Okay here is an idea. All throws are independent. This means if you know that you did not succeeded in the first try, this is equivalent to starting all over. However you already got 'a' points so you have to add them, this means: E(X | Y = a) = a + E(X) for a!=n. It may look confusing but this is due to the fact that overall the expected value does not increase because in the last case a==n you can only add 'a'. It is like a balancing act.
For your second argument: Because there is a probability to throw 'n' in the first try, you can not simply subtract (n+1)/2. I'm not able to give you a good reason, or an alternative, but I hope my answer helps nevertheless.
Answer concerning $k=2$. For $k\geq2$ see edit.
If $X_2$ denotes the number throws needed then $X_2$ only takes positive multiples of $2$ as value, and this with: $$P(X_2=2n)=2^{1-2n}C_{n-1}=2^{1-2n}\frac{(2n-2)!}{n!(n-1)!}$$for $n=1,2,\dots$, leading to:$$\mathbb EX_2=\sum_{n=1}^{\infty}2^{2-2n}\binom{2n-2}{n-1}=\sum_{n=0}^{\infty}2^{-2n}\binom{2n}{n}$$
Here $C_n$ stands for the $n$-th Catalan number.
In this case ($k=2$) we can do it with a fair coin.
Let $H_k$ stand for the number of throws with outcome heads and let $T_k$ for the number of throws with outcome tails by $k$ throws.
Then have a look at the path determined by $(k,H_k-T_k)$ for $k=0,1,\dots,2n$ under condition that $X_2=2n$.
Starting at $(0,0)$ and ending at $(2n,0)$ first WLOG we go to $(1,1)$.
Then with probability $2^{2-2n}C_{n-1}$ from there we go to $(2n-1,1)$ keeping a positive second coordinate.
And then with probability $2^{-1}$ we go to $(2n,0)$.
We have the asymptotic equality: $$2^{-2n}\binom{2n}{n}\sim\frac{1}{n^{\frac12}\sqrt\pi}$$and conclude that: $$\mathbb EX_2=+\infty$$
edit (to make the answer complete):
If $k>2$ let $U$ denote the outcome at first throw and let $V$
be (e.g.) the smallest element of $\left\{ 1,\dots,k\right\} $with $U\neq V$.
Now define $Y$ as the number of throws needed (inclusive the first)
to come back again in the situation that outcome $U$ and outcome
$V$ have appeared the same number of times.
Comparing $Y$ and $X_{2}$ we find easily that $P\left(Y>n\right)\geq P\left(X_{2}>n\right)$ for every nonnegative integer $n$ and consequently: $$\mathbb{E}Y=\sum_{n=0}^{\infty}P\left(Y>n\right)\geq\sum_{n=0}^{\infty}P\left(X_2>n\right)=\mathbb{E}X_{2}$$
Next to that it is evident that $X_{k}\geq Y$ so that $\mathbb{E}X_{k}\geq\mathbb{E}Y$.
Combining this with $\mathbb{E}X_{2}=+\infty$ we conclude that $\mathbb{E}X_{k}=+\infty$
for $k=2,3,\dots$
Best Answer
Consider the expected number of points you will obtain after rolling a particular number.
e.g. Suppose we have just rolled a 6. We need to roll another 6 to keep playing, otherwise we stop. Using conditional expectation, we can compute the number of future points we expect to obtain, $E_6$: $$E_6 = \frac{1}{6}(E_6 + 6) + \frac{5}{6}\times 3.$$
The first term corresponds to rolling a 6, and we are back in the same position as before, just with 6 extra points. The second term gives expected number of points obtained, given that we roll a lower number and stop playing. We solve this to obtain $$E_6 = \frac{21}{5}.$$
Now suppose that, in a new game, we have just rolled a 5. What is the expected number of points from this point, $E_5$? Using the same conditional expectation rules as before: $$E_5 = \frac{1}{6}(E_5+5) + \frac{1}{6}(E_6+6) + \frac{4}{6} \times \frac{5}{2}.$$ Since we know $E_6$ from above, we can now solve for $E_5$.
Repeat this for $E_4, \dots, E_1$, and we know the expected number of future points given the most recent roll number.
Finally, the expected number of points, $E$, will be the conditional sum of these values, i.e.
\begin{equation} E = \sum_{n=1}^6 \frac{1}{6}(E_n + n). \tag{1} \end{equation}
Edit: following an observation by Taner, I thought I'd add a few more lines.
After some generalizing and rearranging, we obtain $$E_n = \frac{1}{5}\left(\sum_{m=n+1}^6 E_m + 21\right),$$ and we can set up a recurrence relation for $E_n$ so we don't have to do this sum every time to compute it. We have
\begin{align} E_{n+1} &= \frac{1}{5}\left(\sum_{m=n+2}^6 E_m + 21\right) \\ &= \frac{1}{5}\left(\sum_{m=n+1}^6 E_m + 21 - E_{n+1}\right) \\ &= E_n - \frac{1}{5}E_{n+1}, \end{align}
which we rearrange to obtain $$E_{n+1} = \frac{5}{6}E_n,$$ which has solution $$E_n = C \left(\frac{5}{6}\right)^n,$$ where $C$ is some constant, for which we can solve by setting $n=6$ and using our value of $E_6$. We obtain (allowing for possible arithmetic errors made by me) $$E_n = \frac{21}{5}\left(\frac{5}{6}\right)^{n-6}.$$
We can substitute this into Equation (1) and use geometric and arithmetic sum formulae to obtain the final answer.
Edit 2: as Alex Zorn pointed out, we don't even need to do this geometric and arithmetic sum stuff in Equation (1). Note that the game is the same before and after rolling a 1, so this tells us straight away that $$E = E_1 = \frac{21}{5}\left(\frac{5}{6}\right)^{-5} = \frac{163296}{15625} \approx 10.45.$$