Yet another explanation for the Monty-Hall problem

I assume that you know what a Monty-Hall problem is. It’s a problem that’s always fun to argue about, especially an uninitiated friend or acquaintance. But each time you discuss the solution with someone who’s adamant that it just doesn’t make sense, just keep in mind that even Paul Erdos got it wrong. We’re mortals, after all. However, here’s a way to think about it that I used to finally convince my Dad by framing my arguments as a series of questions.

Q1. Imagine that 30 people are playing this game simultaneously. Then, in the first phase, how many people will get the correct door when they do a random choice?

Everyone will generally agree that 10 people will get it correct. If not, you may have to invoke dices or cards to give them a basic understanding of probabilistic expectations.

Q2. So how many will get the incorrect door?

Trivially answered as 30-10 when you get Q1.

Q3. So when the host for each of these people opens an unchosen empty door, how many people will get the car if they stay?

Answer should follow from Q1.

Q4. How many people will get the car if they switch?

Answer should follow from Q2.

Q5. So is it preferable to switch? Or to stay?

Answer should follow from Q3 and Q4. Best of luck trying to convince your friends, spouse, coworkers, parents or acquaintances!!