Dice Sum vs. Coin Flip Count
You roll a fair six-sided die 100 times and let $X$ be the total. Separately, you flip a fair coin 600 times and let $Y$ be the number of heads.
What is the probability that $X > Y$?
Hints
- Compare the means of $X$ and $Y$ first. Which one is larger on average, and by how much?
- Both $X$ and $Y$ are sums of many independent trials. Apply the Central Limit Theorem to approximate $W = X - Y$ as a normal random variable.
- Compute $E[W] = 50$ and $\text{Var}(W) = 875/3 + 150 \approx 441.67$. Then $P(W > 0) = \Phi(50/\sqrt{441.67})$.
Worked Solution
How to Think About It: Before touching any formulas, get a feel for the two numbers you are comparing. A single die averages 3.5, so 100 dice average 350. A fair coin in 600 flips gives 300 heads on average. So on average the dice sum beats the head count by 50. The question is whether random fluctuations can close that 50-point gap. Both quantities are sums of independent trials, so CLT applies and each is roughly normal. You just need to check how many standard deviations that 50-point cushion is worth.
Quick Estimate: The dice sum $X$ has mean 350 and variance $100 \times 35/12 \approx 291.7$, so $\sigma_X \approx 17.1$. The head count $Y$ has mean 300 and variance $600 \times 0.25 = 150$, so $\sigma_Y \approx 12.2$. The difference $W = X - Y$ has mean 50 and variance $291.7 + 150 = 441.7$, so $\sigma_W \approx 21.0$. The gap of 50 is about $50/21.0 \approx 2.38$ standard deviations above zero. Glancing at a z-table, $\Phi(2.38) \approx 0.991$. So we expect roughly a 99% chance that the dice sum wins. That is a big cushion -- the 50-point head start is more than two sigma.
Approach: Formalize by writing $W = X - Y$ as a sum of independent random variables, apply CLT, and read off the probability from the standard normal CDF.
Formal Solution:
Let $D_1, \dots, D_{100}$ be i.i.d. $\text{Uniform}\{1,2,3,4,5,6\}$ rolls. Each has:
$$E[D_i] = \frac{7}{2}, \quad \text{Var}(D_i) = \frac{35}{12}$$
So for $X = \sum_{i=1}^{100} D_i$:
$$E[X] = 350, \quad \text{Var}(X) = \frac{3500}{12} = \frac{875}{3} \approx 291.67$$
Let $Y \sim \text{Binomial}(600,\, 1/2)$:
$$E[Y] = 300, \quad \text{Var}(Y) = 150$$
Since $X$ and $Y$ are independent, define $W = X - Y$:
$$E[W] = 350 - 300 = 50$$
$$\text{Var}(W) = \frac{875}{3} + 150 = \frac{875 + 450}{3} = \frac{1325}{3} \approx 441.67$$
$$\sigma_W = \sqrt{\frac{1325}{3}} \approx 21.02$$
By the CLT, $W$ is approximately $N(50,\, 441.67)$. Then:
$$P(X > Y) = P(W > 0) = P\!\left(Z > \frac{0 - 50}{21.02}\right) = P(Z > -2.38) = \Phi(2.38)$$
$$\approx 0.9913$$
Answer: The probability that the dice sum exceeds the head count is $\Phi(2.38) \approx 0.991$, or about $99.1\%$.
Intuition
This problem illustrates why comparing means is not enough -- you need to measure the gap in units of standard deviation. The dice sum leads by 50 points on average, but that only matters relative to the combined variability of both quantities. The key move is forming the difference $W = X - Y$ and recognizing that because $X$ and $Y$ are independent sums of many trials, CLT makes $W$ approximately normal. Once you have that, the entire problem reduces to computing a z-score.
This pattern shows up constantly in quantitative work. Whenever you are comparing two noisy quantities -- P&L of two strategies, fill rates on two venues, signal strength across two models -- the question is never just "which mean is bigger?" but "how many standard deviations apart are they?" A 50-point edge that is 2.4 sigma is very reliable (99%+). The same 50-point edge with ten times the variance would be only 0.75 sigma and far less convincing. Building the habit of immediately converting gaps into z-scores is one of the most useful reflexes in applied probability.