The Monty Hall Problem

The Monty Hall Problem is famous for being counterintuitive, but the more interesting fact is that the standard answer depends on assumptions that the puzzle usually leaves unstated. This is where most arguments about it actually live.

The 2/3 answer requires all of the following:

1. The host always opens a door.
2. The host always opens a door with a goat.
3. The host never opens the door you picked.
4. If the host has a choice of two goat doors, he picks between them at random.
5. The host always offers the switch.

Change any one of these and the answer changes:

• If the host opens a door at random and it happens to reveal a goat, then switching really is 50/50. The extra probability came from the host's knowledge; remove the knowledge and it vanishes. This is sometimes called "Monty Fall".
• If the host only offers a switch when you picked the car ("Monty Hell"), switching loses every time.
• If the host prefers the lower-numbered goat door when he has a choice, then your odds after seeing Door 3 open jump to 100% — because he would have opened Door 2 if he could.

Marilyn vos Savant's critics were mostly wrong, but the small number of them who complained that the problem was underspecified had a real point. The puzzle as usually stated does not tell you rule 4, and the sharper commentators — including statisticians writing in The American Statistician — said so at the time.

The lesson generalises well beyond game shows: when someone shows you evidence, the important question is not "what did I see?" but "under what circumstances would I have been shown this?"

1959. Martin Gardner published the Three Prisoners Problem in Scientific American — structurally identical, and it generated the same volume of angry mail. Gardner called it "a wonderfully confusing little problem".

1889. Earlier still, Joseph Bertrand's box paradox posed the same conditional trap with three boxes of coins.

1975. Steve Selvin, a biostatistician at Berkeley, wrote a short letter to The American Statistician describing the problem in terms of the television game show Let's Make a Deal, hosted by Monty Hall. In a follow-up letter that same year he gave it the name it now carries.

1990. Marilyn vos Savant answered a reader's version of the question in her Parade magazine column "Ask Marilyn" and said to switch. The response was extraordinary: roughly 10,000 letters, the large majority insisting she was wrong, including around 1,000 from people with doctorates. Several were condescending in ways that became famous in their own right. One professor of mathematics wrote, "You blew it, and you blew it big."

She was right. She ran the argument again in three further columns, and eventually asked schoolteachers across the country to run the experiment with their classes. The empirical results came back at roughly two-thirds, and the tide turned.

A note on the real Monty Hall. The actual host of Let's Make a Deal was under no obligation to open a door at all, and in a 1991 New York Times interview he cheerfully demonstrated that a host who chooses when to offer the switch can beat the contestant every time. The puzzle bearing his name describes a host far more rule-bound than he ever was.

Paul Erdős. The mathematician Paul Erdős, one of the most prolific in history, reportedly refused to accept the answer until he was shown a computer simulation. This is the single most reassuring fact about the Monty Hall Problem.

Most famous puzzles are famous because they are hard. This one is famous because it is easy and people still get it wrong — including people who are professionally excellent at exactly this kind of reasoning.

That makes it unusually valuable evidence about how human reasoning fails.

The specific failure has a name: people treat the outcome they observed as the whole of the evidence, and ignore the process that produced it. Seeing a goat feels like the fact. But the fact is "a host who knows where the car is, and is forbidden from opening it, chose to open that particular door." Those are very different pieces of evidence, and only the second one is informative.

This is the same error that makes selection bias so hard to see. It is why a drug trial's protocol matters as much as its results, why "I only hear from satisfied customers" is not evidence of satisfaction, and why the question "under what conditions would I not have been shown this?" is one of the most useful in all of reasoning.

Machine learning and missing data. Whether it is safe to drop rows with missing values depends entirely on why they are missing — the distinction between missing-at-random and missing-not-at-random is the Monty Hall distinction wearing a lab coat. Data that went missing for a reason carries information, exactly as the host's choice does.

A/B testing and peeking. Stopping an experiment when the numbers look good is a host who only opens a door when it suits him. The observed result is real; the inference drawn from it is not.

Medical diagnostics. Interpreting a positive test requires the base rate and the testing policy, not just the result. Physicians shown base-rate problems in the Monty Hall shape make the same error as everyone else, which is why the framing of diagnostic statistics in natural frequencies rather than percentages measurably improves decisions.

Security and information leakage. In cryptography and privacy, what an adversary learns from an oracle depends on the oracle's policy, not only its answers. A system that reveals "this password is wrong" leaks differently from one that reveals "this username does not exist" — same observable, different constraint, different information.

Let $C_i$ be the event that the car is behind door $i$, and $H_j$ the event that the host opens door $j$. Assume you chose Door 1.

Priors: $P(C_1) = P(C_2) = P(C_3) = 1/3$.

The host's behaviour under the standard rules:

• $P(H_3 \mid C_1) = 1/2$ — both remaining doors hide goats, so he picks at random.
• $P(H_3 \mid C_2) = 1$ — he cannot open Door 1 (yours) or Door 2 (the car), so Door 3 is forced.
• $P(H_3 \mid C_3) = 0$ — he never opens the car.

Total probability of the observation:

$$P(H_3) = \tfrac12 \cdot \tfrac13 + 1 \cdot \tfrac13 + 0 \cdot \tfrac13 = \tfrac12$$

Bayes' theorem then gives:

$$P(C_1 \mid H_3) = \frac{P(H_3 \mid C_1)P(C_1)}{P(H_3)} = \frac{\tfrac12 \cdot \tfrac13}{\tfrac12} = \tfrac13$$

$$P(C_2 \mid H_3) = \frac{P(H_3 \mid C_2)P(C_2)}{P(H_3)} = \frac{1 \cdot \tfrac13}{\tfrac12} = \tfrac23$$

The asymmetry sits entirely in the likelihood $P(H_3 \mid C_2) = 1$ versus $P(H_3 \mid C_1) = 1/2$. The observation is twice as likely in the world where the car is behind Door 2, so it doubles that hypothesis's odds.

Notice what happens if the host is ignorant and opens Door 3 at random, conditioned on it happening to show a goat. Then $P(H_3 \mid C_1) = P(H_3 \mid C_2) = 1/2$, the likelihood ratio is 1, and the posterior stays at 1/2 each. The information lives in the host's constraint, not in the goat.

Generalising to $n$ doors where the host opens $k$ goat doors: staying wins with probability $1/n$, and switching wins with probability

$$\frac{n-1}{n} \cdot \frac{1}{n - k - 1}$$

With $n = 100$ and $k = 98$, switching wins with probability $99/100$.