Lesson · Paradoxes and Their Exits · Probability

Monty Hall

A car is behind one of three doors. You choose door 1, and a host who knows where it is opens door 3 to show a goat. The lab updates on the host's act, not on the bare fact that door 3 is empty, and finds that door 2 now holds the car with probability two thirds. A host who opens a door at random gives one half, which is why the host's rule has to be part of the problem.

Open the interactive lesson Practice

car at 1, 2 or 3: 1/3 each; you pick door 1 host opens 3: 1/2 if car at 1, 1 if at 2 P(car at 2 | host opens 3) = 2/3 random host: 1/2 each exit: accept the conclusion

The host's rule is the evidence

Core ideas

What this lesson establishes

Three ideas explain why the obvious answer is wrong and when it is right.

The evidence is what the host did

The fact that door 3 is empty is not all you learned. You learned that the host, following his rule, opened door 3. The likelihood of that act under each hypothesis is what updating needs.

A forced move is stronger evidence than a free one

If the car is behind door 2, the host must open door 3. If it is behind door 1, he could have opened door 2 or door 3 and chose 3 with probability one half. So his act is twice as probable if the car is behind 2.

Change the rule and the answer changes

A host who opens one of the other two doors at random, and happened to show a goat, gives no extra weight to door 2. The same sight, door 3 open and empty, is then worth one half each. The model decides, and the lesson states it.

The material

Updating on the host's act

The game, the common answer, the update, and the host who is not informed.

A game show has three doors. Behind one is a car and behind the other two are goats. You choose a door, say door 1. The host, who knows where the car is, opens one of the other doors to show a goat, and he always does so; if both of the others hide goats he picks at random. He opens door 3. You may stay with door 1 or switch to door 2. Which gives the better chance of the car?

Most people say it makes no difference. Two doors are closed, the car is behind one of them, and so each has chance one half. The reasoning treats the news as “door 3 is empty” and spreads the chance evenly over what remains. That is the premise to look at, because the news was not that, but that the host opened door 3.

Let the hypotheses be the three places for the car, each with prior one third. The evidence is that the host opened door 3. Its likelihood under each hypothesis follows from his rule. If the car is behind door 1, he may open door 2 or door 3, and opens 3 with probability one half. If the car is behind door 2, he cannot open door 2 and cannot open door 1, which you chose, so he must open door 3, with probability 1. If the car is behind door 3, he cannot open it, and the probability is 0.

likelihoods of opening 3: 1/2, 1, 0 prior times likelihood: 1/6, 1/3, 0 total: 1/2 posterior: 1/3, 2/3, 0

So the car is behind door 2 with probability two thirds, and staying wins with probability one third. The lab shows the table and, with the payoffs of staying and switching, names switching as the better act, worth two thirds. The reason is the one in the second concept. Door 1 gave the host a free choice and door 2 gave him none, so seeing him open door 3 is twice as probable if the car is behind door 2.

Example · Counting it another way

Door 1 holds the car with chance one third, and nothing the host does can change that, because he never opens your door and never opens the car. The other two doors together hold two thirds. The host has shown that one of them is empty, and the whole of the two thirds falls on the other. That is the same answer, reached without the table.

Set out as this course sets out a paradox, the argument is short, and every premise of it is part of the setup.

The car was placed at random, so each door holds it with probability one third.
The host knows where the car is, never opens your door or the car's door, and chooses at random when he has a choice.
Under that rule his opening door 3 had probability one half if the car is behind door 1, 1 if it is behind door 2, and 0 if it is behind door 3.
So, by Bayes' rule, the car is behind door 2 with probability two thirds, and switching doubles your chance.

Every step is valid, the premises are the rules of the game, and the conclusion felt wrong. The exit is the third: accept it. The conclusion was unacceptable only because the argument for one half was familiar, and that argument is not the paradox but a mistake, whose false premise, that all you learned is that door 3 is empty, the lab exposes. The price of exit three is small. One loses the habit of treating the sight of an empty door as the whole of the evidence, and gains the habit of asking how the sight came about.

The second preset shows how much the rule matters. Suppose the host does not know where the car is. He opens one of the two other doors at random and, as it happens, it shows a goat. Now the likelihood of opening door 3 and showing a goat is one half if the car is behind door 1, one half if it is behind door 2, and 0 if it is behind door 3, since that would have shown the car. The posterior is one half for each of the first two doors, and switching is worth no more than staying. The lab's best-act tile names both.

The third preset has four doors. You choose door 1, and the host, who knows, opens door 4. Door 1 gave him a choice among three doors, so the likelihood is one third; doors 2 and 3 gave him a choice between two, so it is one half; door 4 gave him none. The posterior is one quarter for door 1 and three eighths for each of door 2 and door 3. Staying still wins one time in four, and each of the two other doors is better.

The lab's limit is that it computes from the rule you type. It does not know which rule the host followed, and nothing in the sight of an open door reveals it. A reader who is not told the rule has to assume one, and the standard statement of the puzzle gives it for that reason. A problem without the rule has two correct answers, one for each rule, and the dispute about which is meant is a dispute about the model.

Interactive

Exact Bayesian updating

prior times likelihood, normalised; the Bayes factor is the weight of the data

The car's place, given what the host did

You chose door 1. The table lists the likelihood of the host's act under each place for the car, then the posterior. The tiles report door 1, the door you hold. The best-act tile compares staying with switching, using the payoffs 1 for the car and 0 for a goat. Then edit the likelihood rows of the first preset toward the second's and watch the best act change.

Worked example

Prior, one per hypothesis

Likelihoods: a row per hypothesis, a column per outcome, rows separated by “;”

Data: outcome names separated by spaces (red*10 is ten reds)

Payoffs, “act: one per hypothesis; …” (empty for none)

The hypothesis the tiles report

Posterior—

Bayes factor—

The data—

Next outcome is the first—

Best act now—

Method

Updating on what an informed agent did

Five steps, for Monty Hall and for any puzzle where someone who knows something acts in front of you.

Fix the hypotheses and the prior

List the places the prize could be, with equal priors if the setup says the placing was at random.

State the agent's rule

Write what the host does under each hypothesis, including what he is forbidden to do. The rule is part of the problem.

Find the likelihood of the act that occurred

For each hypothesis, give the probability that the host did what you saw. A forced act has probability 1 and a free one is divided among the choices he had.

Update and compare the acts

Multiply prior by likelihood, normalise, and compute the chance of winning by staying and by switching.

Try a different rule, then name the exit

Change the host to one who acts at random and see whether the answer moves. If it does, the rule was doing the work. With the rule fixed, the exit is the third: the two thirds is accepted, and what is given up is the habit of counting doors.

The host and the doors

Score 0 / 0

Choose the answer you can justify.

The car's place after the host opens door 3

You hold door 1. The host knows the car, never opens your door or the car's door, and chooses at random when he has a choice.

prior: 1/3 for each door car at 1: host opens 2 or 3, so P(opens 3) = 1/2 car at 2: host must open 3, so P(opens 3) = 1 car at 3: host cannot open 3, so P(opens 3) = 0 products: 1/6, 1/3, 0 with total 1/2 posterior: 1/3, 2/3, 0 so P(car at 2 | host opens 3) = 2/3 random host instead: 1/2, 1/2, 0 and posterior 1/2 each

With an informed host, switching wins with probability two thirds. With a host who opened at random and showed a goat, the same sight gives one half.

Common mistakes

Thinking that two doors remain, so each is one halfThe count of doors would give one half only if the host's act were equally probable under every place for the car. It is not: it was forced if the car is behind door 2 and free if it is behind door 1. The lab's table shows the likelihoods one half, 1 and 0, and the posterior one third and two thirds. Count the host's choices, not the doors.

Treating the sight of an empty door as the whole of the evidenceWhat you learned is that the host, following a rule, opened it. The same sight is worth one half each with a host who opens at random and happens to show a goat. Always ask what the agent would have done under each hypothesis.

Taking the answer to hold without the host's ruleThe two thirds depends on a host who knows the car and never shows it. The lab computes from the rule you type. A story that omits the rule has no single answer, and saying so is correct.

Check your understanding

✓

Finish when you can compute the posterior from the host's rule and show it changes with the rule.

Given a host's rule, you should be able to write the likelihood of his act under each place for the prize, compute the posterior by Bayes' rule, name the better act, show how the answer changes for a host who acts at random or when there are more doors, and say which exit the two-thirds answer takes.

Note: The Monty Hall answer is a case of the third exit: the conclusion is accepted, and the work is to see why it was resisted. The last lesson takes up a paradox of a different kind, in which nothing is computed from chance and a sentence can be true and still cannot be believed.