Probability kth outcome is not a repetition 
A fair die is tossed $k$ times. What is the probability that the kth
outcome is ${\bf not}$ a repetition?

attempt.
First, the size of our sample space is $6^k$. We only have six possible outcomes, so for any outcome, say $O$, that occur at the kth try, we want the previous outcomes (k-1 tries) to be not $O$, so in how many ways can we do this?
Well, $5 \cdot ... \cdot 5 \cdot 1= 5^k$ so
$$ Prob = \frac{5^k}{6^k} $$
is this correct? How would we handle the case when we have $2$ dice instead?
 A: Your approach looks sensible, but would lead to $\dfrac{5^{k-1}}{6^{k-1}}$ for one die
Another way to get there would be consider the possible outcomes $1,2,3,4,5,6$ for the $k$th die and combine the probabilities of these with the conditional probabilities of the previous dies being different so $\frac16\left(\frac56\right)^{k-1} + \frac16\left(\frac56\right)^{k-1} + \frac16\left(\frac56\right)^{k-1} + \frac16\left(\frac56\right)^{k-1} + \frac16\left(\frac56\right)^{k-1} + \frac16\left(\frac56\right)^{k-1}$ to give the same result
So for two dice following a similar approach you would then get $\frac1{36}\left(\frac{36-1}{36}\right)^{k-1} + \frac2{36}\left(\frac{36-2}{36}\right)^{k-1} + \frac3{36}\left(\frac{36-3}{36}\right)^{k-1} + \frac4{36}\left(\frac{36-4}{36}\right)^{k-1} +$ $\frac5{36}\left(\frac{36-5}{36}\right)^{k-1} + \frac6{36}\left(\frac{36-6}{36}\right)^{k-1} + \frac5{36}\left(\frac{36-5}{36}\right)^{k-1} + \frac4{36}\left(\frac{36-4}{36}\right)^{k-1} + $ $\frac3{36}\left(\frac{36-3}{36}\right)^{k-1} + \frac2{36}\left(\frac{36-2}{36}\right)^{k-1} + \frac1{36}\left(\frac{36-1}{36}\right)^{k-1}$ 
A: Your description (in English) of how you wanted to count the possible outcomes is good. You stumbled a bit when writing it as a formula.
You have six different ways for the $k$th throw to come out.
Then for any one of these six outcomes, you have $k - 1$ previous outcomes, each of which could come out any of five ways without causing the $k$th roll to be a repetition.
That gives you
$$
\underbrace{6}_{\text{possible}\\\text{outcomes of}\\\text{the $k$th roll}}
\cdot \underbrace{5 \cdot \cdots \cdot 5}_{\text{$k-1$ rolls,}\\\text{not $k$}} = 6\cdot 5^{k-1}.
$$
Your last step, dividing by $6^k,$ is correct when you count the outcomes this way.
