Comparing a Dice Sum to a Coin Count

Worked Solution

How to Think About It: Before doing any math, think about the means. The dice sum averages $100 \times 50.5 = 5050$, while the coin-flip count averages $600 \times 0.5 = 300$. That is a gap of nearly 5000. The question is whether the tails of these distributions overlap enough for $X$ to fall below $Y$. Given that both have standard deviations in the low hundreds at most, a gap of 5000 is enormous in standard-deviation units. This should be astronomically unlikely.

Quick Estimate: Let $W = X - Y$. By independence:

• $E[W] = 5050 - 300 = 4750$
• $\text{Var}(X) = 100 \times \frac{99 \times 101}{12} = 100 \times 833.25 = 83{,}325$
• $\text{Var}(Y) = 600 \times 0.25 = 150$
• $\text{Var}(W) = 83{,}325 + 150 = 83{,}475$
• $\sigma_W = \sqrt{83{,}475} \approx 289$

By the CLT, $W$ is approximately normal. We need $P(W < 0)$:

$$z = \frac{0 - 4750}{289} \approx -16.4$$

A $z$-score of $-16.4$ corresponds to a probability so small it is effectively zero -- far beyond anything in a standard normal table (which typically stops around $z = -4$ at roughly $3 \times 10^{-5}$). We are talking about $P \approx 10^{-60}$ or less.

Approach: Formalize using the CLT on the difference $W = X - Y$.

Formal Solution:

Each die roll $D_i$ satisfies $E[D_i] = 50.5$ and $\text{Var}(D_i) = \frac{(100^2 - 1)}{12} = \frac{9999}{12} = 833.25$. So:

$$E[X] = 100 \times 50.5 = 5050, \quad \text{Var}(X) = 100 \times 833.25 = 83{,}325$$

For the coin flips, $Y \sim \text{Binomial}(600, 0.5)$:

$$E[Y] = 300, \quad \text{Var}(Y) = 150$$

Since $X$ and $Y$ are independent, $W = X - Y$ has:

$$E[W] = 4750, \quad \text{Var}(W) = 83{,}475, \quad \sigma_W \approx 289$$

By the CLT (both $X$ and $Y$ are sums of many independent terms), $W$ is well-approximated by $N(4750, 83{,}475)$.

$$P(X < Y) = P(W < 0) = \Phi\!\left(\frac{-4750}{289}\right) = \Phi(-16.4) \approx 0$$

For context, $\Phi(-6) \approx 10^{-9}$. At $z = -16.4$, the probability is on the order of $10^{-60}$.

Answer: Yes, $P(X < Y)$ is far less than 1% -- it is effectively zero. The means are separated by over 16 standard deviations of $W$, making the event $X < Y$ essentially impossible.