\frac3{36}$, $P(X=6)=\frac1{36}$, and finally compute $E(X)=\frac{91}{36}$.