Quick probability question on rolling 2 Die If I flip a coin and it lands on heads I roll a fair dice n times. If the coin lands on tails I roll a biased dice n times. Let $X_i$ denote the score of the $i$th roll of the fair die and $Y_i$ the score of the $i$th roll of the biased die.
Would this be how you would denote the sum of the scores on the die?
$$S_n=\frac{1}{2}\sum_{i=1}^n{X_i} + \frac{1}{2}\sum_{i=1}^n{Y_i}$$
 A: Judging from the description of your procedure/game, I take

the sum of the scores on the die  

to mean that after you decide which die you roll, you roll it $n$ times and add up each value rolled.
So, the score should be denoted by 
$$S$$
without the subscript. Your score is given by
$$S = I\sum_{k = 1}^n X_k+(1-I)\sum_{k = 1}^n Y_k$$
where 
$$I = \begin{cases} 1,& \text{flipped heads}\\
0,& \text{otherwise}.\end{cases}$$

Addendum:
I guess you could keep the $n$, it didn't really occur to me.
Regardless, there should be an indicator there:
$$S_n = I\sum_{k = 1}^n X_k+(1-I)\sum_{k = 1}^n Y_k.$$
If we were given the distribution of the $Y_i$, then
$$E[Y_i] = \sum_{k = 1}^6 k P(Y_i = k) = \gamma.$$
Assuming independence across rolls, we have
\begin{align*}
E[S_n] &= E\left[I\sum_{k=1}^nX_k+(1-I)\sum_{k = 1}^nY_i\right] \\
&= E[I]\sum_{k = 1}^nE[X_k]+E[1-I]\sum_{k = 1}^nE[Y_k] \tag1\\
&= \frac{1}{2}\cdot n\cdot \frac 72+ \frac{1}{2}\cdot n\cdot \gamma\tag2\\
&=\frac{n}{2}\cdot\frac{7+2\gamma}{2}
\end{align*}
where in


*

*The indicator is independent of $X_i$ and $Y_i$, and

*We know $E[X_i] = 3.5$ since it is fair and follows a discrete uniform distribution on $\{1,2,\dotsc, 6\}$. 

