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Let $X$ be a random variable representing the roll of one die.

Then,

$$E(X) = \dfrac{1+2+3+4+5+6}{6} = 3.5$$

$$E(X^2) = \dfrac{1^2+2^2+3^2+4^2+5^2+6^2}{6} = \dfrac{91}{6}$$

$$Var(X) = E(X^2)-E(X)^2 = \dfrac{91}{6} - \dfrac{49}{4} = \dfrac{70}{24}$$

$$\sigma = \sqrt{Var(X)} = \sqrt{\dfrac{70}{24}} \approx 1.71$$

Let $Y$ be a random variable representing sum of the rolls of $N$ dice and $S$ representing the set of possible outcomes.

How can I find $P\{ A \leq Y \leq B\}$ for some $A,B \in S, A < B$?

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2 Answers

up vote 2 down vote accepted

One could calculate the probability that the sum is $k$ explicitly, and add up over our interval. Not so pleasant to do by hand, but a task easy to write a program for.

For a possibly crude approximation, note that $Y$ has mean $35$ and variance $10\dfrac{70}{24}$. Although $10$ is not at all a large number, one can approximate $Y$ by a normal with the same mean and variance. It may be useful to cross one's fingers.

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You can look at my comment on this question. I don't have the spreadsheet on this computer, but it isn't hard to reproduce. Then sum for your favorite $A$ and $B$

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That person is probably doing the same homework assignment I am =) haha –  somekindarukus Nov 16 '12 at 5:18
    
@somekindarukus: We have had three questions on throwing 10 dice recently. I gave a description of how to get the exact result on that question. To be fair, the others should be able to see that, too. The normal approximation is not too far off. –  Ross Millikan Nov 16 '12 at 5:22
    
Hm. I had tried searching for it at first but I couldn't find anything. Thaks anyways –  somekindarukus Nov 16 '12 at 5:25
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