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


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.

  1. Approximations for Discrete Distributions, Countinuity Correction Factor: What is it? ↩︎