Setup and assumptions
The monty hall problem has three closed doors. Exactly one door hides a prize and the other two doors hide goats. A contestant chooses one door. A host, who knows where the prize is, opens one of the two unchosen doors and reveals a goat. A switch offer follows, allowing the contestant to keep the original door or switch to the only remaining unopened door.
Standard assumptions used for the monty hall problem in probability theory: the host always opens a door with a goat, the host always offers a switch, and when two goat doors are available to open (only when the contestant initially chose the prize), the host chooses uniformly at random between them.
Switching wins with probability \(2/3\); staying wins with probability \(1/3\). The difference comes from conditioning on the host’s constrained action, which carries information about where the prize can be.
Probability model
Let the doors be labeled \(1,2,3\). The contestant initially selects door \(1\). The prize location is equally likely to be behind any door, so \(P(\text{prize behind }1)=P(\text{prize behind }2)=P(\text{prize behind }3)=1/3\).
| Prize location | Host’s allowable door(s) to open (goat only) | Probability host opens door 3 | Outcome if switching after door 3 opens |
|---|---|---|---|
| Door 1 | Door 2 or Door 3 (both goats) | \(1/2\) | Loss (switching moves to a goat) |
| Door 2 | Door 3 only (door 2 is prize) | \(1\) | Win (switching moves to the prize) |
| Door 3 | Door 2 only (door 3 is prize) | \(0\) | Not applicable (door 3 cannot open) |
Conditional probability calculation
Conditioning makes the logic explicit. Suppose the host opens door \(3\) and reveals a goat. The remaining unopened door besides the original choice is door \(2\). The key quantity is \(P(\text{prize behind }2 \mid \text{host opened }3)\).
\[ P(\text{prize behind }2 \mid H_3) = \frac{P(H_3 \mid \text{prize behind }2)\,P(\text{prize behind }2)}{P(H_3)} \]
The denominator expands over the mutually exclusive prize locations:
\[ P(H_3)= P(H_3 \mid \text{prize behind }1)P(\text{prize behind }1)+ P(H_3 \mid \text{prize behind }2)P(\text{prize behind }2)+ P(H_3 \mid \text{prize behind }3)P(\text{prize behind }3) \]
\[ P(H_3)= \left(\frac{1}{2}\right)\left(\frac{1}{3}\right)+ (1)\left(\frac{1}{3}\right)+ (0)\left(\frac{1}{3}\right) =\frac{1}{6}+\frac{1}{3} =\frac{1}{2} \]
The numerator is \(P(H_3 \mid \text{prize behind }2)P(\text{prize behind }2)=(1)(1/3)=1/3\), so:
\[ P(\text{prize behind }2 \mid H_3)=\frac{1/3}{1/2}=\frac{2}{3} \]
Staying with door \(1\) has the complementary probability: \[ P(\text{prize behind }1 \mid H_3)=1-\frac{2}{3}=\frac{1}{3} \]
Why switching works
The monty hall problem concentrates probability mass onto the single remaining unopened door because the host’s choice is not free. When the initial pick is wrong (probability \(2/3\)), the host is forced to reveal the only goat among the two unchosen doors, leaving the prize behind the remaining closed door. When the initial pick is correct (probability \(1/3\)), switching necessarily moves away from the prize.
Visualization
Variants and pitfalls
Host behavior changes the answer
A different host policy changes the conditional probabilities. A host who sometimes does not offer a switch, sometimes opens a door at random without knowledge, or chooses which door to open using a non-uniform rule can produce posterior probabilities different from \(2/3\) and \(1/3\). The standard monty hall problem result relies on the stated host constraints.
Common misconception
The “\(1/2\) vs \(1/2\)” intuition treats the two remaining doors as symmetric after one door is opened. The symmetry is broken because the host’s opening is not an independent random event; it is conditional on avoiding the prize. That dependence is precisely what conditional probability accounts for.