TITLE: Pigeonhole Principle - Roulette Wheel
QUESTION [11 upvotes]: 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?

REPLY [6 votes]: 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.