Skip to content
DenseChip
Learn
PracticeCode labInterview

DenseChip

Navigate

Menu

LearnPracticeCode labInterview

Tools

All toolsSV editorCode repoResume
DenseChip

Practice, courses, and chip-design tools in one focused workspace.

LearnPracticeCode labInterviewToolsLearning paths

© 2026 DenseChip

AboutRoadmapCareersInternshipsFAQContactPrivacyTerms
Skip to lesson

Course contents

  • How to Use This Puzzle Bank
  • SystemVerilog Language Puzzles25
  • Difficulty Guide
  • Full Topic Map
  • Interview-Day Tips
  • Study Route: One Week Before the Interview
  • Study Route: Two Weeks
  • Study Route: Four Weeks (Full Course)4
  • OOP & Testbench Puzzles25
  • Concurrency & Timing Puzzles25
  • Randomization & Constraint Puzzles25
  • SVA Assertion Puzzles25
  • Functional Coverage Puzzles25
  • Logic & Aptitude Puzzles25
    • Classic Brain Teasers
    • Probability Puzzles
      • Biased Coin to Fair Bit (Von Neumann)
      • Expected Tosses Until First HH
      • Two Children, At Least One Boy
      • Monty Hall Switching Odds
      • Birthday Collision Threshold
      • Bayes on Failing Test Alerts
    • Digital Logic Puzzles
    • Debugging Riddles
    • Logic & Aptitude Puzzles: Tricky Q&A
  • Scenario & Debug Puzzles25
  • UVM Testbench Puzzles25
  • Clock, Reset & CDC Puzzles25
  • Bus and Interface Protocol Puzzles25
  • Interview Prep: Puzzle Rounds7
  • DV Puzzle Cheatsheet
Loading lesson

←Probability Puzzles
  1. Topics
  2. DV Interview Puzzles
  3. Logic & Aptitude Puzzles
  4. Probability Puzzles
  5. Monty Hall Switching Odds
On this page

    VLSI DV Interview PuzzlesAll levels

    Monty Hall Switching Odds

    In a 3-door game, one car and two goats. You pick one door. The host knows where the car is and always opens a different door that hides a goat (choosing at random if he has two goat doors to choose from), then offers you a switch. Should you switch, and what are the win probabilities?

    Puzzle

    Difficulty: Medium · Puzzle 4 of 6 · Topic: Probability Puzzles

    In a 3-door game, one car and two goats. You pick one door. The host knows where the car is and always opens a different door that hides a goat (choosing at random if he has two goat doors to choose from), then offers you a switch. Should you switch, and what are the win probabilities?

    Hint

    Initial pick keeps its original probability.

    Step-by-step solution

    Diagram
    1) Initial chosen door has car with probability 1/3.
    2) Therefore car is in one of the other two doors with probability 2/3.
    3) Host always opens a goat door among those two, never opening the car door.
    4) Host action transfers the full 2/3 mass to the single unopened alternate door.
    5) Staying wins with 1/3; switching wins with 2/3.

    Answer

    Answer: Switching doubles win probability from 1/3 to 2/3.

    How this was checked

    Checked by running a model of the rule in a Python model. A computational check: exact enumeration of car position and host choice gives stay 1/3 and switch 2/3 for a host who avoids the car, and 1/2 each (with a 1/3 chance of the car being revealed) for a random host.

    Why candidates get it wrong

    Counting the two closed doors as 50-50. The host's choice is not random: he must open a goat door, so his action carries information and shifts the whole 2/3 onto the other door.

    Interviewer follow-up

    What changes if host sometimes opens a random door without knowing? Model answer: if the host opens a random door and it happens to show a goat, switching and staying are both 1/2. The 2/3 only appears when the host always avoids the car on purpose. (If he opens the car by accident the game just ends, with probability 1/3.)

    Say it out loud

    My first pick is the car with probability one third, so the car is behind one of the other two doors with probability two thirds. The host knows where the car is and always opens a goat door among those two, so he never reveals the car. That means the whole two thirds lands on the one remaining closed door. So I should switch, and switching wins with probability two thirds while staying wins one third. The trap is calling the last two doors fifty-fifty, when the host's choice is forced and carries information.

    Related topics

    • Back to Probability Puzzles

    • Logic & Aptitude Puzzles section

    Previous

    170 / 293

    Next: Birthday Collision Threshold

    Up next

    Up nextBirthday Collision Threshold

    Sign in to save progress

    Practice this lessonPrevious lessonReport an error

    See also

    • Biased Coin to Fair Bit (Von Neumann)
    • Expected Tosses Until First HH
    • Bayes on Failing Test Alerts