monty hall question with 4 doors

Question

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

Answer 1

Assuming 2 doors are open after first pick:

With the non-switching strategy:

You need to pick the right door from the start with probability $\frac{1}{4}$

with the switching strategy:

You need to pick a non-correct door at first with probability $\frac{3}{4}$

After that ther are two doors remaining, the one you picked first (wrong) and the one unpicked remaining (the winner) so after switching you have the winning one.

So probability with switching is $\frac{3}{4}>\frac{1}{4}$, so you should switch.

Answer 2

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!