The Monty Hall Problem

• Problem:

  • There are 3 doors: behind one is a car, and behind the other two are goats.
  • The Twist: You pick a door, then the host (who knows what’s behind the doors) opens one of the other doors to reveal a goat and asks if you want to switch.
  • Which has a higher chance of getting the car: switching or not switching?

• Scenarios & the possibilities of getting the car

  • $P(\text{Total Win})=P(\text{Initial Pick})\times P(\text{Win given that pick})$
  • Staying
    • Pick car & staying (win) → $\frac{1}{3}\ast 1=\frac{1}{3}$
    • Pick goat & staying (lose) → $\frac{2}{3}\ast 0=0$
    • Total = $\frac{1}{3}+0=\frac{1}{3}$
  • Switching
    • Pick car & switching (lose) → $\frac{1}{3}\ast 0=0$
    • Pick goat & switching (win) → $\frac{2}{3}\ast 1=\frac{2}{3}$
    • Total = $0+\frac{2}{3}=\frac{2}{3}$
  • So since $\frac{2}{3}>\frac{1}{3}$, switching is better!