Note that the answer to your problem in general, can be given as follows.
Let us define:
$$
\eqalign{
& {\rm No}{\rm .}\,{\rm of}\,{\rm solutions}\,{\rm to}\;\left\{ \matrix{
{\rm 1} \le {\rm integer}\;y_{\,j} \le r + 1 \hfill \cr
y_{\,1} + y_{\,2} + \; \cdots \; + y_{\,m} = s + m \hfill \cr} \right. = \cr
& = {\rm No}{\rm .}\,{\rm of}\,{\rm solutions}\,{\rm to}\;\left\{ \matrix{
{\rm 0} \le {\rm integer}\;x_{\,j} \le r \hfill \cr
x_{\,1} + x_{\,2} + \; \cdots \; + x_{\,m} = s \hfill \cr} \right. = \cr
& = N_b (s,r,m) \cr}
$$
then we have the formula
$$
N_b (s,r,m)\quad \left| {\;0 \le {\rm integers }\;s,m,r} \right.\quad = \sum\limits_{\left( {0\, \le } \right)\,\,k\,\,\left( { \le \,{s \over r}\, \le \,m} \right)} {\left( { - 1} \right)^k \left( \matrix{
m \hfill \cr
k \hfill \cr} \right)\left( \matrix{
s + m - 1 - k\left( {r + 1} \right) \cr
s - k\left( {r + 1} \right) \cr} \right)}
$$
Also consider that the o.g.f. on the parameter $s$ is:
$$
F_b (x,r,m) = \sum\limits_{0\, \le \,\,s\,\,\left( { \le \,mr} \right)} {N_b (s,r,m)\;x^{\,s} } = \left( {1 + x^{\,1} + x^{\,2} + \; \cdots \; + x^{\,r} } \right)^{\,m} = \left( {{{1 - x^{\,r + 1} } \over {1 - x}}} \right)^{\,m}
$$
For more details have a look to the answers to this other post.
When the number of rolls ($m$) takes large values, the formula above becomes impractical and we shall resort to an asymptotic approximation.
To this regard note that each variable $x_j$ is a discrete uniform variable with support $[0,r]$, therefore with mean and variance given by $$
\mu = {r \over 2}\quad \sigma ^{\,2} = {{\left( {r + 1} \right)^{\,2} - 1} \over {12}}
$$
The sum of $m$ such variables tends very quickly to be Normally distributed with mean $m \mu$ and variance $m\sigma ^2$, that is
$$
p(s,r,m) = {{N_{\,b} (s,r,m)} \over {\left( {r + 1} \right)^{\,m} }}\;\; \to \;{\cal N}\left( {m{r \over 2},\;m{{\left( {r + 1} \right)^{\,2} - 1} \over {12}}} \right)
$$
Refer also to this related post.
Best Answer
You’re right that the first roll can be anything and that the probability of getting something different on the second roll is $1-\frac16=\frac56$. However, the third roll has to differ from both of the first two in order for you to get three different numbers, and that happens with probability $\frac46=\frac23$. Moreover, you need to multiply the probabilities, so that even if the correct probability for the third roll were $1-\frac16$, your answer would be $\left(1-\frac16\right)^2=\left(\frac56\right)^2$, not $1-\left(\frac16\right)^2$.
Thus, the actual probability of getting three different numbers is $$\frac56\cdot\frac23=\frac59\;.$$