Monty Hall With 100 Doors
There are 100 doors. Behind one door is a car; behind each of the other 99 doors is a goat. You pick one door at random. The host, who knows exactly which door hides the car, then opens 98 of the remaining 99 doors, every one of which reveals a goat. Two doors are now closed: yours and one other.
- Should you switch to the remaining unopened door?
- What is the probability you win the car if you switch? What if you stick with your original choice?
- Generalize: with $n$ doors, the host opens $n - 2$ goat doors. What is your win probability if you switch?
Hints
- Think about what your original pick tells you versus what the host's deliberate choices tell you.
- If you always switch, you win exactly when your original pick was wrong. How often is that?
- The host is forced to keep the car door closed. So among the 99 doors you didn't pick, 98 goat doors get eliminated, concentrating all $99/100$ of the probability onto the one surviving door.
Worked Solution
How to Think About It: This is the classic Monty Hall problem scaled up to make the logic unmistakable. The key is that the host is not acting randomly -- he *knows* where the car is and is forced to open only goat doors. That means his action carries information. When you first picked, you had a 1-in-100 shot. The host then systematically eliminated 98 losing doors from the other 99. All of the probability mass that was spread across those 99 doors gets funneled into the single door the host chose *not* to open. Your door is still stuck at its original 1/100.
Quick Estimate: Before any math, think about it this way: if you always switch, you win whenever your original pick was wrong. Your original pick is wrong 99 times out of 100. So switching wins with probability 99/100 = 0.99. Done. No Bayes' theorem needed for the answer -- the logic is that clean.
Approach: We can formalize this with conditional probability to confirm the intuition.
Formal Solution:
Label the doors $1, 2, \ldots, 100$. Suppose you pick door 1. Let $C$ be the door hiding the car, uniformly distributed: $P(C = i) = 1/100$ for each $i$.
- Case 1: Car is behind door 1 (your pick). This happens with probability $1/100$. The host opens 98 of the remaining 99 doors (all goats). If you switch, you lose.
- Case 2: Car is behind some other door $j \neq 1$. This happens with probability $99/100$. The host *must* open all 98 remaining goat doors, leaving door $j$ as the only other closed door. If you switch, you land on door $j$ and win.
So the probability of winning by switching is:
$$P(\text{win} \mid \text{switch}) = P(C \neq 1) = \frac{99}{100}$$
And sticking gives:
$$P(\text{win} \mid \text{stick}) = P(C = 1) = \frac{1}{100}$$
Generalization to $n$ doors: The same argument works. You pick one door out of $n$. The host opens $n - 2$ goat doors. Switching wins whenever your original pick was wrong:
$$P(\text{win} \mid \text{switch}) = \frac{n - 1}{n}$$
For the standard 3-door Monty Hall, this gives $2/3$. For 100 doors, $99/100$. As $n \to \infty$, switching is essentially guaranteed to win.
Answer: Always switch. The probability of winning by switching is $99/100 = 0.99$, while sticking wins with probability $1/100 = 0.01$. In the general $n$-door version, switching wins with probability $(n-1)/n$.
Intuition
The reason switching is so powerful is that the host is not a random agent -- he has perfect information and is constrained to only open goat doors. His actions are a giant information funnel: he takes the 99% probability mass spread across the doors you didn't pick and compresses it into a single door. Your original pick, made in ignorance, is frozen at 1%. Scaling up from 3 doors to 100 makes this crystal clear, which is why this variant is such a great teaching tool.
In quant interviews, this problem tests whether you understand conditional probability at a gut level. The broader lesson is about information asymmetry: when someone with superior information takes a constrained action (like a market maker posting a quote, or an informed trader choosing *not* to trade certain names), you can infer a lot from what they did and what they chose not to do. The Monty Hall problem is the simplest possible model of learning from the actions of an informed agent.