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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?

$$S_n=\frac{1}{2}\sum_{i=1}^n{X_i} + \frac{1}{2}\sum_{i=1}^n{Y_i}$$

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    $\begingroup$ Not sure what this represents. The sum of the scores is either $\sum X_i$ or $\sum Y_i$, probaiblity $\frac 12$ for either. Are you trying to compute an expectation? $\endgroup$ – lulu Jun 1 '16 at 11:22
  • $\begingroup$ yes I'm asked to compute $E(S_n)$ $\endgroup$ – Jordan Jun 1 '16 at 13:13
  • $\begingroup$ would it be $E(S_n) = \frac{1}{2}\sum_{i=1}^n{E(X_i)} + \frac{1}{2}\sum_{i=1}^n{E(Y_i)}$ $\endgroup$ – Jordan Jun 1 '16 at 13:15
  • $\begingroup$ Ok, well then it is true that $E[S_n]=\frac 12\times \sum E[X_i]+\frac 12\times \sum E[Y_i]$. Of course all the $E[X_i]= 3.5$ but we don't know any $E[Y_i]$ from the given information. $\endgroup$ – lulu Jun 1 '16 at 13:16
  • $\begingroup$ Crossed comments. Yes, that formula is correct. $\endgroup$ – lulu Jun 1 '16 at 13:16
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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

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