Monty Hall With Envelopes

Worked Solution

How to Think About It: This is the Monty Hall problem dressed up with envelopes. The key is that the interviewer is not opening a random envelope -- they *know* which one has the money and they always reveal an empty one. That asymmetry is everything. Before you do any math, think about it this way: when you first picked, you had a 1-in-3 shot. The interviewer's reveal does not change what is inside your envelope. It just concentrates all the remaining probability onto one specific envelope. If you can internalize that the interviewer's action is informative (it tells you something about the envelope you did *not* pick), the answer falls out immediately.

Quick Estimate: You picked 1 of 3 envelopes, so your chance of having the money is $1/3$. The other two envelopes collectively hold $2/3$ of the probability. The interviewer removes one empty envelope from that pair, so the surviving envelope inherits the full $2/3$. Switching wins with probability $2/3$, staying wins with probability $1/3$. Switching doubles your odds.

Approach: We can confirm this with a direct application of Bayes' theorem, conditioning on which envelope the interviewer opens.

Formal Solution:

Label the envelopes 1, 2, 3. You pick envelope 1. Let $C_i$ denote the event that envelope $i$ contains the $\$2$. Each has prior probability $1/3$.

Suppose the interviewer opens envelope 3 and shows it is empty. We want $P(C_1 \mid \text{open 3})$ and $P(C_2 \mid \text{open 3})$.

First, compute the likelihood of the interviewer opening envelope 3 under each hypothesis:

• If the money is in envelope 1 (your pick): the interviewer can open either 2 or 3. Assuming they choose uniformly at random among valid options, $P(\text{open 3} \mid C_1) = 1/2$.
• If the money is in envelope 2: the interviewer *must* open envelope 3 (they cannot open 2, which has the money). So $P(\text{open 3} \mid C_2) = 1$.
• If the money is in envelope 3: the interviewer cannot open envelope 3. So $P(\text{open 3} \mid C_3) = 0$.

The total probability of the interviewer opening envelope 3:

$$P(\text{open 3}) = \frac{1}{2} \cdot \frac{1}{3} + 1 \cdot \frac{1}{3} + 0 \cdot \frac{1}{3} = \frac{1}{6} + \frac{1}{3} = \frac{1}{2}$$

Now apply Bayes' theorem:

$$P(C_1 \mid \text{open 3}) = \frac{P(\text{open 3} \mid C_1) \cdot P(C_1)}{P(\text{open 3})} = \frac{(1/2)(1/3)}{1/2} = \frac{1}{3}$$

$$P(C_2 \mid \text{open 3}) = \frac{P(\text{open 3} \mid C_2) \cdot P(C_2)}{P(\text{open 3})} = \frac{(1)(1/3)}{1/2} = \frac{2}{3}$$

The posterior probability that the money is in your original envelope is still $1/3$. The posterior probability that it is in the other unopened envelope is $2/3$.

Answer: You should always switch. Switching wins with probability $2/3$, while staying wins with probability $1/3$. The interviewer's forced reveal of an empty envelope concentrates the $2/3$ prior mass of the two unchosen envelopes onto the single remaining one.

Follow-up: Uninformed Interviewer

This variant changes everything. When the interviewer does not know where the money is and opens a random remaining envelope:

• There is a $1/3$ chance the interviewer reveals the money (game over — you lose or restart).
• Conditional on the interviewer revealing an empty envelope, the probability the money is in your original envelope is:

$$P(\text{your envelope} \mid \text{revealed empty}) = \frac{P(\text{revealed empty} \mid \text{your envelope has money}) \cdot P(\text{your envelope has money})}{P(\text{revealed empty})}$$

$$= \frac{1 \cdot 1/3}{2/3} = \frac{1}{2}$$

So switching gives no advantage — the probability is $1/2$ either way.

Why the difference? In the original problem, the interviewer's knowledge creates an asymmetry: they are *forced* to reveal an empty envelope, which transfers information. When the interviewer acts randomly, the reveal carries no strategic information — it is like a coin flip that happened to come up "empty." The posterior is symmetric.

This distinction (informed vs. uninformed host) is the most common follow-up in quant interviews and tests whether the candidate truly understands conditional probability vs. just memorizing "always switch."