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Just started a course in probability when I stumbled upon this question. Now, I know that one formula for variance is: $$\text{Var}(X)= E(X^2) - (E(X))^2$$

I understand that, in general, the probability of rolling an outcome that hasnt been rolled is $(6-n)/6$, and thus the $E(X)= \sum_{i=1}^6 6/i = 14.7 $

My question is regarding:

a. The validity of this approach

b. How exactly to calculate $E(X^2)$ or if there is some other approach to calculating variance.

I apologize if this seems a basic question, but I have searched online and through my book and cannot really find any help.

Of course, a point in the right direction would much appreciated (even more so then just an answer).

Thank you!

Em.
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Aric
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1 Answers1

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Yes, that is what the expectation simplifies to.

In other words $$E[X] = \sum_{k = 1}^6kP(X=k) = \sum_{k = 1}^6\frac{6}{k} = 14.7.$$ Then to find $E[X^2]$, you could approach in the usual way $$E[X^2] = \sum_{k=1}^6 k^2P(X=k).$$ However, if you model rolling a die until a new face is shown as a geometric distribution with $p_k = \frac{7-k}{6}$ where $k = 1,\dotsc,6$ is the number of faces you have seen, then $X_k$ denotes rolling until you see $k$th different face, and $X_k\sim\text{Geom}(p_k)$ on $\{1,2,3,\dots\}$.

Hence we have that $$\text{Var}(X) = \text{Var}(X_1)+\dotsb+\text{Var}(X_6)$$ since the $X_i$ are independent, and the variance of a geometric random variable is well known.

Em.
  • 16,196
  • Hey, thanks for the response. I thought about a geometric distribution but I was not sure how independent the rolls would be in the context of the distribution (also, I never would have thought of $(7-k)/6$ . However, I like the latter solution! I appreciate your help. – Aric Mar 28 '16 at 21:18