The easiest and most elegant way to find a formula for this is to show you an equivalent problem. This is the case of indistinguishable objects in distinguishable boxes.
Rather than thinking that a Skittle is a particular color, let's have identical beads be placed in one of $5$ boxes, and depending on which box it is in determines its color of the Skittle.
Say we have boxes labeled "red", "green", "yellow", "orange", and "purple". If there were two red Skittles, that's the same as saying both beads were in the red box. If there was one green and one orange Skittle, then one bead was in the green box and the other in the orange box.
So we can count up the ways of putting the beads in boxes and it would be the same thing.
Now, instead of $5$ boxes, let's just split an area into $5$ regions using dividers. Let $|$ be a divider, like so:
$$
\square | \square | \square | \square | \square
$$
Now let's use stars for our beads, so the two red Skittles example could be represented with:
$$
\star \star | | | |
$$
Finally, we count up the ways of ordering the stars and stripes, which is the binomial coefficient $\binom{6}{2}$. In general, for $n$ Skittles and $k$ colors, the formula is $\binom{n+k-1}{k}$.
This kind of question is very sensitive to how the roller decides what to say. Maybe the dice are different colors and he will tell you the number on the green die. You know nothing about the other die, so all numbers are equally probable. Maybe he will tell you the higher number. Now the only rolls possible are $13,23,33,31,32$ so you should bet on $1$ or $2$. Maybe he will pick a random die and tell you the number on it. Now he could have rolled $33$ and been forced to say $3$ or rolled $3x$ or $x3$ and chosen to say $3$. You have the same chance of $x$ as $3$ for the other die so bet on anything you like. If you don't know how the roller chooses what to say it is not a mathematical question. There is not enough information to answer.
Best Answer
For your first guess you would have been right if your friend had rolled a $1$ with the $2$ and $4$. Because you have three different dice, there are six ways this could have happened: $$1,2,4\quad\hbox{or}\quad 1,4,2\quad\hbox{or}\quad 2,1,4\quad\hbox{or}\quad 4,1,2\quad\hbox{or}\quad 2,4,1\quad\hbox{or}\quad 4,2,1\,.$$ The same thing happens if he had rolled a $3,5$ or $6$. If the rolls were $2,2,4$ it's a bit different and there are only three possibilities: $$2,2,4\quad\hbox{or}\quad 2,4,2\quad\hbox{or}\quad 4,2,2\,.$$ And the same for $2,4,4$. So altogether you would have been right in $30$ cases and the chance of your first guess being right is $30/6^3$. You can figure out the probability of the second guess being right in a similar way - it's actually easier than the first one - and then combine them to get your final answer.