If we have three dice: red, green, blue dice. Why is the sample space $6^3$? What I think is, it should be $3! \times 6^3$. Because we can have top sides show 3R, 2G, 1B and also 3G, 2B, 1R.
And if I want to calculate the probability of getting two sixes, should I multiply it by $3!$? the reason is the same, it can be 6R, 1G, 6B or 6B, 1G, 6R?
Best Answer
Each element in the sample space can be represented by the ordered triple $(b, g, r)$, where $b$ represents the outcome on the blue die, $g$ represents the outcome on the green die, and $r$ represents the outcome on the red die. Since $b, g, r \in \{1, 2, 3, 4, 5, 6\}$, there are six possible values for each of the three entries. Therefore, the size of the sample space is $6^3 = 216$.
In the comments, you asked why we use different colors for the dice. If we used three white dice, there would be $$\binom{6}{1} + \binom{2}{1}\binom{6}{2} + \binom{6}{3}$$ distinguishable outcomes. The first term represents the number of outcomes in which all three dice show the same value. The second term represents the number of outcomes in which two dice show one value and the other die shows a different value. The factor of $\binom{6}{2}$ represents the two values shown by the three dice. The factor $\binom{2}{1}$ represents the two ways one of those two values can appear twice. The third term represents the number of outcomes in which the three dice show different values.
We do not use this sample space since the outcomes are not equally likely to occur. Each of the six outcomes in which all three dice show the same value can occur in only one way. However, each of the $30$ outcomes in which two of the dice show one value and the other die shows a different value can occur in three ways and each of the $20$ outcomes in which the dice show three different values can occur in $3! = 6$ ways. Notice that $$\binom{6}{1} + 3 \cdot \binom{2}{1}\binom{6}{2} + 6 \cdot \binom{6}{3} = 216$$
As for the probability that exactly two sixes occur when the three different color dice are shown, observe that there are $\binom{3}{2}$ ways to choose which two of the three distinguishable dice show a $6$ and $5$ ways for the third die to show a value other than $6$. Hence, the number of favorable cases is $$\binom{3}{2} \cdot 5$$ so the desired probability is $$\frac{\binom{3}{2} \cdot 5}{6^3} = \frac{3 \cdot 5}{216} = \frac{5}{72}$$ If you instead meant at least two sixes, we gain the additional favorable outcome that three sixes occur, in which case the probability is $$\frac{\binom{3}{2} \cdot 5 + 1}{6^3} = \frac{3 \cdot 5 + 1}{216} = \frac{16}{216} = \frac{2}{27}$$