Generalized Monty Hall: n Doors
You are a contestant on a game show. There are $n \geq 3$ doors. Behind one door is a prize; the other $n - 1$ doors are empty. The prize is placed uniformly at random.
You pick a door. The host -- who knows where the prize is -- then opens one of the other $n - 1$ doors, always revealing an empty one. You are now offered a chance to switch to any one of the remaining $n - 2$ unopened doors (not your original pick, not the one the host opened).
(a) What is the probability of winning if you switch?
(b) For $n = 8$, what is the probability you win by switching? Express as a fraction.
Hints
- To win by switching, you need two things: your original pick was wrong, AND you switch to the correct door. These are not independent -- condition on them in sequence.
- After the host reveals an empty door, there are $n - 2$ remaining doors (excluding yours and the opened one). If your original pick was wrong, the prize is uniformly among those $n - 2$ doors.
- Multiply: $P(\text{original wrong}) \times P(\text{switch to correct} \mid \text{original wrong}) = \frac{n-1}{n} \cdot \frac{1}{n-2}$.
Worked Solution
How to Think About It: The classic Monty Hall trap is forgetting that the host's action is not random -- he always reveals an empty door. This means opening a door carries information. Your initial pick is correct with probability $1/n$. The remaining $n - 1$ probability mass was "spread" across the other doors, and the host just collapsed it down to $n - 2$ doors. Switching distributes that $1/n$ mass over $n - 2$ remaining doors, giving you a better-than-$1/n$ chance. The rule of thumb: switching is always better, regardless of $n$.
Quick Estimate: For $n = 8$: staying wins with probability $1/8 = 12.5\%$. Switching should win more often. Quick upper bound: at most $1/(n-2) = 1/6 \approx 16.7\%$ if the whole conditional probability landed on the remaining doors. The exact answer is somewhere in that range.
Formal Solution:
To win by switching, two things must happen: 1. Your initial pick was wrong (probability $(n-1)/n$). 2. Given your pick was wrong, the prize is behind the specific door you switch to. After the host opens one empty door, there are $n - 2$ remaining doors (all non-initial, non-opened), and the prize is uniformly among them. So the conditional probability of picking it is $1/(n-2)$.
These events are independent given the setup, so: $$P(\text{win by switching}) = \frac{n-1}{n} \cdot \frac{1}{n-2} = \frac{n-1}{n(n-2)}$$
Comparison to staying: $P(\text{win by staying}) = 1/n$. Since $\frac{n-1}{n(n-2)} = \frac{1}{n} \cdot \frac{n-1}{n-2} > \frac{1}{n}$ for all $n \geq 3$, switching is always strictly better.
For $n = 8$: $$P(\text{win by switching}) = \frac{7}{8 \cdot 6} = \frac{7}{48}$$
Answer: Switching wins with probability $\dfrac{n-1}{n(n-2)}$. For $n = 8$, this is $\dfrac{7}{48} \approx 14.6\%$, compared to $12.5\%$ for staying.
Intuition
The Monty Hall problem is a lesson in conditional probability and the value of information. The host's action is not random -- he always opens an empty door -- and that constraint is what makes switching valuable. If the host opened a door at random and happened to reveal an empty one, switching would provide no benefit (the probabilities would equalize). But since the host is constrained to reveal only empty doors, the opened door is information-bearing: it concentrates the probability mass you did not initially capture.
For large $n$, the switching advantage becomes more dramatic in absolute terms. With $n = 3$, switching gives $2/3$ vs. $1/3$ -- a $2\times$ edge. With $n = 100$, staying gives $1\%$, switching gives $99/(100 \times 98) \approx 1.01\%$ -- barely any edge per switch. This is because with many doors and one host reveal, you still have 98 doors to switch among, so each individual door gets very little of the redistributed probability. The practical upshot: the Monty Hall intuition -- "the host's constraint forces you to update" -- is a building block for understanding how constrained observations carry information, which is fundamental in Bayesian reasoning and market microstructure.