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I am wondering if there is a fast way to tell which expected value is higher: "Roll 1 die and take the square of the number that comes up or roll 2 dice and multiply them". Thank you!

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Let $\xi$ and $\eta$ be values of the two die rolls. The expected value of squaring the the first roll is $\operatorname{E}(\xi^2)$, and the expected value of multiplying the two values is $\operatorname{E}(\xi\eta)$. Since we can assume the two die rolls are iid, we have $$\operatorname{E}(\xi\eta)=\operatorname{E}(\xi)\operatorname{E}(\eta)=\operatorname{E}(\xi)^2$$

Since the difference between these two expected values is equal to the variance, $$\operatorname{Var}(\xi)=\operatorname{E}(\xi^2)-\operatorname{E}(\xi)^2$$ it must be nonnegative. Therefore, we have $$\operatorname{E}(\xi^2)\ge\operatorname{E}(\xi)^2$$ so the expected value of squaring the result of one die roll is higher.

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  • $\begingroup$ perfect! I was trying to do something like this. Thanks a lot $\endgroup$ – Kaan Yolsever Nov 15 '20 at 20:40
  • $\begingroup$ Nice and general. +1 $\endgroup$ – RobPratt Nov 15 '20 at 20:47
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Not sure whether you consider this faster. The first probability is $$\frac{1}{n}\sum_{k=1}^n k^2 = \frac{1}{n}\cdot\frac{n(n+1)(2n+1)}{6}=\frac{(n+1)(2n+1)}{6}.$$ The second probability is $$\left(\frac{1}{n}\sum_{k=1}^n k\right)^2 = \left(\frac{1}{n}\cdot\frac{n(n+1)}{2}\right)^2 = \frac{(n+1)^2}{4}.$$ The first one is larger when $(2n+1)/6 > (n+1)/4$, that is when $n > 1$.

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  • $\begingroup$ yeah this is definitely faster. Thanks a lot! $\endgroup$ – Kaan Yolsever Nov 15 '20 at 20:40

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