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)

$n/4$

(c)

None of the above

(d)

$n/3$

(e)

$n/2$

Solution

Define an Indicator Random Variable

For each friend $ i $ (where $ i = 1, 2, \ldots, n $):

Define an indicator random variable $ X_i $ such that: $X_i =$
- $1 \text{ if friend $ i $ shouts “Huzzah!“}$
- $0 \text{ otherwise}$

The total number of friends who shout “Huzzah!” can be expressed as: $X = \sum_{i=1}^{n} X_i$. The expected value of $X$ is $E(X) = E(\sum_{i=1}^{n} X_i)$.

Using the linearity of expectation, we can simplify the expected value to the following: $E(X) = \sum_{i=1}^{n} E(X_i)$

Since each friend has the same probability of shouting “Huzzah!” (by symmetry), we know: $E(X) = n \cdot E(X_1)$

Our task now reduces to finding $ E(X_1) $, the probability that a given friend shouts “Huzzah!“.

Probability of Shouting “Huzzah!”

A friend shouts “Huzzah!” if:

Their coin flip is different from their left neighbor’s flip.
Their coin flip is different from their right neighbor’s flip.

For three coin flips (Left Player, Current Player, Right Player), the probability distributions are:

There are $ 2^3 = 8 $ possible outcomes for these three flips.
Out of these, only 2 outcomes satisfy the “Huzzah!” condition:
- HTH
- THT

So the probability is: $P(X_1 = 1) = \frac{2}{8} = \frac{1}{4}$

Thus: $E(X_1) = \frac{1}{4}$

Step 3: Substitute into the Expectation Formula

Substituting into the linearity formula we calculated above:

$E(X) = n \cdot E(X_1) = n \cdot \frac{1}{4} = \frac{n}{4}$