Posted on September 11, 2026
Uh-oh, Charlie’s teacher is giving a pop quiz today! The quiz consists of three multiple choice questions, each with four choices (A, B, C, and D). Each correct answer is worth 1 point. Charlie didn’t study, and has absolutely no idea what the answers might be. That said, he still wants to maximize his score on the quiz.
Importantly, Charlie happens to know that his teacher is a little obsessive when it comes to creating quizzes. In particular, consecutive questions never have the same correct answer. For example, if the answer to one question is C, then the answer to the following question cannot be C.
Also Charlie is a bit of a, well, cheat. He glances over at the quiz of his neighbor, who definitely studied and knows the answer to every question. Charlie spots the correct answer to the first question on his neighbor’s quiz.
At this point, the neighbor carefully covers their quiz, meaning Charlie must rely on his own wits for the remaining two questions.
Fiddler: On average, what score can Charlie expect to earn on this quiz?
Extra Credit: Oh no, Charlie’s teacher is giving another pop quiz! This time, there are seven multiple choice questions, each with four choices (A, B, C, and D). Charlie still hasn’t studied, and his teacher remains obsessive about avoiding any repeating correct answers.
Charlie has one chance to glance over at his neighbor’s test and spot a single correct answer before the neighbor notices and prevents Charlie from seeing any others.
Which of the seven answers should Charlie look at? And, if he does so, then what score should he expect to earn on this quiz, on average?
Let \(C_k \in \{\text{A}, \text{B}, \text{C}, \text{D}\}\) be the correct answer to question \(k\). Because the teacher strictly avoids repeating the same answer on consecutive questions, each subsequent answer is chosen uniformly at random from the 3 remaining choices:
\[ P(C_{k+1} = j \mid C_k = i) = \begin{cases} 0 & \text{if } j = i, \\ \frac{1}{3} & \text{if } j \neq i. \end{cases} \]
This is a discrete-time Markov chain on 4 states with transition matrix:
\[ \mathbf{P} = \begin{pmatrix} 0 & 1/3 & 1/3 & 1/3 \\ 1/3 & 0 & 1/3 & 1/3 \\ 1/3 & 1/3 & 0 & 1/3 \\ 1/3 & 1/3 & 1/3 & 0 \end{pmatrix} = \frac{1}{3}(\mathbf{J} - \mathbf{I}), \]
where \(\mathbf{J}\) is the \(4 \times 4\) all-ones matrix and \(\mathbf{I}\) is the identity matrix. Notice that \(\mathbf{P}\) is symmetric and doubly stochastic with uniform stationary distribution \(\pi = (1/4, 1/4, 1/4, 1/4)\). Because the transition rule is symmetric across time, the distribution of answers at a graph distance \(d = |k - m|\) from an observed answer \(C_m\) depends purely on \(d\).
By diagonalizing \(\mathbf{P}\) (or setting up the simple recurrence for returning to the starting state), the eigenvalues of \(\mathbf{P}\) are \(\lambda_1 = 1\) (with multiplicity 1) and \(\lambda_2 = -1/3\) (with multiplicity 3). Therefore, after \(d\) transitions:
\[ \mathbf{P}^d = \frac{1}{4}\mathbf{J} + \left(-\frac{1}{3}\right)^d \left(\mathbf{I} - \frac{1}{4}\mathbf{J}\right). \]
Concretely, conditioned on seeing answer \(C_m\) at question \(m\), the probability that question \(k\) (at distance \(d = |k - m|\)) has answer \(j\) is:
To maximize his expected score, Charlie simply picks whichever answer choice has the highest probability under the conditional distribution. The sign of \((-1/3)^d\) alternates with the parity of \(d\):
Evaluating the optimal success probabilities for distances \(d = 0, 1, \dots, 6\):
| Distance \(d\) | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|---|
| Optimal Guess | \(C_m\) | Any \(\neq C_m\) | \(C_m\) | Any \(\neq C_m\) | \(C_m\) | Any \(\neq C_m\) | \(C_m\) |
| Success Prob \(p(d)\) | \(1\) | \(\frac{1}{3}\) | \(\frac{1}{3}\) | \(\frac{7}{27}\) | \(\frac{7}{27}\) | \(\frac{61}{243}\) | \(\frac{61}{243}\) |
| Decimal | 1.0000 | 0.3333 | 0.3333 | 0.2593 | 0.2593 | 0.2510 | 0.2510 |
Charlie peeks at Question 1, so \(m = 1\). The distances to the three questions are \(d = 0, 1, 2\).
By linearity of expectation, Charlie's total expected score is the sum of his success probabilities across all three questions:
\[ \mathbb{E}[\text{Score}] = p(0) + p(1) + p(2) = 1 + \frac{1}{3} + \frac{1}{3} = \frac{5}{3} \approx 1.6667 \text{ points}. \]
Fiddler Answer: Charlie can expect to earn \(\displaystyle \frac{5}{3} = \mathbf{1\tfrac{2}{3}} \approx 1.667\) points on average.
Now there are 7 questions, and Charlie gets to choose which single question \(m \in \{1, 2, \dots, 7\}\) to observe. His expected score if he observes question \(m\) is:
\[ S(m) = \sum_{k=1}^7 p(|k - m|). \]
Let's evaluate \(S(m)\) for each candidate index:
| Observed Question \(m\) | Distances \(\{|k-m|\}\) | Sum of Probabilities | Exact Expected Score | Decimal |
|---|---|---|---|---|
| \(m = 1\) or \(m = 7\) | \(0, 1, 2, 3, 4, 5, 6\) | \(1 + \frac{1}{3} + \frac{1}{3} + \frac{7}{27} + \frac{7}{27} + \frac{61}{243} + \frac{61}{243}\) | \(\frac{653}{243}\) | \(\approx 2.6872\) |
| \(m = 2\) or \(m = 6\) | \(1, 0, 1, 2, 3, 4, 5\) | \(1 + 3\left(\frac{1}{3}\right) + 2\left(\frac{7}{27}\right) + \frac{61}{243}\) | \(\frac{673}{243}\) | \(\approx 2.7695\) |
| \(m = 3\) or \(m = 5\) | \(2, 1, 0, 1, 2, 3, 4\) | \(1 + 4\left(\frac{1}{3}\right) + 2\left(\frac{7}{27}\right)\) | \(\frac{77}{27} = \frac{693}{243}\) | \(\approx 2.8519\) |
| \(m = 4\) | \(3, 2, 1, 0, 1, 2, 3\) | \(1 + 4\left(\frac{1}{3}\right) + 2\left(\frac{7}{27}\right)\) | \(\frac{77}{27} = \frac{693}{243}\) | \(\approx 2.8519\) |
Intuition might suggest that looking dead-center at Question 4 should be strictly optimal because it minimizes the maximum distance to any other question. However, moving from \(m=4\) (distances \(\{3, 2, 1, 0, 1, 2, 3\}\)) to \(m=3\) (distances \(\{2, 1, 0, 1, 2, 3, 4\}\)) simply trades one question at distance 3 for one question at distance 4.
Because \(p(3) = p(4) = 7/27\), this trade costs Charlie zero expected points! Thus, Questions 3, 4, and 5 all share the exact same maximal expected score of \(\frac{77}{27}\).
Extra Credit Answer:
Charlie should look at Question 3, 4, or 5 (any of these three choices is optimal).
His maximum expected score is \(\displaystyle \frac{77}{27} = \mathbf{2\tfrac{23}{27}} \approx 2.8519\) points.
Explore Charlie's expected score for any quiz length \(N\) and choice of observed question \(m\). Run thousands of simulated quizzes generated under the teacher's no-repeat rule to verify empirical scores against theoretical expectations.
The chart below plots Charlie's expected score \(S(m)\) as a function of the glanced question index \(m\).