# Normal Distribution with dice probability

Suppose a “fair” die is rolled 100 times. Let $$X_i$$ be the value obtained in the i-th roll. Calculate an approximation for:

$$P( \displaystyle\prod_{i=1}^{100} X_i \leq a^{100})$$, where $$1 < a < 6$$

I know that $$X_i$$~Unif{1,2,3,4,5,6}, and I think I should use some kind of normal distribution to approximate the probability, but I don't know what to do with the product. Can someone help me?

## 1 Answer

Hint: try taking the log of both sides of the inequality.

$$P(\prod_{i=1}^{100} X_i \le a^{100}) = P(\frac{1}{100} \sum_{i=1}^n \log X_i \le \log a)$$ and $$\frac{1}{100} \sum_{i=1}^n \log X_i$$ is approximately normal (you need to find the mean and variance).

Mean: $$E \frac{1}{100} \sum_{i=1}^{100} \log X_i = E \log X_1 = \frac{1}{6} \log 720.$$

Variance: $$\text{Var}(\frac{1}{100} \sum_{i=1}^{100} \log X_i) = \frac{1}{100} \text{Var}(\log X_1) = \frac{1}{600} \sum_{x=1}^6 (\log x - \frac{1}{6} \log 720)^2.$$

• But $Z = \frac{X-E(X)}{Var(X)}$, how does $\frac{1}{100} \sum_{i=1}^n \log X_i$ is equal to $Z$? – Mystery Nov 23 '19 at 21:16
• @Mystery I only said that its distribution is approximately normal (not standard normal). If you find the mean and variance of $\frac{1}{100} \sum_{i=1}^n \log X_i$, then you can standardize it by subtracting the mean and dividing by the standard deviation (not the variance) to get a random variable that is approximately standard normal. – angryavian Nov 24 '19 at 0:11