# Wheel of Fortune Problem

The Summation formula is $$\sum_{i=1}^ni =\frac{n(n+1)}2$$

How is it that we know the integers $1,2,...36$ appear exactly $3$ times. And why do we multiply the sum by $3$ in the last part of the proof?

Source of Question: Discrete and Combinatorial Mathematics: An Applied Introduction by Ralph P. Grimaldi, 5th Edition.

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You should acknowledge the source of that lengthy quotation and, ideally, transcribe it into text so the site's search function can help people find it. –  David Richerby Jul 12 '14 at 7:00
Gladly. Is there a special way to do this, or can I just put it right in the original question as an edit? –  Dunka Jul 12 '14 at 7:10

You have $36$ inequalities of the form

$$x_i + x_{i + 1} + x_{i + 2} < 55$$

as $i$ ranges from $1$ to $36$, with the interpretations that $x_{37} = x_1$ and $x_{38} = x_2$. When you add all of these inequalities together, each $x_i$ appears in the sum exactly $3$ times - once per each of the three inequalities that involves it. Hence, we can write

$$\sum_{i = 1}^{36} \left(x_i + x_{i + 1} + x_{i + 2}\right) = 3 \sum_{i = 1}^{36} x_i = 3 \sum_{i = 1}^{36} i$$

where the final equality is due to the choices of $x_i$ allowed.

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There is a little person sitting on each number. Each little person adds together her number, and the numbers of her two neighbours. So we end up with $36$ sums. We are asked to show that one at least of the little people will end up with a sum of $55$ or more,

Note that each number on the wheel is part of the sum of $3$ people: the person sitting on the number, the person immediately to the right of the number, and the person immediately to the left of the number.

So if you take the sum of all the sums found by the little people, each number from $1$ to $36$ will have appeared $3$ times, will have been counted $3$ times.

In symbols, if the little people are called $P_1$ to $P_{36}$, and $P_i$'s sum is called $S_i$, then $$S_1+S_2+\cdots +S_{36}=3(1+2+\cdots+36)=(3)(18))(37).$$ Divide by $36$. We get $55.5$.

So the average of the sums of the little people is $55.5$. Their sums can't all be $\le 55$, they can't all be below average.

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Thanks to everyone for the help. If anyone else needs this question answered, and isn't helped by the excellent answers given, I found the following graphic very clear:

The circle is the wheel of fortune. $x_i$ is an arbitrary number on the wheel of fortune, you can write it as one of the inequalities by using the two other numbers to its right, the number closest to its left, or the two numbers on its right.

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