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We were given this seatwork:

With four independent dice:

a) the expected value of the sum of the rolls,

b) the expected value of the product of the rolls, and

c) the variance of the sum of the rolls .

I was able to answer a and b, but I don't know how to get the variance. Here's my attempt:

\begin{align*} S = \dfrac{1 + 2 + 3 + 4 + 5 + 6}{6} = 3.5 \end{align*} This is the expectation of one die. For the expectation of four dice, we could assume the expectation of the sum four dice is equal to the sum of the expectations of a die: \begin{align*} &= S + S + S + S\\ &= 4S\\ &= 4(3.5)\\ &= 14 \end{align*}

Similarly, we could also do this for the products. The expected product of four dice rolls is: \begin{align*} &= S \cdot S \cdot S \cdot S\\ &= S^4\\ &= 3.5^4\\ &= 150.06 \end{align*} Are these assumptions correct?

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    $\begingroup$ $3.5^4=150.0625\ne256$. But your assumptions are correct. $\endgroup$ Dec 20, 2020 at 12:25
  • $\begingroup$ My bad, that's a typo. $\endgroup$
    – muw
    Dec 20, 2020 at 12:28
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    $\begingroup$ you work is correct, great stuff! $\endgroup$
    – Sergio
    Dec 20, 2020 at 12:29
  • $\begingroup$ @Sergio thanks! However, I do not know how to get the variance of 4 dice. $\endgroup$
    – muw
    Dec 20, 2020 at 12:35
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    $\begingroup$ I will post an answer $\endgroup$
    – Sergio
    Dec 20, 2020 at 12:38

1 Answer 1

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You have found the expected value of a dice roll to be $E(X)=3.5$, thus the variance of one roll is equal to:

\begin{align} \operatorname{Var}(X) &= \sum_{k=1}^6 \frac{1}{6}\left(k - 3.5\right)^2 \\ &= \frac{35}{12} \approx 2.92. \end{align}

Since the rolls of the 4 dies are independent the variance of their sum is equal to $4 \cdot \frac{35}{12}\approx 11.66$

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