Monty Hall When the Host Behaves Differently

Everyone remembers the punchline of Monty Hall — switch and you win the car two-thirds of the time — but the proof depends on three quiet assumptions about the host: he knows where the car is, he always opens a door, and he always opens a goat door. Interviewers love changing exactly one of those and watching whether you notice.

The baseline, briefly

Three doors, one car. You pick door 1. The host, who knows where the car is and must open a losing door other than yours, opens door 2 or 3 to reveal a goat, then offers a switch. Your original pick was right with probability $1/3$; that number can't change just because the host, who was always going to find a goat, found one. So the remaining unopened door absorbs the other $2/3$. Switching wins with probability $2/3$.

Worked variant: the host doesn't know

Now the host opens a random door, not knowing where the car is, and it happens to reveal a goat. This looks identical to the baseline from where you're sitting, but it isn't. Condition on the actual event: "random door opened, goat shown." Using Bayes' rule on the two cases —

• Car behind your door (prior $1/3$): host opens a random other door, always a goat. $P(\text{goat shown} \mid \text{your door}) = 1$.
• Car behind another door (prior $2/3$): host opens one of the two remaining doors at random; it shows a goat only half the time, since half the time it reveals the car instead. $P(\text{goat shown} \mid \text{other}) = 1/2$.

$$P(\text{your door} \mid \text{goat shown}) = \frac{\frac13 \cdot 1}{\frac13\cdot 1 + \frac23 \cdot \frac12} = \frac{1/3}{2/3} = \frac12.$$

Once you condition on actually seeing a goat, the asymmetry that powered the original result is gone: switching and staying are now a coin flip. The host's ignorance is what erased the information.

The generalizable technique

Don't compute a probability from what happened; compute it from the rule that generated what happened. Write down the host's decision procedure as a conditional probability for every case (car behind your door vs. not), then run Bayes' rule across those cases — the arithmetic is only ever a weighted average, weighted by how likely each case was to produce the observed action. This is the same move needed for the two-children problem and for reading DNA-match evidence correctly: the observation alone is never enough, you also need the process that could have produced it.