Question

I am in a game show.

In this game show there are 3 doors. Behind one of these doors lies a brand new and spectacular Ferrari, however, behind two of these doors lie nothing.

These doors are represented by the letters A, B and C respectively.

The host of this game show gives me the opportunity to win a Ferrari, and tells me that if I open the door which has this Ferrari sitting behind it, I will get to keep it. The problem is, I will only get to open one door. If it turns out to have nothing behind it, I will go home empty handed.

So, the game show host asks me which door I'd like to open. I select the door B.

Before the host opens the door B for me, he opens up the door C and shows me that nothing lies behind it.

This means that the Ferrari is either behind door A or door B.

Now, the host asks me if I'd like to open door A instead of door B. He gives me the chance to open door A. Should I open door A or should I open door B?

Would my chances of winning a Ferrari increase if I were to open door A instead of door B?

Explain your answer and justify why you are correct...

Answer 1

if you switch after the host shows you the door without the ferrari, your chance of getting the car is greater.

we have 3 hypothetical situations.

if the car is in door A, but and then you pick door A, the host will open either door B or C and you will pick the other remaining door, resulting in a loss.

if the car is in door B, and you pick door A, the host will show you door C, and if you switch to door B, you will win.

if the car is in door C, and you pick door A, the host will show you door B, and if you switch to door C, you will win.

There is a 2/3 chance of winning a ferrari if you switch after the host shows you the door without.

:D

Answer 2

This is the well known Monty Hall problem.

You have more chances of winning by switching doors.

Let A,B,C denote the events "The ferrari is behind A" and so on
Let HA,HB,HC denote the events "The host opens A" and so on

By Baye's formula,
`P(B)_(HC)=(P(HC)_BP(B))/(P(HC))=\frac{\frac{1}2\cdot\frac{1}3}{\frac{1}2}=\frac{1}3` whereas
`P(A)_(HC)=(P(HC)_AP(A))/(P(HC))=\frac{1\cdot\frac{1}3}{\frac{1}2}=\boxed{\frac{2}3}` !!