Scotty's Car Between Vacant Spots
Scotty parks his car in a row of $N = 15$ spots. He never parks at either endpoint -- his car is always somewhere in the interior. He then leaves, and when he returns, he finds that $k = 6$ total cars are in the lot (including his own), with the other $5$ having parked uniformly at random among the remaining open spots (at most one car per spot).
What is the probability that Scotty's car has an empty spot on both its left and its right? Round to the nearest thousandth.
Hints
- Scotty is in an interior spot -- does the exact position matter, or does every interior spot look the same in terms of adjacency?
- Count favorable arrangements as those where the $k-1$ other cars avoid the $2$ spots neighboring Scotty's car. Use combinations, since parking order doesn't matter.
- The probability is $\binom{N-3}{k-1} / \binom{N-1}{k-1}$. With $N=15$ and $k=6$, evaluate $\binom{12}{5} / \binom{14}{5}$.
Worked Solution
How to Think About It: The key move here is to recognize that Scotty's specific interior position doesn't matter. Any interior spot has exactly two neighbors. So the question reduces entirely to: of all the ways the other $5$ cars can fill the remaining $14$ spots, in how many arrangements do both of Scotty's neighbors stay empty? Once you see that framing, it's a clean combinatorics ratio -- favorable arrangements over total arrangements.
Quick Estimate: There are $14$ spots for the other $5$ cars. Scotty's car occupies $1$ interior spot, so there are $2$ "blocked" spots (his immediate neighbors) and $12$ free spots. Rough sanity check: each of the $5$ other cars independently avoids the $2$ blocked spots with probability $12/14 \approx 0.857$. If cars were placed independently (they are not -- no replacement -- but this is our ballpark), the probability would be roughly $0.857^5 \approx 0.46$. The actual answer will be somewhat lower because placing $5$ cars without replacement concentrates more mass, but 0.40-0.45 is a reasonable bracket to check against.
Approach: Straightforward combinatorial counting. Total arrangements vs. favorable arrangements, both as combinations.
Formal Solution:
Since Scotty is in an interior spot, the specific position is irrelevant -- any interior spot has exactly $2$ adjacent spots. There are $N - 1 = 14$ remaining spots for the other $k - 1 = 5$ cars.
*Total arrangements:* The $5$ other cars choose $5$ spots from the $14$ available: $$\binom{14}{5}$$
*Favorable arrangements:* Both spots adjacent to Scotty's car must be empty, so the $5$ other cars must park in the $N - 3 = 12$ spots that are not Scotty's spot and not either neighbor: $$\binom{12}{5}$$
*Probability:* $$P = \frac{\binom{N-3}{k-1}}{\binom{N-1}{k-1}} = \frac{\binom{12}{5}}{\binom{14}{5}}$$
Computing: $$\binom{12}{5} = \frac{12!}{5! \cdot 7!} = 792, \qquad \binom{14}{5} = \frac{14!}{5! \cdot 9!} = 2002$$
$$P = \frac{792}{2002} = \frac{36}{91} \approx 0.396$$
Answer: $\dfrac{\binom{12}{5}}{\binom{14}{5}} = \dfrac{792}{2002} \approx \boxed{0.396}$
Intuition
The cleanest move in this problem is the symmetry argument: because Scotty is in the interior, every interior spot looks identical from his car's perspective -- two neighbors, surrounded by spots to either side. You don't need to track where he is. This lets you collapse a seemingly position-dependent problem into a single combinatorial ratio.
The general formula $\binom{N-3}{k-1} / \binom{N-1}{k-1}$ has a nice interpretation: it's a hypergeometric probability. Imagine the $N-1$ non-Scotty spots labeled "blocked" (the 2 neighbors) or "safe" (the other $N-3$). Placing $k-1$ cars uniformly at random is equivalent to drawing $k-1$ items without replacement. The probability that none fall in the 2 blocked spots is exactly the hypergeometric tail -- no items from the "bad" group of size 2. This structure appears whenever you condition on a random subset avoiding a specific region, which shows up in things like order statistics, gap probabilities, and spacing distributions in queuing and auction models.