Monty Hall problem variation

[estellashaw]

Hello,
I've been really struggling with probability. Imagine I have three doors, and behind one of them is a million dollars. I can open two doors—no catches, just pick and open. I choose the first door, and it’s empty. My initial chances were 1/3. Now, I have one pick left and two doors remaining.

Here’s where I might be confused: with two doors left and one pick available, are my chances now 1/2? Does the empty door still factor into the equation? And if it doesn’t, would opening two doors at once change my chances to 2/3, or would it still just be 1/2? Sorry if my explanation is a bit unclear; I’m not great at articulating my thoughts!

 

[Dr.Peterson]

Yes, your probability of success now (given that the first door has already been opened) is 1/2.

If you had opened two at once, your probability of success (from the start) would be 2/3.

Now, we could calculate the latter probability using the former. Your first door could be either success or failure. In the former case, you are finished; in the latter case, the probability of success on the second pick is 1/2. So the total probability of success is

P(first or second pick correct) = P(first correct) + P( first wrong and second correct)​

= P(first correct) + P( first wrong)*P(second correct | first wrong)​

=1/3 + 2/3*1/2​

= 2/3​

In the actual Monty Hall problem, Monty has extra knowledge and manipulates things. Here, everything is straightforward.