Rolling dice Probability that Sum

If 10 pairs of fair dice are rolled, approximate the probability that the sum of the values obtained (which ranges from 20 to 120) is between 30 and 40 inclusive.

I dont know where to start with this one. I have been looking all over the web for example, but nothing i find is applicable for finding the sum of numbers.

• Start by listing the combinations of numbers that result in a sum between 30 and 40. – Eckhard Apr 5 '13 at 16:06
• i dont think so man, the combination would be almost endless. – KevinCameron1337 Apr 5 '13 at 16:58
• That depends on one's understanding of almost endless. I admit that I overlooked the word 'approximate' in your question, though. – Eckhard Apr 5 '13 at 17:05

We are tossing $20$ dice. Let $X_i$ be the number showing on the $i$-th die. We are interested in the random variable $X_1+\cdots+X_{20}$.

The $X_i$ are independent, mean $\frac{7}{2}$, variance $\frac{35}{12}$ (please verify).

So $Y$ has mean $\mu=20\cdot\dfrac{7}{2}$, variance $\sigma^2=20\cdot \dfrac{35}{12}$.

We cross our fingers and use the normal approximation. This is because $Y$ is the sum of a not terribly small number of independent identically distributed respectable random variables $X_i$.

So we want the probability that if $W$ is normal with mean $\mu$ and variance $\sigma^2$, then $30\le W\le 40$.

The rest depends to some degree on whether you are expected to use the continuity correction.

Without the continuity correction, we want the probability that a standard normal $Z$ satisfies $\frac{30-\mu}{\sigma}\le Z\le \frac{40-\mu}{\sigma}$.

With continuity correction, replace $40$ by $40.5$, and $35$ by $34.5$.

• When we are using the normal to approximate, for example, a binomial $X$, the probability that $X\le k$ is often better approximated by the probability that $W\le k+\frac{1}{2}$, where $W$ is the normal with same mean and variance. (The no continuity correction uses $\Pr(W\le k)$.) Makes no significant difference if $n$ is large, like $400$. But for smallish $n$, the correction often leads to improved estimates. If your course has not discussed it, you are not expected to use it! I was just being cautious. – André Nicolas Apr 5 '13 at 16:26
• Should the Variance be $sqrt(20)*35/12 – KevinCameron1337 Apr 5 '13 at 16:43 • The standard deviation of$Y$is about$7.6376$. The prob. that$Y\le 40$is roughly the probability that$Z\le \frac{40-70}{7.6376}$. This is virtually$0$. I suspect there is a typo in the problem. – André Nicolas Apr 5 '13 at 16:43 Let's$x_1,\ldots x_{20}\in\{1,2,3,4,5,6\}\$. For a exact solution $$P\left(30\leq x_1+\ldots+x_{20}\leq 40 \right)=\frac{1}{20^6}\sum_{30\leq N\leq 40}\quad\sum_{x_1+\ldots+x_{20}\leq N} \frac{N!}{x_1!\cdot \ldots \cdot x_{10}!}$$