Five doors, two different hosts
A game show has five closed doors; exactly one hides a car and the rest hide goats. You pick door 1. In version A, the host knows where the car is, always opens three of the other four doors to reveal goats (never the car, never your door), and then offers you the single remaining unopened door. In version B, the host is replaced by an audience member who has not seen the setup, does not know where the car is, and opens three of the other four doors at random; on this occasion all three happen to be goats, and the same offer is made. A friend says the two versions are identical because 'in both cases you have seen three goats and two doors remain, so it is 50–50 either way.' You are asked to decide, for each version, whether switching is better than staying, and to say how the answer would change with n doors and k opened.
1. For version A, compute the probability of winning by switching, with an explicit case analysis.
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Split on the location of the car.
- Car behind door 1 (probability 1/5). The other four doors are all goats. The host opens any three goats and leaves one goat behind the remaining door. Switching loses.
- Car not behind door 1 (probability 4/5). Among the four unchosen doors there are three goats and the car. The host must open goats and cannot open the car, so the only door he can leave closed is the car door. Switching wins with certainty in this branch.
P(win by switching) = 4/5 = 0.80. P(win by staying) = 1/5 = 0.20.
The key is that the host's action is constrained by the car's location, so 'three goats revealed' carries almost no new information about door 1; the 4/5 probability mass originally spread over the four unchosen doors is compressed onto one door.
2. For version B, compute the probability of winning by switching after the random opener happens to reveal three goats.
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Condition on the observed event G = 'the three opened doors were all goats'.
- If the car is behind door 1 (prior 1/5), the other four doors are all goats, so any three opened are goats: P(G | car at 1) = 1.
- If the car is not behind door 1 (prior 4/5), the four unchosen doors contain the car and three goats. A random choice of three doors reveals only goats only if it picks exactly the three goats: P(G | car not at 1) = 1/C(4,3) = 1/4.
Bayes: P(car at 1 | G) = (1/5 × 1) / (1/5 × 1 + 4/5 × 1/4) = 0.2 / (0.2 + 0.2) = 1/2.
So switching wins 1/2 and staying wins 1/2: no advantage. The random opener's success is evidence that door 1 hides the car, because a random opener is more likely to survive without revealing the car when the car is already yours. The friend's 50–50 claim is right here but wrong for version A.
3. Generalise: with n doors, you pick one, and k of the other n−1 doors are opened to reveal goats. Give the switch probability for the informed host and for the random opener, and check both against n = 5, k = 3.
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Informed host. The car is among the unchosen doors with probability (n−1)/n. After k goat doors are opened, n−1−k alternative doors remain, and the host's constraints distribute that probability equally among them: P(switch wins) = (n−1)/[n(n−1−k)]. With n = 5, k = 3: 4/(5 × 1) = 4/5, as found. With n = 3, k = 1: 2/(3 × 1) = 2/3, the classic result.
Random opener, conditioned on k goats. If the car is at your door, the event has probability 1. If not, the opener must pick exactly the k goats out of the n−1 unchosen doors, probability 1/C(n−1,k). Then P(car at your door | k goats) = 1/[1 + (n−1)/C(n−1,k)], so P(switch wins) = 1 − that. With n = 5, k = 3: C(4,3) = 4, so P(stay wins) = 1/(1 + 4/4) = 1/2.
Note the structural difference: the informed host's probability depends on k only through how many alternatives survive, and switching always beats staying for k ≥ 1. The random opener's probability depends on C(n−1,k); when k = n−2 it is always 1/2, and for small k the random opener is strongly informative in favour of your original door.