Question

I am in a game show.

Answer 1

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, $P(B)_{HC}=\frac{P(HC)_BP(B)}{P(HC)}=\frac{\frac{1}2\cdot\frac{1}3}{\frac{1}2}=\frac{1}3$ whereas $P(A)_{HC}=\frac{P(HC)_AP(A)}{P(HC)}=\frac{1\cdot\frac{1}3}{\frac{1}2}=\boxed{\frac{2}3}$ !!

Answer 2

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