No.
Throwing two dice is the same as throwing a dice twice.
Let's think naturally:
What is the set of outcomes when a die is thrown twice?
It is just the set of all ordered pairs such that, the first of the two entries tell you the outcome on the first throw and the second of the entries tell you the outcome on the second throw.
What is the set of outcomes when two dice are thrown?
Without loss of generality, I can label the dice as die A and die B. So, now, the set of outcomes will be the set of ordered pairs such that he first of the two entries tell you the outcome on die A and the second of the entries tell you the outcome on the die B.
So, radically, the set of outcomes, technically called the sample space, is the same.
To conduct an experiment with $100$ dice, if you have the means, you could buy $100$ dice and use a method that is reasonably unbiased in generating outcomes, save time or throw a single die $100$ times, the outcomes will be unbiased but you'll waste time.
The point is both methods have their own practical merits and demerits.
Here's a heavy-handed approach. After zero or more rolls, you are in one of four situations:
$\begin{align}
\emptyset:&\qquad\textrm{No 5 or even rolled yet.}\\
E:&\qquad\textrm{Even was rolled, but no 5 yet.}\\
5:&\qquad\textrm{A 5 was rolled, but no even yet.}\\
*:&\qquad\textrm{Both 5 and even have been rolled. Game over.}\\
\end{align}$
The transition matrix of probabilities between each pair of situations is easy to compute:
$\begin{array}{l|cccc}
\nearrow&\emptyset&E&5&*\\
\hline
\emptyset&\frac{1}{3}&\frac{1}{2}&\frac{1}{6}&0\\
E&0&\frac{5}{6}&0&\frac{1}{6}\\
5&0&0&\frac{1}{2}&\frac{1}{2}\\
*&0&0&0&1\\
\end{array}$
So this is now modeled as a absorbing Markov chain with transition matrix
$\left({\begin{array}{cccc}
\frac{1}{3}&\frac{1}{2}&\frac{1}{6}&0\\
0&\frac{5}{6}&0&\frac{1}{6}\\
0&0&\frac{1}{2}&\frac{1}{2}\\
0&0&0&1\\
\end{array}}\right)$
The final state being listed last, the behavior is characterized by the $3\times3$ matrix in the upper left, which is the transition matrix for the non-final states.
$Q=\left({\begin{array}{ccc}
\frac{1}{3}&\frac{1}{2}&\frac{1}{6}\\
0&\frac{5}{6}&0\\
0&0&\frac{1}{2}\\
\end{array}}\right)$
The so-called fundamental matrix $N$ for this chain is
$N=(I-Q)^{-1}
=\left({\begin{array}{ccc}
\frac{3}{2}&\frac{9}{2}&\frac{1}{2}\\
0&6&0\\
0&0&2\\
\end{array}}\right)
$.
The expected number of steps from the $i$-th state to the final one is the sum of the entries of the $i$-th row of $N$, or equivalently the $i$-th entry of the matrix
${\bf t}=N\mathbb{1}=\left({\begin{array}{c}
\frac{13}{2}\\
6\\
2\\
\end{array}}\right)$,
so for the starting state $\emptyset$, it's $\frac{13}{2}$ steps.
The variance of the number of steps from the $i$-th state is the $i$-th entry in the matrix
$(2N-I){\bf t-t_{\textrm sq}}$,
where $t_{\textrm sq}$ is the matrix $\bf t$ with each entry squared. If I didn't slip up with Mathematica,
$(2N-I){\bf t-t_{\textrm sq}}=\left({\begin{array}{c}
\frac{107}{4}\\
30\\
2\\
\end{array}}\right)$,
and the variance you want is $\frac{107}{4}$
Best Answer
I think you need to work through it again. If getting an even number is twice as likely as getting an odd number, then it's as if you had a 9-sided dice with two 2's, two 4's and two 6's.