Pigeonhole Principle - Roulette Wheel

Question

Question - A Roulette wheel is divided into 36 sectors. Each sector is assigned a random number from 1 to 36. Show that there are three consecutive sectors such that the sum of their assigned numbers is at least 56.

My approach -> (Proof by Contradiction)

Let $a_1, a_2, \ldots, a_n$ be the numbers assigned to the 1st, 2nd, ..., nth sectors respectively. Assume that the sum of all three consequtive sectors is less than 56.

Thus,

$$ a_1 + a_2 + a_3 < 56 \\ a_2 + a_3 + a_4 < 56 \\ \vdots \\ a_{36} + a_1 + a_2 < 56 $$ Adding them all, $$ \Longrightarrow 3(a_1 + a_2 + \ldots + a_{36}) < 56 \cdot 36 \\ \Longrightarrow 3 \frac{36 \cdot 37}{2} < 56 \cdot 36 \\ \Longrightarrow 111 < 112 $$ (which is true and hence, unable to prove by contradiction)

Where am I going wrong?

Answer 1

You were on the right track. It is correct that if the statement

There are three consecutive sectors such that the sum of their assigned numbers is at least 56.

does not hold then $$ a_i + a_{i+1} + a_{i+2} < 56 $$ for $i = 1, 2, 3, \ldots$. However, this estimate is not strong enough to obtain a contradiction. Since all $a_i$ are integers we have in fact the better estimate $$ a_i + a_{i+1} + a_{i+2} \le 55 $$ for all $i$. Adding these inequalities now gives a contradiction, as desired: $$ 3 \frac{36 \cdot 37}{2} \le 55 \cdot 36 \Longrightarrow 1998 \le 1980 \, . $$

---

Remark: Actually there must be three consecutive sectors on the Roulette wheel whose number add up to at least 57, see

A066385 Smallest maximum of sum of 3 consecutive terms in any arrangement of [1..n] in a circle.

in the On-Line Encyclopedia of Integer Sequences for the general question and more information. In particular it is listed that $a(36) = 57$.

Answer 2

Since the average value of $a_i=18.5$ the average value of the sum of triples is 55.5 thus exists at least one triple such that the sum is at least 56.