Dice rolls probability Given a six-faced die with six distinctly marked faces, what is the probability that a toss will turn-up any one of the six faces. If three of the six faces are red and three blue, what is the likelihood that a toss shall turn-up red? Shall turn-up blue? Replace ‘red’ by the number ‘1’, blue by the number ‘2’. Give the expectation value (mean) to be associated with tosses of this die. What is the expectation value for the cube of the values marked on the faces? I'm pretty certain the first part is 1/6 because there are 6 faces. Then it would be 1/2 red and 1/2 blue. Then I'm not sure how to find the expectation value of the tosses. Some help would be appreciated. Thank you.

To find the expected value, take the sum of the products of the value of an outcome and the probability of that outcome. Intuitively, this is a weighted average of the outcomes. If you were to roll this dice some large amount of times, add 1 or 2 to the sum depending on whether it landed on red or blue, and then take the average, you should _expect_ an answer close to the expected value.

In this case, "red" has a value of $1$ and "blue" has a value of $2$. Each has probability of $\frac{1}{2}$, so this gives

$$\text{Expected value }= \frac{1}{2} (1) + \frac{1}{2} (2) = \frac{3}{2}.$$