Wednesday, April 13, 2016

The Curious Incident from Let's Make a Deal

The Curious Incident from Let’s Make a Deal

Statistical analyses depend on probability distributions.  So it is not surprising that rudiments of probability theory are emphasized in statistics courses.  The theme of this post is that, for many, probability theory is not being taught well, not being understood, or both.

Some years ago, I read a book called The Curious Incident of the Dog in the Nighttime, by Mark Haddon.  Haddon, and his protagonist in the book, are fascinated with mathematics.  One of the reasons given for this fascination is that math can explain things that otherwise might seem counterintuitive.  An example of this power of math, given in the book, is the “Let’s Make a Deal” scenario.  Perhaps you know of this scenario and it implications, but reading this book was my first exposure to it.  It goes like this:

The host of Let’s Make a Deal has great news for you.  You could win money!  All you must do is choose one of the three doors: A, B, or C.  Behind one of those doors is money.  Behind the other two doors, goats.  Now goats are cool and all, but let’s assume you want the money.  So you choose a door, say A.  Before the host opens door A, he first opens one of the other doors.  The host, of course, knows which door hides the money, so will only open a door hiding goats.  So the host opens a door (B or C) to reveal goats.  Now, the host says you can keep door A, or switch to the other unopened door.  What do you do?  One door must hide the money.  The other must hide goats.  50/50, right?  So it doesn’t matter if you switch or keep door A, right?.

Probability theory, however, shows that given this scenario, switching increases the chances of getting the money beyond 50%.  To be exact, switching gets the money 2 in 3 times, or about 66.7% of the time.  Not switching, then, gets the money 1/3 times.  When the host asked you to pick among 3 doors, that choice created a 1/3 chance of picking the money and 2/3 chance of picking goats.   The act of opening another door that hides goats does not change your winning probability (and surely does not raise it to 50%).  However, 2/3 of the time (i.e., when you first chose a door that hides goats), there is only one door the host can open that hides goats.  Thus, 2/3 of the time, the remaining door (i.e., not the one you originally chose, nor the door the host just opened) hides the money.

This Let’s Make a Deal exercise in probability is not very complicated, but I have been amazed by the number of people, many with advanced degrees, who have argued with me that there is no benefit to switching.  My explanation usually falls on deaf ears, suggesting I did not explain well, or the message just is not being received.  At that point,  I turn to Google, or this page:


  1. Reading your post I was a little confused. When I looked at the wikipedia page I understood immediately. By the host not opening door 2, you gain the information that the host would prefer not to reveal to you what is behind door 2. This additional knowledge is what changes the probabilities.

  2. Reading your post I was a little confused. When I looked at the wikipedia page I understood immediately. By the host not opening door 2, you gain the information that the host would prefer not to reveal to you what is behind door 2. This additional knowledge is what changes the probabilities.

  3. This exact example was discussed in out textbook, Intuitive Biostats, on pages 6-7. And I was immensely confused about the logic behind this, and to some degree I still am.
    I understand why the 66% vs. 33% dichotomy exists because of the initial circumstances of the choice, but I really have to wrestle with myself to assert that there still isn't a 50/50 shot after the first door gets open. The wiki link you provided is super helpful, but provides WAY more analysis and explanations than I was expecting.
    Something I did find very interesting was a paper linked in the Wikipedia page about "The Collapsing Choice Theory". In short, the paper discusses the idea that exceeding someone's working memory limitations, or giving them so many choices they can't possibly keep up with all of them, actually improves decision making abilities. In the Monty Hall problem you've discussed, they propose that using 1,000,000 (or any number greater than 7 choices) starts to overload our memory responses. It also gives the chooser an added advantage of perspective in regards to the likelihood that their door does contain the prize and not a goat.
    Consider if the Monty Hall problem were set up with 1 million doors. You then choose one door, and the host opens all of them except your door and one other door. How do you feel about the chances that you managed to pick the correct door out of 1 million, and that the one door the host left closed has a goat behind it?
    This theory focuses on the idea that when we can't logically keep up with all out choices, we start to collapse our logic to handle the situation better, which is exactly what is call for in probability: a simplification of choices.
    Regardless, this is definitely a useful probability to remember if anyone is ever a contestant on a game show.

    1. Thanks, Jacob. Shows how closely I read the text!

      I was just going to add the 1 in 1,000,000 example you alluded to above. It really helps drive home the point. I like to think of this scenario as a continuum. The 1/3 example is the point where the probability of getting the prize upon switching is at a minimum. However, even when I have tried this line of explanation with some of my post-doc friends, they still insist on 50/50 in the 3 door scenario.

  4. As I understand it, the reason that people (myself included) struggle with this problem is that we instinctively break the scenario up into multiple independent problems instead of treating it as a continuous statistical scenario in which additional information is provided that changes the probability. I understand why, mathematically, the chances of selecting the correct door are better if the contestant switches. To arrive at this conclusion, though, you have to account for the probability of a positive outcome and a negative outcome for each door and carry those probabilities through the entire scenario. You essentially have to perform arithmetic with probabilities, and that is not at all instinctive for people, even well-educated people. People misunderstand this problem because they treat it as two separate, independent scenarios. To begin with, you have three doors, and one is correct; probability of choosing correctly - 1/3. When one door is taken away, the previous scenario has been altered; thus we remove it from our immediate consciousness and start fresh with a new scenario. There are two doors, and one is correct; probability of choosing correctly - 1/2. I admit, even though I understand how the numbers really shake out for this problem and have seen the math multiple times, I couldn’t reproduce it without careful thought, and I highly doubt I’d recognize a similar probability trick if I encountered one. That’s the power of instinct. It makes me wonder what else we get wrong because we act on instinct that is mathematically fallacious.

    1. Good points, Madelyn. Especially your last.

      Since our instincts seem to be faulty, we can assume safely the founders of probability theory probably suffered the same affliction.

      Which makes their accomplishments all the more remarkable.

  5. Great post. This is really quite an interesting dilemma. I'm glad I have read this in case I ever end up on a game show. Personally I, like one the masses, would not have come to conclusion that choosing again increases your odds to 50%. My logic would be that 1/3 of time you have already chosen the right door and the 2 other doors both have goats behind them. So 1/3 of the time changing your choice yields a 0% probability of winning. Then the other 2/3 of the time you chose wrong initially and the next choice provides a 50% chance of winning. This would lead you to believe that your chances are always at 1/3. Ohh but I see my mistake now. He will never open the door with the money behind it. If he always picks a door with a goat and a door you haven't picked, then regardless of your first pick, his reveal leaves one door with a goat and one with the money thus a 50% chance.