Discussion on the Monty Hall problem.
The MHP
- A player stands in front of three closed doors. behind one of them is a car, behind the others goats.
- The player may choose a door and have what's behind it.
- From this setting ordinary people conclude that the car is placed randomly. Call C the door with the car: P(C=c)=1/3.
- And furthermore that the player is completely unaware of the position of the car. Call X the chosen door: C and X are statistically independent (of each other).
- Suppose the player chooses door x; as typical example take x=1.
- What is the probability the chosen door hides the car?
- This is a tricky question: many people will think it is the probability the car is behind door 1: P(C=1)=1/3.
- The actual meaning must be:P(C=1|X=1), accounting for the event X=1.
- Due to the independency: P(C=1|X=1)=P(C=1)=1/3.
- Before the chosen door is opened, the host opens one of the two other doors, and exhibits a goat. Call H the opened door.
- Strategy of the host: always opens one of the remaining doors, always showing a goat.P(H=x}X=x)=0; P(H=c|C=c)=0.
- As an example take H=3.
- The player is offered to switch to the other closed door.
- The player now sees only two closed doors, between she has to choose.
- At this point many people think the odds are equal.
- The player has to base her decision on the probability in this situation the car is behind door 1 (or door 2).
- This probability is: P(C=1|X=1,H=3)
- To find the numerical value of this probability we need info about what the host will do when he has to choose between two doors with a goat.
- It seems likely the host then chooses randomly: P(H=2|X=1,C=1)=P(H=3|X=1,C=1), etc.
- Then with Bayes' formula or using arguments of symmetry, it follows: P(C=1|X=1,H=3)=1/3.
- This is the same value as the unconditional probability.