Expectation of a die roll with a coin toss

Question

Let's say we throw a coin and pick a number from [0,1] uniformly. Then we roll a die, and if the coin is heads, we add the number we picked to the number on the die, otherwise we subtract it. What is the expected value of a random variable which is the output of this routine? I have an intuition that it is the same as just rolling a die but cannot prove it.

Answer 1

Let's say $C = 1$ if coin was heads, or $-1$ otherwise and your number is $u \sim \mathcal{U}(0,1)$, and $D$ is the die roll outcome. You are looking for $$ \begin{split} \mathbb{E}[D + Cu] &= \mathbb{E}[D] + \mathbb{E}[Cu] &\quad \text{by linearity of expectation}\\ &= \mathbb{E}[D] + \mathbb{E}[C] \cdot \mathbb{E}[u] &\quad \text{by independence of } C,u \\ &= \mathbb{E}[D] \end{split} $$ since $\mathbb{E}[C] = 0$ if the coin was fair...