Making Sense of the Monty Hall Problem
The Monty Hall Problem is one of those proof points that illustrates how bad humans (myself included) are at statistical thinking. The quick version is this: you are playing a game with three options, often illustrated as doors. One of the options is a winning option, and the other two options are losing options (one door has a car behind it and the other two have nothing.)
Most everyone would agree that your probability of picking the correct door at this point is 33.33%. So far, no controversy. Let’s say you choose door number 1.
At the next step in the game, the host reveals one of the doors that you *did not* pick as being one of the losing doors.
You are then given the option to either keep your original selection, or switch to door number 2. The question is, “Should you do it?”
This is where the controversy kicks in (or did, when this was first publicized back in the 90's.) Most of us naturally now remove door number 3 from our evaluation, which means we judge there to be a 50% chance of the prize being behind either door number 1 or door number 2. Thus, statistically speaking, it doesn’t matter whether you switch doors or not — your probability of winning remains the same. HOWEVER, it turns out that switching doors as a strategy ends up winning 2/3 of the time, not the 50% of the time you might expect. In short, for this scenario, switching doors would be the right thing to do every single time, from a statistical perspective.
I’m From Missouri, So…
This melted my brain, so in order to wrap my head around it, I constructed an experiment in Python to simulate the game. I made it more complex than I needed to, just to make sure the actual steps of the game would be followed. (Missouri is known as the “Show Me” state, so the inference is I need to see it in order to believe it.)
I was honestly a little surprised to see that the results worked just like the explanations described.
I performed some additional tests just to make sure this wasn’t some weird artifact about my environment:
- Played the game without switching (resulted in winning about 1 in 3 times, definitely not 50%).
- Played the game with an initial option of only 2 doors (resulted in winning about 50% of the time, as expected, regardless of switching or not.)
- Played an “Interrupted” version of the 3-door game, where the first function stopped after the reveal and produced outputs that were relabeled as A and B (both door numbers, winning door, and initial guess):
No matter what I did to interrupt the process, if I started with initial conditions of a choice of three doors, the switching strategy always ended up winning about 2/3 of the time.
Additional experiments were performed, to no avail (if the goal was to break the system.) No matter what, the numbers held.
It took me a bit, but I eventually figured it out when I thought through the game in reverse: What if, instead of picking one door at the beginning of the game, you got to pick two doors? Let’s say you picked doors 2 and 3.
What is the probability that the prize is behind one of those selected doors? Again, non-controversially, most of us would agree we now have a 2/3 chance of winning, or a 66.66% probability.
What Do We Already Know?
This is where it gets interesting, and for me was the “a-ha!” moment. When we made this selection of two doors, we already knew that one of them was not a winner. This knowledge does not change our probability of winning. It’s still 66.66%. Thus, learning that one of our initially selected doors in this version of the game was not the winning door does nothing to add information to the selection process. And that’s the key. We already knew one of them was a losing door, and our probability of winning was still 66.66%. Learning **which one** of our selections was the losing door does not change the probability that we picked one of the right doors in the first place! Our probability of winning, based on our initial selection, is still 66.66%.
To further drive the point home, assume there is no reveal stage. The two doors you selected in the game will be opened at the same time. What is your probability of winning? Again, no matter how you look at it, it will always be 66.66%.
Bringing This Back
If we take our knowledge of how the inverse version of the game would work and our probability of winning if we got to choose the two doors in the first place and apply it to the single door with an option to switch version of the game, the reason this works this way comes into better focus. Forget the reveal (since it adds no information that we did not already have.) Let’s assume you initially pick door number 1 again.
Now, instead of revealing one of the doors you did not select, the host gives you the opportunity to right away trade your one selected door for the two you did not select. You’d be crazy not to take the offer — your probability of winning doubles from 33.33% to 66.66%! And this is why the Monty Hall Problem works the way it does. Our human brains are tricked into thinking that the reveal has provided additional information, and somehow changed the game and probability that we picked the right door in the first place. In reality, the reveal doesn’t tell us anything we didn’t already know, and the option to switch is really the option to choose every door **except** for the one you initially picked, which in every scenario with three doors or more, increases your probability of selecting the winning door.
I fully admit this is a little weird, but our universe is full of little weirdnesses that on the surface don’t make sense to our brains but work out quite nicely from a mathematical perspective. Hopefully this blog post has helped you wrap your head around the Monty Hall Problem a little better, and maybe helps you challenge what we commonly refer to as “comon sense” thinking. Ain’t statistics neat?