تعاریف:

$n friends ([math]i\in\{1,...,n\}[/math]) each toss a [math]k[/math]-sided die (whose values take a uniform distribution over [math]\{1,...,k\}[/math]). Define each of the [math]n[/math] tosses' results to be a random variable [math]X_i\in\{1,..,k\}[/math]. Define the random variables [math]X_{max} = \max{(X_1, ..., X_n)}[/math] and [math]X_{min} = \min{(X_1, ..., X_n)}[/math]. We want to find [math]E[X_{max}-X_{min}][/math]. For our specific question, [math]n=4[/math] and [math]k=6[/math].$

پاسخ ها:

$E[X_{max}-X_{min}] = \boxed{k-1-\frac{2}{k^n}[\zeta(-n)-\zeta(-n,k)]}, where [math]\zeta(i)[/math] is Riemann Zeta Function and [math]\zeta(i,j)[/math] is Hurwitz Zeta Function.$

$For our specific question, 6 - 1 - \frac{2}{6^4}[\zeta(-4) - \zeta(-4,6)] = \boxed{\frac{2261}{648}} \approx 3.49.$

استدلال:

$First, we get P(X_{max} \leq x) = P(X_1 \leq x, ..., X_n \leq x). Assuming that each die toss is independent of the other tosses, [math]P(X_1 \leq x, ..., X_n \leq x) = \prod\limits_{i=1}^{n}{P(X_i \leq x)} = (\frac{x}{k})^n[/math]. Then, noting that [math]P(X_{max} = x) = P(X_{max} \leq x) - P(X_{max} \leq x-1)[/math], we get the probability mass function [math]P(X_{max} = x) = (\frac{x}{k})^n - (\frac{x - 1}{k})^n[/math].$

$Similarly, we get P(X_{min} \geq x) = P(X_1 \geq x, ..., X_n \geq x). Again assuming the independence of die tosses, [math]P(X_1 \geq x, ..., X_n \geq x) = \prod\limits_{i=1}^{n}{P(X_i \geq x)} = (\frac{k-x+1}{k})^n[/math]. Then, noting that [math]P(X_{min} = x) = P(X_{min} \geq x) - P(X_{min} \geq x+1)[/math], we get the probability mass function [math]P(X_{min} = x) = (\frac{k - x + 1}{k})^n - (\frac{k - x}{k})^n[/math].$

$Now that we have both random variables' probability mass functions, we can simply use linearity of expected value to compute E[X_{max}]-E[X_{min}] instead of [math]E[X_{max}-X_{min}][/math]. By the definition of expected value for a discrete random variable, [math]E[X_{max}] = \sum\limits_{x=1}^{k}{xP(X_{max} = x)} = \sum\limits_{x=1}^{k}{x[(\frac{x}{k})^n - (\frac{x - 1}{k})^n]}[/math]. Similarly, [math]E[X_{min}] = \sum\limits_{x=1}^{k}{xP(X_{min} = x)} = \sum\limits_{x=1}^{k}{x[(\frac{k - x + 1}{k})^n-(\frac{k - x}{k})^n]}[/math]. Subtracting [math]E[X_{min}][/math] from [math]E[X_{max}][/math]and simplifying, we get [math]\frac{k^{n+1}-k^n-2\sum\limits_{x=1}^{k-1}{x^n}}{k^n}[/math], which simplifies to the expression in Answer (since [math]\sum\limits_{x=1}^{k-1}{x^n} = \zeta(-n)-\zeta(-n,k)[/math]).$