Solution: 2022 Winter Final - 19

Author: Michiel Smid

Question

A group of $n\ge 3$ friends stand around in a circle and each friend tosses a coin. If the result of a friend's coin toss is the different from the result of the coin tossed by their left neighbour and different from result of the coin tossed by their right neighbour, then the friend shouts Huzzah! Let $X$ be the number of friends who shout Huzzah!. What is $ E(X)$?

(a)

$n/8$

(b)

$n/3$

(c)

$n/4$

(d)

$n/2$

(e)

None of the above

Solution

ngl, I have no idea how the right answer is the right answer. Let’s just do this and let the math take us to heaven or hell

Since they’re standing in a cirlce, the last person has the first person to their right. That means we consider every single person

Let S be the set of all possible coin tosses for $p_i$ and his/her neighbours
$ |S| = 2^3 = 8 $
Let A be the event that $p_i$ says Huzzah when they get a head $p_i$ gets a head: 1
$p_{i-1}$ must get a tail: 1
$p_{i+1}$ must get a tail: 1
$ |A| = 1 $
$ \Pr(A) = \frac{1}{8} $
Let B be the event that $p_i$ says Huzzah when they get a tail
$p_i$ gets a tail: 1
$p_{i-1}$ must get a head: 1
$p_{i+1}$ must get a head: 1
$ |B| = 1 $
$ \Pr(B) = \frac{1}{8} $
A and B are mutually exclusive, so let's just calculate their union
$ \Pr(A \cup B) = \Pr(A) + \Pr(B) $
$ \Pr(A \cup B) = \frac{1}{8} + \frac{1}{8} $
$ \Pr(A \cup B) = \frac{2}{8} $
$ \Pr(A \cup B) = \frac{1}{4} $

$E(X) = \sum_{i=1}^{n} 1 \cdot \Pr(A \cup B) $

$E(X) = n \cdot \Pr(A \cup B) $

$E(X) = n \cdot \frac{1}{4} $

$E(X) = \frac{n}{4} $