A wheel-of-fortune has the numbers from 1 to 2n arranged in a circle. The wheel has a spinner, and a spin randomly determines the two numbers at the opposite ends of the spinner. How would you arrange the numbers on the wheel to maximize the expected value of the sum of the numbers chosen? What is this maximum?
2026-04-12 15:07:45.1776006465
Maximizing expectation (probability stuff)
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The expectation is fixed no matter how you arrange the numbers.
Given any arrangement, there are $2n$ possible output pairs, each with probability $\frac{1}{2n}$. For convenience, let these $2n$ pairs be $(a_1, b_1)$, $(a_2, b_2)$, $\cdots$, $(a_{2n}, b_{2n})$. Then the expected output is $$ \frac{1}{2n} \sum_{i=1}^{2n}(a_i + b_i) $$ Note that $\sum\limits_{i=1}^{2n}(a_i + b_i)$ is double the sum of all $2n$ numbers, which is fixed.