What is the probablity that the sum of two dice is 4 or 6?

In developing an understanding of the sample space, you should think of the two dice as being distinct. Yes, in principle, the two dice may be indistinguishable. However, from the point of view of probability theory, the behaviour of each die is described by a different random variable. Hence the two dice are not, really, indistinguishable. Maybe one of them is blue and the other red, or one is rolled before the other, or one has a cute little smiley face drawn in place of the $3$.

From this point of view, the outcome $1+3$ represent rolling a $1$ and then a $3$, while the outcome $3+1$ represents rolling a $3$ and then a $1$. In terms of the ultimate nature of the events being modeled, these two outcomes are the same (both yield a $4$), but as outcomes (or elementary events) in the underlying sample space, they are distinct. To make this more concrete, imagine that the dice are rolled one after the other, and suppose that they are colored differently. For example, if one of the dice is blue and the other red, then the sample space looks like

$$ begin{array}{c|cccccc} & color{red}{1} & color{red}{2} & color{red}{3} & color{red}{4} & color{red}{5} & color{red}{6} \hline color{blue}{1} & (color{blue}{1}, color{red}{1}) & (color{blue}{1}, color{red}{2}) & (color{blue}{1}, color{red}{3}) & (color{blue}{1}, color{red}{4}) & (color{blue}{1}, color{red}{5}) & (color{blue}{1}, color{red}{6}) \ color{blue}{2} & (color{blue}{2}, color{red}{1}) & (color{blue}{2}, color{red}{2}) & (color{blue}{2}, color{red}{3}) & (color{blue}{2}, color{red}{4}) & (color{blue}{2}, color{red}{5}) & (color{blue}{2}, color{red}{6}) \ color{blue}{3} & (color{blue}{3}, color{red}{1}) & (color{blue}{3}, color{red}{2}) & (color{blue}{3}, color{red}{3}) & (color{blue}{3}, color{red}{4}) & (color{blue}{3}, color{red}{5}) & (color{blue}{3}, color{red}{6}) \ color{blue}{4} & (color{blue}{4}, color{red}{1}) & (color{blue}{4}, color{red}{2}) & (color{blue}{4}, color{red}{3}) & (color{blue}{4}, color{red}{4}) & (color{blue}{4}, color{red}{5}) & (color{blue}{4}, color{red}{6}) \ color{blue}{5} & (color{blue}{5}, color{red}{1}) & (color{blue}{5}, color{red}{2}) & (color{blue}{5}, color{red}{3}) & (color{blue}{5}, color{red}{4}) & (color{blue}{5}, color{red}{5}) & (color{blue}{5}, color{red}{6}) \ color{blue}{6} & (color{blue}{6}, color{red}{1}) & (color{blue}{6}, color{red}{2}) & (color{blue}{6}, color{red}{3}) & (color{blue}{6}, color{red}{4}) & (color{blue}{6}, color{red}{5}) & (color{blue}{6}, color{red}{6}) \ end{array} $$ This sample space gives all of the possible outcomes, of which there are $36$. Moreover, and of crucial importance, is that every one of these outcomes is equally likely. Showing that all of these outcomes are equally likely requires a little bit of work, but the essential ideas are that

• each die is modeled by a uniform variable on the set ${1,2,3,4,5,6}$, which means that the probability of rolling any particular number on either die is $1/6$, and

• the two die rolls are independent, which means (more or less) that the number rolled on one of the two dice does not depend on the result of the other.

These two observations are sufficient to show that all $36$ outcomes are equally probable.

However, we are only interested in the event which describes a sum of either $4$ or $6$. Replacing the specific die rolls with the corresponding sums in the table above gives $$ begin{array}{c|cccccc} & color{red}{1} & color{red}{2} & color{red}{3} & color{red}{4} & color{red}{5} & color{red}{6} \hline color{blue}{1} & 2 & 3 & boxed{4} & 5 & boxed{6} & 7 \ color{blue}{2} & 3 & boxed{4} & 5 & boxed{6} & 7 & 8 \ color{blue}{3} & boxed{4} & 5 & boxed{6} & 7 & 8 & 9 \ color{blue}{4} & 5 & boxed{6} & 7 & 8 & 9 & 10 \ color{blue}{5} & boxed{6} & 7 & 8 & 9 & 10 & 11 \ color{blue}{6} & 7 & 8 & 9 & 10 & 11 & 12\ end{array} $$ From this presentation, it can be seen that there are $8$ "good" outcomes (i.e. there are $8$ elementary events giving a sum of either $4$ or $6$), out of a total of $36$ equiprobable outcomes. Thus $$ P(text{two dice sum to either $4$ or $6$}) = frac{8}{36} = frac{2}{9}. $$

Here's another way to think of this question.

Take a coin. There are two sides, Heads and Tails. With your logic then HT and TH is the same. Does that mean the probability of getting one head and one tail is $frac{1}{4}$? No, it's $frac{1}{2}$ because you can arrange the ways.

-FruDe

Here is a simple way to visualize the sample space: begin{array}{c|llllll} + & 1 & 2 & 3 & 4 & 5 & 6 \ hline 1 & 2 & 3 & color{blue}{4} & 5 & color{blue}{6} & 7\ 2 & 3 & color{blue}{4} & 5 & color{blue}{6} & 7 & 8\ 3 & color{blue}{4} & 5 & color{blue}{6} & 7 & 8 & 9\ 4 & 5 & color{blue}{6} & 7 & 8 & 9 & 10\ 5 & color{blue}{6} & 7 & 8 & 9 & 10& 11\ 6 & 7 & 8 & 9 & 10& 11 & 12\ end{array} In blue we have the favourable outcomes. Hence, the probability of getting a sum of $4$ or $6$ is $8/36$, or $2/9$, as desired.

Write down $Omega$

$Omega={(1;1),(1;2),...,(6;6)}$

And count the favourable pairs...