Expected number of cards that are larger than both of their neighbors

Question

Seven cards numbered from $1$ to $7$ are randomly shuffled and then arranged in a circle. Given that the $1$ and $7$ are adjacent to each other, what is the expected number of cards that are larger than both of their neighbors?

I am not sure I understand the question correctly. But here's my thought process:

If $1$ and $7$ are adjacent, their neighbors can be chosen in $5C_2 = 10$ ways, therefore, if the neighbors are $2$ and $3$, for instance, these can come in $2$ ways $(2,3)$ and $(3,2)$, hence it's probability is $\frac{2}{10}$ and the numbers larger than $2$ and $3$ are $4,5,$ and $6$. Working on individual cases like this I arrived at the following:

$\frac{2}{10} \times 3 + \frac{2}{10} \times 2 + \frac{2}{10} \times 1 + \frac{2}{10} \times 2 + \frac{2}{10} \times 1 + \frac{2}{10} \times 1 = 2$

Therefore, the answer is $2$, but this is wrong. I simulated it and I got $\approx 2.5$. How can I solve this question?

Answer 1

Consider the position for the 7 cards as (WLOG) $ 1, 7, A, B, C, D, E$.

Let I be the random variable for the number of larger-than-neighbors (LTN for short).
We apply the indicator variables to calculate the expected value that each position is LTN, so $I = I_1 + I_7 + I_A + I_B + I_C + I_D + I_E$

• 1: $E(I_1) = 0$ clearly.
• 7: $E(I_7) = 1 $ clearly.
• A: $E(I_A) = 0 $ clearly.
• B: $E(I_B) = 1/3$, since there are 6 ways to arrange A, B, C with 2 ways resulting in $ A < B > C$.
• C: $E(I_C) = 1/3$ for the same reason.
• D: $E(I_D) = 1/3$ for the same reason.
• E: $E(I_E) = 1/2$. Can you explain why?

Then $E(I) = E(I_1) + E(I_7) + E(I_A) + E(I_B) + E(I_C) + E(I_D) + E(I_E) = 2.5$. This matches OP's simulation.

---

Note

• Without the condition that 1 is next to 7, a similar indicator variable analysis gives that the expected number of larger cards is $ \frac{7}{3}$.

Answer 2

For lack of a better place, here's the solution to the analogous problem for arbitrary number $N$ of cards: The expected number $E_n$ of cards larger than both neighboring cards is $$ \begin{align*} E_n = & 1 + \frac{1}{(N-2)!} \sum_{j=3}^{N-1} \left[ (j-2) (N-4)! + N (j-2) (j-3) (N-5)!\right] \\ = & 1 + \frac{1}{(N-2)(N-3)} \sum_{j=3}^{N-1} (j-2)^2 \\ = & \frac{2N+1}{6} \end{align*} $$ The $1$ is for the card labelled $N$. The first term in the sum counts the number of configurations with card $j$ adjacent to card $1$ and larger than its other neighbor. The second term in the sum counts all other configurations where card $j$ is larger than its neighbors.

If you remove the constraint on card $1$ being next to card $N$, then the analogous expression becomes

$$ \begin{align*} E_n = & \frac{1}{N!} \sum_{j=1}^{N} N (j-1) (j-2) (N-3)! \\ = & \frac{1}{(N-1)(N-2)} \sum_{j=1}^{N} (j-1) (j-2) \\ = & \frac{N}{3} \end{align*} $$

This is obvious if you phrase it this way: Given three random cards displayed from left to right, the odds that the largest one is in the middle is $1/3$.

---

Some informal analysis:

The fact that the asymptotic behavior of the two problems is the same says that boundary conditions don't matter in the large $N$ limit. The difference between $E_n$ in the two cases, $1/6$, can be interpreted like this:

• Call a card "hot" if it is bigger than its two neighbors, and "cold" otherwise.
• For large $N$, the highest card $N$ always makes its two neighbors cold. These neighbors would otherwise have a $1/3$ chance each of being hot. So the $N$ card on average cools $1/3$ of a hot card for each of its two neighbors.
• Similarly, the $1$ card increases the chance that each of its neighbors is hot from $1/3$ to $1/2$, since a neighbor to $1$ has $1/2$ odds of being bigger than its other neighbor. So on average, the $1$ card heats up each neighbor by $\left(\frac{1}{2}-\frac{1}{3}\right) = 1/6$.
• When you put $1$ next to $N$, both lose one neighbor that would otherwise be affected as above. That means the other cards are on average $1/3 - 1/6 = 1/6$ hotter.