## Understanding the Monty Hall Problem

When I first heard about the Monty Hall Problem, the answer seemed unintuitive to me, as it did to famous mathematicians. At that time, I dismissed it without much thought.

When I encountered it again in a the book The Drunkard’s Walk: How Randomness Rules Our Lives, I decided I’d better come up with my own way to think about this problem so I could wrap my head around it.

The traditional statement of the problem is based on the old TV game show Let’s Make a Deal. The show’s host, Monty Hall, presents you with three closed doors. Behind one of the doors is a real prize (say, a car), and behind the other two are joke prizes (as the problem is typically stated, goats). Without being able to see behind any of the doors, you must choose one. Without any clue of what’s behind any door, whichever door you choose has a 1/3 chance of having the car behind it. For the sake of presentation, let’s say you choose door #1.

Now, Monty Hall opens one of the other doors, let’s say #2, revealing a goat. Then he offers you the chance to change your choice to the other unopened door.

Here’s the question: Are you better off switching doors, or does it matter? I, like many others, initially thought no, the odds are still 1/3 that the car is behind either unopened door, so it makes no difference if you switch.  The real answer, though, is that it’s to your advantage to switch: the probability that the car is behind the door you originally chose is still 1/3, but the probability it’s behind the other door is now 2/3.

Here’s the reformulation I came up with which makes it clear to me (and hopefully, you) that this is correct.

Instead of prizes behind doors, let’s say instead that Monty Hall has three balls in an opaque bag — one red and two white. The object of the game is to end up with the red ball.  First, you get to reach into the bag without looking and choose one ball, which you put in your own opaque bag.  Now, the probability that you have the red ball in your own bag is 1/3.

In our first version of this scenario, let’s now imagine that Monty Hall offers to switch bags with you. In this version, it’s irrelevant if you happen to end up with an extra white ball; the only thing that counts is whether you have the red one. Do you switch bags? Duh! It’s pretty obvious that, with two balls in Monty Hall’s bag, the odds are 2/3 that it contains the red ball. So of course you switch.

Now consider the second version: Before offering to switch bags with you, Monty Hall looks in his bag and removes a white ball. What does this change? You both knew that his bag contained at least one white ball, so it doesn’t change anything. The odds that the red ball is in Monty’s bag is still 2/3, and you’re still better off switching.

The set of two doors you didn’t choose is like the balls in Monty’s bag in this reformulation, and his opening of one of those doors to reveal a joke prize is like him removing a white ball from that bag.

Now it all makes perfect sense.

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### 2 Responses to Understanding the Monty Hall Problem

1. Chaos says:

I’m still not grasping this. I don’t see this as two bags, one for you, one for him. I see 3 bags, one with a ball each. He has two of the bags, and you have one. Before anything is revealed, we know he has at least one white ball. He then proceeds to remove the bag with the white ball. Now you know he has a 50/50 chance of having the red ball. You also know you have a 50/50 chance of having the red ball. If you swap with him, you’ll still have the same chances.

• By making it three bags, you’re making it just like the three doors of the original problem statement, and reaching the same erroneous conclusion that many others have (including me, when I first heard the problem).

This reformulation is supposed to make it clear. The key fact is that Monty has access to information that you don’t have. The two balls in one bag is like the two doors you didn’t choose — Monty knows which one to open, but you don’t.

Another way I’ve seen it explained is, imagine that instead of 3 doors, there are 100. You choose one and Monty opens 98 of the other doors, leaving only one unopened. Then do you still think the odds are 50/50 between your door and the one Monty has left unopened? At this the intuition (of many people) starts to say, ok, maybe there’s something to this.

I liked my version better though, because it’s still a choice among 3 things, and yet — I thought — shows the principle at work.