Five game-show doors with x-ray goggles
Question
A game show has five closed doors; $10,000 is behind exactly one of them and the rest are empty.
(a) If you simply pick one door, what are your expected winnings?
(b) Before choosing, you may buy goggles that let you see behind exactly one door of your choice. What is the most these goggles are worth to you?
(c) You use the goggles and the door you inspect is empty. You then pick one of the other four doors. The host, who knows where the money is, opens a different, empty door among the rest and lets you keep your pick or move to any other unopened door. Playing optimally, what is your chance of winning?
Short answer
(a) $2,000; (b) $2,000; (c) . (a) One door in five wins. (b) With the goggles you either see the money () or rule out a door and then win with probability , so the win chance doubles to , worth $4,000. (c) Your pick holds the prize with probability ; the remaining is shared by the two other unopened doors, so switching gives .
Adapted from OpenQuant; question and answer rewritten by SelfMadeQuant.com.
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Open the worked solutionLessons behind this question
- Expectation, Mean, and Variance
- Counting and Combinatorics
- Conditional Expectation and Variance Revisited
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