Responding to the Monty Hall Problem

It’s been a while since I got round to writing anything, so I’m easing myself back in by avoiding anything too mathematically strenuous. Shortly after my last post, there was a nice clip on Horizon featuring Marcus du Sautoy explaining the famous Monty Hall Problem to the ever-incredulous Alan Davies. The BBC article here shows the clip, and then an excellent account by probabilist John Moriarty explaining the supposed paradox.

Recall the situation. A game show is hosted by Monty Hall. At some point, to decide the grand prize, or whatever, a contestant plays the following lottery. There are three doors, and three prizes, one behind each. One of the prizes is worthwhile, normally a car, while the other two are rubbish. For reasons I don’t understand, it is normally said that behind the other two doors are goats. Anyway, MH knows where the car is from the beginning. Then the contestant picks a door, but doesn’t open it yet. MH then dramatically opens a different door, revealing a goat. The contestant then has the option to stick with his original choice, or switch to the third door. What should he or she do?

So the natural error to make is that since one possibility has been eliminated, since the remaining two were a priori equally likely, they should still be equally likely, and hence each occur with probability 1/2. So it makes no difference whether you stay or switch. But, of course, as well as removing information (ruling out one door), we have also gained information. The fact that MH chose to reveal that door, rather than the potential third door is important.

In my opinion, the best way to think about this is to imagine you are the host. The nice thing about being in this position is that then you can imagine you know everything about the configuration in advance, so are less likely to fall foul of counter-intuitive conditional probabilities. Anyway, there’s loads of symmetry, so from my point of view as the host, I only care whether the contestant initially selects the right door or the wrong door. If he selects the wrong door, then I need to open the ONLY OTHER wrong door. In other words, I have no choice in what I do next. I open the other wrong door. Therefore in this case the contestant should switch. If the contestant in fact selects the right door initially, I then have a choice about which door to open. However, it doesn’t really matter which I pick, by symmetry, and in either case the contestant would do better to stay with his original choice. Note that we have defined the outcome entirely by the initial choice. And as the host, I knew where the car was at the start, so from my point of view, unless he had somehow cheated, the contestant always had a 1/3 chance of guessing correctly. So it is better to switch 2/3 of the time.

The BBC’s weekly collation feature, The Loop, printed a couple of responses to the article. This is one:

“I believe that the problem is a game of two halves. In the first instance you have a 1:3 chance of getting the prize. Once you’ve made your initial choice, even though the ‘box’ contents remain the same, one ‘false’ choice is removed and you are then offered the chance to choose again. There is a play on words here, perhaps, with the option to ‘switch’. Effectively you are being asked to make a choice of 1:2. Now, if this is always going to happen then the odds of you winning the prize in the first instance is 1:2, as it doesn’t really matter whether or not you choose correctly the first time – one ‘false’ is going to be removed and you’ll be presented with just two options. It’s mind trickery. The first choice is totally irrelevant because you will always end up having to pick one option from two in order to attain the prize. In my humble opinion.”

I highlight this not to inspire contempt, since it is an easy mistake to make, even after reading as nice an explanation as Moriarty’s. Rather it is an excellent example of how using too many words can convince you an argument is complete before you’ve really made any concrete statement at all. Note that the sentences beginning ‘Effectively…’, ‘Now, if this…’ and ‘The first choice…’ all say exactly the same thing, which is, to use inappropriately formal terminology, exactly what the writer is trying to prove. The padding of ‘mind trickery’ and ‘humble opinions’ is as close as a maths argument can ever really come to an ad hominem.

So I was trying to come to a clean conclusion about exactly why the Monty Hall problem does such a good job at generating confusion. I think the following aspects of the set-up are important:

1) In general, people often get confused about the meaning of ‘random’. In some sense, the sequence 416235 is more random than the sequence 123456. However, typically, we use ‘random’ to mean ‘not determined in advance’. The key is that the bit where MH reveals a goat is not a random action. Whatever you do initially, you know that this will happen. So the fact that it does actually happen should not add extra information.

2) This fact is compounded by the time-steps at which decisions have to be made. It feels like the arc of the narrative is directing everything towards this final choice, which feels very separate from the initial choice. You can imagine the dramatic music that undoubtedly plays as MH opens the appropriate door, and the camera pans back to the contestant making their decision. All of which is an excellent way to suggest that something complicated has happened, when in fact in terms of what we are interested in overall, nothing has happened.

A point I hadn’t thought about before was that it is important that MH knows what happens in advance. Let’s develop this point and consider what happens if MH in fact has no knowledge of where the car is, and chooses a door at random. We will still assume that he must choose a different door to the one that the contestant originally chose.

Then, 1/3 of the time the contestant picks a goat, and MH picks the car, so he gets a goat whether he stays or switches. 1/3 of the time, the contestant picks a goat, and MH picks the other goat, so he should switch. And 1/3 of the time, the contestant originally picks the car, and then by default MH picks a goat, so he should stay. So, overall, it is equally preferable to stay as to switch.

I guess the morals of this story are: “be careful with conditional probabilities; sometimes apparently new information doesn’t change anything; and that it’s probably not a good idea to get into a discussion about the Monty Hall problem with a skeptic unless you have a piece of paper to hand on which to draw a probability table.”

PS. A final moral is that, based on the links WordPress suggested, at least 10% of the internet is devoted to explanations of the Monty Hall problem…

Advertisements

5 thoughts on “Responding to the Monty Hall Problem

  1. I don’t think this observation is original, but there’s a way to develop a strong intuition without doing any maths, by varying the parameters. Suppose there’s 100 doors: 99 goats and 1 car. You pick door #82. The host opens 98 of the remaining doors and shows you 98 bleating goats. Suspiciously, the host leaves door #47 bolted shut. Do you swap to door #47? You bet your life you do.

  2. A good explanation (Bayesian-esque one could say) solves the Monty Hall problem in my view. Good knowledge of probability and conditional probability and not thinking fast, rather thinking slow, once handed this problem, solves it quickly.

  3. James’ analogy is exactly the one I use to explain this to people, except I extend it to 1 million doors. Sadly, some people still think of the 2 doors at the end being a 50/50 choice.

    In my opinion, many people have poor intuition with conditional probabilities because they are not used to working on a restriction of the state space. How we are taught probability seems to enforce (in my opinion) an intuition based on only a priori calculations on the whole space.

    Another well known problem that seems to catch people out is this: I have two cards face down on the table.

    i) One of them is an ace. What’s the probability that both are aces?
    ii) Now I tell you the left card is an ace. What’s the probability that both are aces?

    i) is 1/33, ii) is 1/17.

  4. I usually find the clearest way to explain things like this is with a tree diagram. In this case, I draw it with the first level for the chosen door, second level for the door opened by MH. Then the leaves have two sets of labels showing the prizes for stick and switch.

    I actually once wrote a simulator to help convince somebody of this. Here’s the source if anyone is interested (compiles with g++ on linux, not sure about other operating systems, sorry): http://pastebin.com/YK0wPvCc

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s