Consider a game that uses a 5-sided die. Each die is numbered 1-5 with an equal probability of rolling each number. The die is rolled 10 times, with each roll adding to the player's points. After 10 turns the points are added up.

Assume the CLT applies here. Which R code would calculate the chance of a player's points being between 25 and 35?

Question

Consider a game that uses a 5-sided die. Each die is numbered 1-5 with an equal probability of rolling each number. The die is rolled 10 times, with each roll adding to the player's points. After 10 turns the points are added up.

Assume the CLT applies here. Which R code would calculate the chance of a player's points being between 25 and 35?

BlackTom AI · Accepted Answer

The sum of 10 independent rolls has mean 10·E[X] and standard deviation sqrt(10)·SD(X). For a fair 5-sided die with faces 1–5, E[X] = (1+2+3+4+5)/5 = 3 and Var(X) = E[X^2] − E[X]^2 = 11 − 9 = 2, so SD(X) = sqrt(2).

Step-by-step reasoning and calculations:
1. Mean of one roll: mean(box) = 3. Therefore expected total after 10 rolls: ev = 10 * 3 = 30.
2. Variance of one roll: Var(X) = 2, so SD(X) = sqrt(2) ≈ 1.4142136.
3. SD of the sum of 10 rolls: se = sqrt(10) * SD(X) = sqrt(10)*sqrt(2) = sqrt(20) ≈ 4.472136.
4. Using the Central Limit Theorem, sum S ≈ Normal(mean = 30, sd = 4.472136).
5. Desired probability P(25 ≤ S ≤ 35) = Φ((35 − 30)/se) − Φ((25 − 30)/se).
   Compute z = 5 / 4.472136 ≈ 1.1180. Then probability = Φ(1.1180) − Φ(−1.1180) = 2Φ(1.1180) − 1 ≈ 2(0.8686) − 1 ≈ 0.7372.

Why the provided R code is correct:
- box = 1:5 sets the population of outcomes.
- ev = 10 * mean(box) computes the sum's mean (10 rolls).
- se = sqrt(10)*popsd(box) computes sqrt(10) times the population standard deviation of a single roll; popsd from rafalib gives the population SD (divide by n), which matches SD(X)=sqrt(2). This yields the SD of the sum.
- pnorm(35,ev,se)-pnorm(25,ev,se) computes P(S ≤ 35) − P(S ≤ 25), which equals P(25 ≤ S ≤ 35) under the continuous normal approximation.

Why other common variants would be wrong:
- Using sd(box) instead of popsd(box) uses the sample (n−1) denominator and slightly misestimates the true SD for the known population.
- Using se = sd(box)/sqrt(10) would be the standard error of the sample mean, not the SD of the sum, and so is incorrect here.
- Forgetting to multiply mean(box) by 10 would give the mean per roll, not the sum.

Conclusion: The provided R code correctly implements the CLT-based approximation; it yields the probability P(25 ≤ S ≤ 35) and matches the authoritative answer given.

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