Explain the Monty Hall problem in the case of 4 doors computing specific probabilities.
I got that you have 1/4 chance of picking the door with the goat.
1/4 chance to pick the door with the prize and so on.
if I pick an empty door you have a 1/2 chance of doing this in this case you have 1/2 chance of winning the prize. if you don't switch
if you don't switch your chance is 1/4
switching is better
I think Im missing something
Best Answer
I assume there are 4 doors - 3 with goats and a single one with the prize:
FIRST CASE: Show host opens 2 doors for you:
If you had picked the right door to begin with, you would lose from changing. However, this will only be the case 1/4 of the time. So 1/4 of the time you will win by staying.
Then show opens two doors with goats -
3/4 of the time, you will be at a goat to begin with, but the showing of two goat makes it so that the remaining one (if you are on a goat) is the prize. Now since 3/4 of the time you will be on a goat - 3/4 of the time you will win by changing.
SECOND CASE: Show host opens just 1 door for you:
If you had picked the right door to begin with, you would lose from changing. However, this will only be the case 1/4 of the time. So 1/4 of the time you will win by staying.
Then show opens a door with a goat -
3/4 of the time, you will choose a door with a goat to begin with. The revealing of a goat makes it so that 3/4 of the time the remaining 2 doors must hold the prize! But wait, now it's down to pure luck, good old 50/50. So your chance of winning in this case will be (3/4) * (1/2) or 3/8.
So overall your three options: there is a 2/8 (1/4) chance of winning by not changing, a 3/8 chance of losing by changing (moving to lets say door x) and a 3/8 chance of winning from changing (lets say moving to door y). The sum is fortunately =1.
Cheers!