Monty Hall paradox in statistical terms
The monty hall paradox describes a probability update caused by an informed action. Three doors hide one car and two goats. A door is chosen. The host, who knows where the car is, opens one of the two unchosen doors that has a goat and offers a switch to the remaining unopened door. The core statistical issue is conditioning on the host’s constrained choice.
Host behavior assumptions
A complete probability model requires explicit assumptions about the host: the host always opens a door different from the initial choice, always reveals a goat, and always offers a switch. When two goat doors are available to open, the host selects between them uniformly at random.
Statistical framing
The initial choice carries a prior probability \( \tfrac{1}{3} \) of being correct. The host’s reveal is not an independent random event; it is conditioned on the car’s location and the initial choice. The remaining unopened door inherits the probability mass of the two doors not originally chosen.
Case partition using the law of total probability
Let \(C\) be the event that the initially chosen door hides the car. Then \(P(C)=\tfrac{1}{3}\) and \(P(C^{c})=\tfrac{2}{3}\). Two mutually exclusive cases control the outcome:
| Underlying case | Probability | Host action | Outcome if staying | Outcome if switching |
|---|---|---|---|---|
| Car behind the initially chosen door (\(C\)) | \( \tfrac{1}{3} \) | Host must open one of the two goat doors | Win | Lose |
| Car not behind the initially chosen door (\(C^{c}\)) | \( \tfrac{2}{3} \) | Host must open the only goat door among the two unchosen doors | Lose | Win |
Staying wins only in the first case, so \( P(\text{win by staying}) = P(C) = \tfrac{1}{3}. \) Switching wins exactly in the second case, so
\[ P(\text{win by switching}) = P(C^{c}) = \frac{2}{3}. \]
Conditional probability derivation with Bayes’ theorem
A Bayes calculation makes the conditioning explicit. Label doors \(1,2,3\). Suppose the initial choice is door \(1\), and the host opens door \(3\) showing a goat. Let \(H_3\) be the event “host opens door 3”. The posterior probability that the car is behind door \(2\) is \(P(\text{car at 2}\mid H_3)\).
\[ P(\text{car at 2}\mid H_3) = \frac{P(H_3\mid \text{car at 2})\,P(\text{car at 2})}{P(H_3)}. \]
The priors are \(P(\text{car at 1})=P(\text{car at 2})=P(\text{car at 3})=\tfrac{1}{3}\). Under the host assumptions:
| Car location | \(P(H_3\mid \text{car location})\) | Reason |
|---|---|---|
| Car at 1 (initial door) | \( \tfrac{1}{2} \) | Two goat doors available; random choice between doors 2 and 3 |
| Car at 2 | \( 1 \) | Door 3 is the only goat door among unchosen doors |
| Car at 3 | \( 0 \) | Door 3 cannot be opened because it hides the car |
The total probability of opening door 3 is \[ P(H_3) = \left(\tfrac{1}{2}\right)\left(\tfrac{1}{3}\right) + (1)\left(\tfrac{1}{3}\right) + (0)\left(\tfrac{1}{3}\right) = \tfrac{1}{6} + \tfrac{1}{3} = \tfrac{1}{2}. \]
Therefore \[ P(\text{car at 2}\mid H_3) = \frac{(1)\left(\tfrac{1}{3}\right)}{\tfrac{1}{2}} = \frac{2}{3}, \qquad P(\text{car at 1}\mid H_3) = \frac{\left(\tfrac{1}{2}\right)\left(\tfrac{1}{3}\right)}{\tfrac{1}{2}} = \frac{1}{3}. \]
Visualization: probability tree for switching vs staying
Common misconceptions in the monty hall paradox
“Two doors remain, so 1/2 each”
Equal probability would require the opened door to be chosen without regard to the car’s location. In this setting, the opened door is selected with knowledge and constraints, so the remaining doors are not symmetric.
Independence assumption
The event “host opens a particular door” depends on the hidden placement of the car. Conditioning on the host’s action is the statistical mechanism that changes the odds.
Generalization to \(n\) doors
With \(n\) doors, one car, and a host who opens \(n-2\) goat doors after the initial choice, the same logic applies. The initial choice has probability \( \tfrac{1}{n} \) of being correct, so switching wins with probability \[ P(\text{win by switching}) = 1 - \frac{1}{n} = \frac{n-1}{n}. \]
Simulation interpretation
Repeated trials approximate the theoretical probabilities by the law of large numbers. The observed switching win rate tends toward \( \tfrac{2}{3} \) as the number of games increases, provided the host behavior matches the stated assumptions.