## Ask Professor Puzzler

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Yesterday, I answered a question about the Monty Hall Three-Door Game, which you can read about in the previous blog post. After posting the article, I shared it on social media, and commented that talking about this problem feels like wading into a murky swamp, because everyone brings their own assumptions into the problem, and it's tough to guess what those assumptions are.

But I realized, too, that it's not just this problem; it's probability in general. I love probability, and I hate probability. Whenever a district/county/state asks me to write competition math problems for their league or math meet, I know that I need to give some probability problems - everyone expects it! But there's no kind of math problem that I'm more afraid that I'll mess up. I'm grateful for a proofreader to give my problems a second pair of eyes (although my proofreader shares my feelings about probability). More often than not, I'll figure out a way to write a program to function as an electronic simulation of my problem, to verify empirically that I have arrived at the right solution.

Anyway, one of the reasons that probability feels so murky to me is that gut reactions can lead you astray. The Three-Door game is a prime example of how those gut reactions can mess you up. Monty asks you to pick a door, knowing that one of those doors has a prize behind it, and the other two have nothing. After you pick, Monty opens one of the *other* doors to show you that it's empty, and asks you if you'd like to keep your original guess, or switch to the other unopened door.

Most people's gut reaction is that it doesn't matter whether you keep your original guess or switch to a different guess. This gut reaction is wrong, as you can verify by playing the simulation I built here: The Monty Hall Simulation. For those who are still struggling with this, I'd like to offer a couple different ways of looking at the problem.

## The "Not" Probability

We tend to focus on the probability of guessing correctly the first time. Instead, let's focus on the probability of NOT guessing correctly. If you picked door A, then the probability that you were correct is 1/3. This means that the probability you did *not* guess correctly is 2/3. But really, what is that? It's the probability that either door B or door C is correct.

So if P_{B} is the probability that door B holds the prize, and P_{C} is the probability for door C, we can write the following equation:

P_{B} + P_{C} = 2/3.

Now Monty opens one of those two doors (we'll say C), that he knows is empty. This action tells you absolutely *nothing *about the door you opened, but it does tell you something about door C - it tells you that P_{C} = 0.

Since P_{B} + P_{C} = 2/3, and P_{C} = 0, we can conclude that P_{B} = 2/3. This is exactly the result which the simulation linked above gives.

## Two Doors vs. One Door Choice

Related to the above way of looking at it, try looking at it with a set of slightly modified rules:

- You pick a door
- Monty says to you: "I'll let you keep your guess, or I'll let you switch your guess to
*both*of the other two doors."

Under these circumstances, of *course* you're going to switch. Why? Because if you keep your guess, you only have one door, but the choice Monty is offering you is to have *two *doors, so your probability of winning is twice as great. In a sense, that's actually what you're doing in the real game, even though it doesn't appear that way. You're choosing two doors over one, and the fact that Monty knows which one of those two doors is empty (and can even show you that one is empty) doesn't change the fact that you're better off having two doors than one.