## Ask Professor Puzzler

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Before getting to the question for this morning, I'd like to direct your attention to our simulation of a game show with three doors. It's called the "Monty Hall Game." Understanding this game will be important to thinking about Tracie's question below.

The short summary of the game: There are three doors, and only one of them has a prize behind it. You pick a door. Then Monty opens another door to show you that it's empty, and asks you if you want to switch your guess. What do you do?

Tracie from South Dakota asks, "So if the game show host opened 2 doors for me, would I still have 1/3 probability of winning?"

Hi Tracie, your question highlights a couple of the things about the Monty Hall game that often causes confusion.

- Monty Hall knows in advance where the pot of gold (or new car, or whatever) is. The rules of the game are that he opens an empty door. But if he
*always*opens an empty door, that means he*knows*where the empty doors are. That's an important concept to understand. He's not being entirely random in his choice of doors to open. There are two empty doors, so if you pick one of them, he picks the other. The only time he's being random is when you pick correctly; in that circumstance he randomly picks one of the empty doors to open. - This is not a static problem - the circumstances change, and because the circumstances change, the probabilities could change as well. Let me give you an example. Suppose that during Christmas vacation, we decide we want to take our kids sledding. Since I'm not teaching that week, I can pick any day of the week to go. So we would say that my probability of picking Tuesday is 1/7 (there are seven days in a week, and Tuesday is one of them). Now suppose that my wife looks at a weather forecast, and says, "Oh! Thursday, Friday and Saturday are supposed to be cold, with a wind chill of -20 F!" If I use that information in making my choice, what is my probability of picking Tuesday now? It's 1/4, because we've eliminated 3 of the 7 days. That doesn't mean the original 1/7 is
*wrong*- it means that the conditions of the problem have been changed, and so we have to calculate a new probability. This is not, by the way, a perfect analogy to the Monty Hall problem, so please don't try to make it match. The differences are:- We're talking about the probability that I'll pick a certain day, not the probability that the day I pick is a good one.
- We don't know that there's only one good sledding day.
- My wife is not deviously hiding known information from me.
- Weather forecasts are not 100% accurate anyway.
- There is no #5, but at the end of this post, I'll - just for fun - turn this example into a better analogy for the Monty Hall Game.

Okay, so how does this relate to your question? Number two should be fairly obvious; we have a change in conditions, so there's no reason to assume that the probability will stay the same. The probability of you winning WAS 1/3, but the changing conditions mean we have to recalculate the probability.

But here's the more important issue. Monty Hall *can't* play the game the way you suggest. Why not? Because if the rules of the game are "Monty will open two empty doors," Monty is going to run into trouble every time you *don't* pick the right door. Because if you pick the wrong door, how many empty doors are there left for Monty to open? Only one! If you pick one of the empty doors, then there's only one other empty door for him to open.

If he opens two doors, that means you've picked the correct door. So even though your original choice was 1/3 probability of winning, under the new circumstances, you have a probability of 1 (100% chance) of winning.

If you're wondering how that number 1/3 fits into the solution it fits like this: The probability that Monty can open two doors is 1/3 (the same as the probability that you selected correctly). So the number 1/3 is still in there - just in a different place!

If you're wondering why the probability changes in this case, but doesn't when he opens one door, the answer is this: When he opens just one door, he has not given you any information about the door you opened. When he opens two doors, he *has* (indirectly) given you information about your door.

To add to this, the only way Monty could do the two-door rule would be to have to have a pair of rules:

- If you pick the right door, he will open two doors. (this happens 1/3 of the time)
- If you pick the wrong door, he will open one door. (this happens 2/3 of the time)

The problem with this is: if you know the rules, you can do an always-winning strategy. If he opens two doors, you've automatically won, and if he only opens one door, that means you picked the wrong door, so you must swap. Monty would DEFINITELY not want to play by those rules!

Addendum: If you would like to look at a couple different ways of understanding the Monty Hall problem, you can find more write-up here: The Murky Swamp of Probability.

**Sledding in December**. Let's say my wife looked at the weather forecast and told me, "Every day but one this week is going to have a wind chill of -20 F, and there will be one day that's going to be sunny and warm. So, randomly pick a day for sledding." (My wife hates the cold, so she would definitely not do something as ridiculous as that!) We'll make the wild assumption for this example that the National Weather Service has 100% predictive accuracy.

So I pick Tuesday. I have a one-in-seven chance of "winning," based on the information I have. Of course, my wife has a different perspective, because she has more information than I do. She knows with 100% certainty whether I've picked correctly or not!

Now my wife says, "Okay, I'll tell you that the following days are going to be -20 F: Sunday, Monday, Thursday, Friday, Saturday."

She's narrowed my choices down to *two* possibilities. But it's important to remember that this was not a completely random choice. She didn't pick 5 out of 7 days to eliminate; she picked 5 out of *6*! She couldn't eliminate Tuesday, because that was the day I had picked, and if she eliminated it, I would be *forced *to change.

My reasoning now goes as follows: Based on my original information I was given, Tuesday had a 1/7 probability of being the best day for sledding. That means there was a 6/7 probability that one of the other days was the good sledding day. So there's a 6/7 probability that the good sledding day is: Sunday, Monday, Wednesday, Thursday, Friday or Saturday. As a probability equation, I could write:

P_{su }+ P_{m }+ P_{w }+ P_{th }+ P_{f }+ P_{sa }= 6/7

But now my wife has changed the problem. She's changed it by giving me more information: She's told me the values of all but one of those probabilities is zero!

That means: 0 + 0 + P_{w} + 0 + 0 + 0 = 6/7, which means that P_{w} = 6/7!

So do I switch my decision to Wednesday? You bet I do! The key to understanding this is that my wife's choice of information to give me is *not random*, and it therefore significantly alters the shape of the problem.

I hope that this is a helpful answer. The Monty Hall problem is one that repeatedly trips people up. I wonder if my blog post will increase the probability of people *not *getting tripped up by it!

PS - if you go to the Monty Hall game and change the game settings to 7 doors, 5 hints, that will be the same as my sledding example. Turn on the automation, and watch the percent. If you let it run long enough, you'll see that it eventually settles down to about 85.7%, which is the 6/7 probability we calculated.