Find the question you need
You pick a door. A host opens another and reveals a goat. Should you switch? The answer can be two-thirds or one-half—depending on the rule the host followed before you saw the goat.
The decisive point is not the number of closed doors. It is how the evidence was generated. We construct two fully specified games, calculate both answers, show an exact outcome table and turn the puzzle into a method for judging real-world claims.
Define the question
Start at chapter 1 · Name what the host knows and is required to do.
Work the mathematical model
Start at chapter 3 · Count outcomes and write conditional probabilities.
Compare what changes
Start at chapter 6 · Calculate the classic and random-host versions.
Check evidence and limits
Start at chapter 8 · Audit ambiguous wording and selection bias.
Practise and continue
Start at chapter 11 · Test altered games and use the connected Probability library.
Full chapter index · Worked checks · Probability subject root
Chapter index
Define the question · Chapters 1–2
Work the mathematical model · Chapters 3–5
Compare what changes · Chapters 6–7
Check evidence and limits · Chapters 8–10
Practise and continue · Chapters 11–12
Imagine three doors, one car and two goats. You choose Door 1 without knowing the car’s location. Before anybody opens a door, the car is behind your choice with probability 1/3 and behind one of the other two doors with probability 2/3. That initial split is common to both versions of our game.
After a goat appears, two doors remain shut. It is tempting to divide probability evenly between them. But a visible outcome does not tell us the mechanism that generated it. Did the host know where the prize was? Was the host required to expose a goat? What happens if the host accidentally exposes the car? A trustworthy answer records those rules first.
CHAPTER 2 OF 12 · Define the question
2. Define the informed-host game without loopholes
In the classic game, the host knows the car’s location, never opens your chosen door, never opens the car door, always opens exactly one goat door among the two unchosen doors and always offers the option to switch. When two goat doors are available, suppose the host chooses between them fairly. The last assumption makes door-specific observations unambiguous; it is not needed for the overall two-thirds success of always switching.
This model must not be silently replaced by one where the host sometimes refuses an offer, behaves differently when your first choice is correct or selects doors without knowledge. Each change can alter the information contained in the reveal and therefore the conditional answer.
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CHAPTER 3 OF 12 · Work the mathematical model
3. Enumerate car positions before using a formula
Hold your original choice at Door 1 and let the car be uniformly distributed across Door 1, Door 2 and Door 3. If the car is behind Door 1, switching to the only other closed door loses. If it is behind Door 2 or Door 3, the knowledgeable host must reveal the goat on the other unchosen door, so switching wins.
The exhaustive count is two switching wins out of three equally likely initial car placements. The host’s reveal does not turn your original one-third chance into one-half. It removes one losing door from the two-thirds probability that your initial choice was wrong.
| Car behind | Host shows | Stay wins? | Switch wins? |
|---|---|---|---|
| Door 1 | Door 2 or Door 3, both goats | Yes | No |
| Door 2 | Door 3, goat | No | Yes |
| Door 3 | Door 2, goat | No | Yes |
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CHAPTER 4 OF 12 · Work the mathematical model
4. Understand where the other two-thirds went
Write C for the event that your first choice is correct. Under the informed-host rules, P(C) = 1/3 and P(not C) = 2/3. Staying wins exactly on C. Switching wins exactly on not C. Therefore P(stay wins) = 1/3 and P(switch wins) = 2/3. These are unconditional success rates for the specified game.
The host’s action is constrained information, not a fresh lottery between remaining doors. Consider a hundred-door variation: you pick one, a knowledgeable host reveals 98 goats and leaves only one unchosen door closed. With the same rules, staying wins 1/100 and switching wins 99/100. The three-door case uses precisely the same allocation.
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CHAPTER 5 OF 12 · Work the mathematical model
5. Use Bayes when the identity of the opened door matters
Suppose you chose Door 1, the host opened Door 3 and displayed a goat. Under the fair tie-break, the probability of that precise reveal is 1/2 if the car is behind Door 1, is 1 if the car is behind Door 2, and is 0 if the car is behind Door 3.
Bayes’ rule weights each prior 1/3 by that likelihood. The unnormalised weights for car at Doors 1, 2 and 3 are 1/6, 1/3 and 0. Their sum is 1/2. After dividing by 1/2, the probabilities are 1/3, 2/3 and 0. Door 2—the available switch—is still twice as likely as Door 1. The host’s specific reveal can be analysed without treating it as an unconditioned random exclusion.
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CHAPTER 6 OF 12 · Compare what changes
6. Change only one rule: the host opens at random
Now imagine an ignorant host who does not know where the car is. After you choose Door 1, the host randomly opens Door 2 or Door 3 with equal probability. Sometimes this mistakenly reveals the car. We study only the occasions when the opened door happens to contain a goat and a choice to switch remains possible.
This is a different experiment from the classic game. When your initial choice is correct, both other doors contain goats; a random reveal succeeds with probability 1. When your initial choice is wrong, one other door holds the car, so the random reveal shows a goat with probability 1/2. The observed goat now selects disproportionately from games where your first choice was correct.
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CHAPTER 7 OF 12 · Compare what changes
7. Calculate the conditional answer in the random-host game
Let G mean ‘a random host happened to reveal a goat’. The two ways to obtain G are: initial choice correct, probability (1/3)×1 = 1/3; or initial choice wrong, probability (2/3)×(1/2) = 1/3. Thus P(G) = 2/3.
Among the games with G, the two possibilities have equal weight. Therefore P(first choice correct | G) = (1/3)/(2/3) = 1/2, and P(first choice wrong | G) = 1/2. Staying and switching each win half the qualifying games. This does not contradict the classic two-thirds result: one underlying selection rule changed, and so did the conditional population being analysed.
| Initial position | Share of all games | Chance random host shows goat | Share of all games with this outcome |
|---|---|---|---|
| Your door has car | 1/3 | 1 | 1/3 |
| Your door has goat | 2/3 | 1/2 | 1/3 |
| Total | 1 | — | 2/3 |
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CHAPTER 8 OF 12 · Check evidence and limits
8. Do not misread the missing third of random-host games
In one-third of all random-host games the host inadvertently opens the car door. Those games are not part of the post-goat switching comparison, because the premise ‘host reveals a goat and offers two doors’ failed. Counting them as switching losses or silently dropping them from the numerator only can produce a different and misleading answer.
For example, think of 600 hypothetical games in the exact theoretical proportions: 200 initial choices correct and 400 incorrect. The ignorant host shows a goat in all 200 correct-choice games and in half of the 400 incorrect-choice games. There are 400 qualifying goat reveals, with 200 switching wins and 200 staying wins. This is a frequency illustration of the conditional calculation, not an observed experiment.
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CHAPTER 9 OF 12 · Check evidence and limits
9. Test the host’s offer as an additional source of selection
If a host is allowed to offer a switch only on selected occasions, learning that an offer was made becomes another event to condition on. For example, suppose the knowledgeable host offers the switch only when the contestant’s initial choice is wrong. Then, given a genuine offer, switching wins with certainty under that constructed rule. The result says more about the host’s selective offer than about the two remaining doors.
A statistical report should name the complete protocol: how the prize was placed, how the player chose, what information the host possessed, what door was opened, whether the car could be revealed, and which rounds entered the analysis. Without that, the phrase ‘after a goat appeared’ does not uniquely determine the probability.
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CHAPTER 10 OF 12 · Check evidence and limits
10. Recognise the everyday pattern without overclaiming
The puzzle models a broad reasoning problem: observations produced by a decision or screening procedure carry the signature of that procedure. A selection rule can change the composition of the cases we see. Similar questions arise in sampling, diagnostic testing, recruitment, auditing and algorithm evaluation, although their probabilities require their own models.
The transferable habit is straightforward: distinguish prior probabilities, a process that generates evidence, and the condition defining which observations are reported. Then compute conditional probabilities from all plausible paths. A surprising result is often evidence that one of those stages was skipped, not that ordinary arithmetic stopped working.
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Try these before reading the answers: (A) A knowledgeable host follows the classic rules for 100 doors and opens 98 goats. (B) A random host opens one of two unchosen doors and you condition on seeing a goat. (C) A knowledgeable host always opens a goat, but the player is allowed to stay without any information about which goat door the host will open.
Answers: (A) switching wins with probability 99/100; (B) switching wins with conditional probability 1/2; (C) staying has probability 1/3 and switching 2/3. For (C), the host’s tie-breaking rule can affect a posterior conditioned on which particular door was opened, so specify it before answering a more detailed door-specific question.
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CHAPTER 12 OF 12 · Practise and continue
12. Build a durable Probability checking method
Before accepting a conditional-probability argument, write a compact audit: What is sampled? What are the prior probabilities? Who knows what? Which actions are mandatory? Which are random or selective? What event is being conditioned on? Are any failed or excluded rounds missing? Is the claim about all rounds or only a subset?
The car-and-goat story is memorable precisely because two nearly identical pictures can emerge from two different mechanisms. Keep the mechanism attached to the answer. Continue with the inspection paradox and length-biased sampling to see another case where how observations are obtained changes what a typical observation looks like.
Sources and connected reading
This is an original educational worked explanation. Illustrative numbers and scenarios are invented for teaching unless a source is explicitly identified. The assumptions and limits stated in each chapter determine whether a result applies.