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We roll a dice exactly $n$ times. What is the expected value of the sum of the outocomes?

I have given this problem some thought, but I don't know how to tackle this. The problem is, that each sum has a different probability of coming out. For example, there is only one way to achieve the sum $6n$, $5n$ etc, but more than one way to achieve $4n + 4$.

Is there a clever way to tackle this?

Aemilius
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    Have you ever heard of "linearity of expectation?" The average result on each die is $3.5$. The average sum $n$ dice is quite simply then $3.5\times n$. – JMoravitz Feb 26 '19 at 20:08
  • What is the expected value for $n=1$? What for $n=2$? – Sqyuli Feb 26 '19 at 20:09

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Because of the linearity of expectations, the expectation of $n$ dice rolls is simply $n\times E(X)$ where $E(X)$ is the expectation of a single die.

Peter Foreman
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  • I can't believe that this answer is so simple.. I thought about this solution, but have never heard of the "linearity of expectation" and the fact that some sums are more likely than others advised me against this. – Aemilius Feb 26 '19 at 20:12