Expected value to get number 1 when roll the dice three times

Question

I fighting with one practise. Questions is what is expected number of 1s when tossing a die (6 faces) three times?

For one die it is $(1/6)\times(1+2+3+4+5+6)=3.5$.

Answer 1

Since you give the expected value of one die, are you asking about the expected total of three dice? If so, this is just three times the expected total for one die, by linearity of expectation, so $10.5$.

Similarly, the expected number of 1s is three times what it would be for one die, if that is what you are looking for.

Answer 2

The expected number of $1$s from one dice roll is $\frac16$

So, by linearity of expectation, the expected number of $1$s from three dice rolls is three times that, i.e. $\frac12=0.5$

Answer 3

Let us introduce a random variable $$X=\text{number of 1s within 3 dice tossings}.$$

The possible values of $X$ are 0, 1, 2, 3.

$X$ follows binmial distribution with parameters $n = 3$ and $p = \frac{1}{6}$, i.e. $X\sim Bin\left(n=3, p = \frac{1}{6}\right)$.

Then, the expected value of $X$ is $$E\left[X\right] = n\cdot p = 3\cdot \frac{1}{6} = 0.5$$

Answer 4

You can use Indicator random variable for this, the solution is quite elegant.

Consider $I_i$ an indicator variable that is 1 if the ith dice rolls 1 or 0 otherwise. In this case a r.v. X that represents the number of ones out of the 3 rolls can be written as:

$$X = I_1 + I_2 + I_3$$

And the expectation: $$E(X) = E(I_1) + E(I_2) + E(I_3) = P(I_1) + P(I_2) + P(I_3) = 3 \times \frac{1}{6} = \frac{1}{2}$$