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๐Ÿšช The Monty Hall Problem

A probability puzzle that fooled thousands of mathematicians

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๐ŸŽฒ The Setup
Topics: conditional probability, tree diagrams, common misconceptions
You are a contestant on a game show. There are 3 doors. Behind one door is a car. Behind the other two are goats.

Step 1: You pick a door โ€” say, Door 1.
Step 2: The host (who always knows where the car is) opens a different door to reveal a goat โ€” say, Door 3.
Step 3: The host offers you a choice: stick with Door 1, or switch to Door 2.

Question: Does it matter whether you switch? Is one choice better than the other?
Tree of possibilities (you always pick Door 1):

Car behind Door 1 (prob 1/3): Host opens Door 2 or 3 โ†’ Stay wins, Switch loses
Car behind Door 2 (prob 1/3): Host must open Door 3 โ†’ Stay loses, Switch wins
Car behind Door 3 (prob 1/3): Host must open Door 2 โ†’ Stay loses, Switch wins

Stay wins in 1 out of 3 cases. Switch wins in 2 out of 3 cases.
Why switching helps: When you first pick, you have a 1/3 chance of being right. The other two doors together hold a 2/3 chance. The host's reveal doesn't change your original odds โ€” it just concentrates the 2/3 probability onto the one remaining door. Switching is like betting on both other doors at once.
๐Ÿ“ Questions
1
When you first pick Door 1, what is the probability the car is behind your door? What is the probability it is behind one of the other two doors combined? 2 marks
2
The host opens Door 3 to reveal a goat. The host always reveals a goat โ€” this is guaranteed. Does this new information change the probability that the car is behind Door 1? Explain. 2 marks
3
If P(car behind Door 1) is unchanged after the reveal, and the car must be behind Door 1 or Door 2 (Door 3 is open), what is P(car behind Door 2) now? 2 marks
4
Complete the table below showing all scenarios where the car is behind each door and you always pick Door 1. Tick whether staying or switching wins each time. 3 marks
Car behindHost opensStay wins?Switch wins?
Door 1Door 2 or 3
Door 2
Door 3
5
Using your table, what fraction of the time does staying win? What fraction of the time does switching win? What is the optimal strategy? 3 marks
6
Many people believe it "doesn't matter" because they see two remaining doors and assume 50/50. Explain clearly why this reasoning is wrong. 3 marks
7
Extension: Suppose there were 100 doors. You pick one. The host opens 98 doors, all revealing goats, leaving your door and one other. What is the probability of winning if you switch? What if you stay? 5 marks

Answer Key

๐Ÿšช Monty Hall

1.P(car behind Door 1) = 1/3. P(car behind Door 2 or 3) = 2/3.
2.No. The host's action was predetermined โ€” he always reveals a goat. It gives you no new information about Door 1. Your door's probability stays at 1/3.
3.P(car behind Door 2) = 1 โˆ’ 1/3 = 2/3. All the probability from Door 3 transfers to Door 2.
4.Car behind Door 1: host opens 2 or 3, Stay wins, Switch loses. | Car behind Door 2: host opens Door 3, Stay loses, Switch wins. | Car behind Door 3: host opens Door 2, Stay loses, Switch wins.
5.Staying wins 1/3 of the time. Switching wins 2/3 of the time. Always switch.
6.The two remaining doors are NOT equally likely. Door 1 was locked in at 1/3 before the host acted. The host's reveal was not random โ€” it was constrained to always show a goat. The 50/50 assumption ignores this asymmetry.
7.P(win by switching) = 99/100. P(win by staying) = 1/100. The 1/100 chance of your original pick being right is unchanged; switching concentrates the remaining 99/100 onto the one other door.