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Q: Roll 100 dice together, what is the probability that the sum of all dice is 400?

A

Central Limit Theorem: Let $$X_1, X_2,... X_n$$ be independent and identically distributed random variables. The sum of these random variables approaches a normal distribution as $$n \rightarrow \infty$$

$$\sum_{i=1}^n X_i \sim N(n \cdot \mu, n \cdot \sigma^2)$$

, where $$\mu = E[X_i]$$ and $$\sigma^2 = Var(X_i)$$.

Let $$X$$ be the value of a die. We have $$E[X_i] = \frac{1+2+...+6}{6} = 3.5$$ and the variance $$Var(X_i) = \frac{35}{12}$$.

Rolling 100 dice, the sum of 100 values follows the normal distribution:

$$\sum_{i=1}^n X_i \sim N(n \cdot \mu, n \cdot \sigma^2) = N(350, \frac{3500}{12})$$

Let S be the sum of all dice values. Apply Continuity Correction, $$P(S = 400) \approx P(S < 400.5) - P(S < 399.5)$$

Now that we know the mean and variance, we can calculate the z-score value to look up the normal CDF. $$P(S = 400) \approx P(S < 400.5) - P(S < 399.5) = \Phi(\frac{400.5 - 350}{\sqrt{3500/12}}) - \Phi(\frac{399.5-350}{\sqrt{3500/12}}) \approx \Phi(2.96) - \Phi(2.90) \approx 0.0004$$ Given that this uses Continuity Correction1 method, the actual probability should be lower than 0.0004.