Question

Suppose you are on a TV show. There are three doors: A, B, and C. Behind only one of the door, there is a new car. If you pick the correct door, the car belongs to you. Suppose you pick a door, then the compere opens another door and shows that there is no car behind that door. The compere then asks you if you want to stick to your door, or switch to the other remaining unopened door. Does it matter? (Hint: Well define each event and use Bayes’ rule.)

Answer 1

This is similar to famous monty hall problem .

And making a decision of changing the doors is very much beneficial. So according to situation we can write

P(choosing door 1)=P(choosing door 2)=P(choosing door 3)=P(A)= 1/3 ( all are likely to choose ) where A is event of choosing a door

The compare will always open an door having no car in it and it will not depend on and of the probability of choosing any door . So

P(he opens door with no car=B)=1

So we can write

P(he opens door with no car(B) | choosing a door(A))

=1

By Bayes law, P(choosing a door(A)|he opens door with no car(B)) = P(A)*P(B|A)/P(B).

Since P(B|A) = P(B) = 1, therefore P(A|B) = P(A) = 1/3

Thus, the contestant has a 1/3 chance of winning if he sticks to his first choice choice.

And the contestant has a very much better chance

That is :-

(1–1/3 = 2/3) of winning if he switches.