My Dominant Hemisphere

The Official Weblog of 'The Basilic Insula'

Does Changing Your Anwer In The Exam Help? The Monty Hall Paradox

One of the 3 doors hides a car. The other two hide a goat each. In search of a new car, the player picks a door, say 1. The game host then opens one of the other doors, say 3, to reveal a goat and offers to let the player pick door 2 instead of door 1. Is there an advantage if the the player decides to switch? (Courtesy: Wikipedia)

Hola amigos! Yes, I’m back! It’s been eons and I’m sure many of you may have been wondering why I was MIA. Let’s just say it was academia as usual.

So what the heck are decision trees anyway? They are basically conceptual tools that help you make the right decisions given a couple of known probabilities. You draw a line to represent a decision, and explicitly label it with a corresponding probability. To find the final probability for a number of decisions (or lines) in sequence, you multiply or add their individual probabilities. It takes skill and a critical mind to build a correct tree, as I learned. But once you have a tree in front of you, its easier to see the whole picture.

Let’s just ignore decision trees completely for the moment and think in the usual sense. How good an idea is it to change an answer on an MCQ exam such as the USMLE? The Kaplan lecture notes will tell you that your chances of being correct are better off if you don’t. Let’s analyze this. If every question has 1 correct option and 4 incorrect options (the total number of options being 5), then any single try on a random choice gives you a probability of 20% for the correct choice and 80% for the incorrect choice. The odds are higher that on any given attempt, you’ll get the answer wrong. If your choice was correct the first time, it still doesn’t change these basic odds. You are still likely to pick the incorrect choice 80% of the time. Borrowing from the concept of “regression towards the mean” (repeated measurements of something, yield values closer to said thing’s mean), we can apply the same reasoning to this problem. Since the outcomes in question are categorical (binomial to be exact), the measure of central tendency used is the Mode (defined as the most commonly or frequently occurring thing in a series). In a categorical series – cat, dog, dog, dog, cat – the mode is ‘dog’. Since the Mode in this case happens to be the category “incorrect”, if you pick a random answer and repeat this multiple times, you are more likely to pick an incorrect answer! See, it all make sense 🙂 ! It’s not voodoo after all 😀 !

Coming back to decision analysis, just as there’s a way to prove the solution to the Monty Hall Paradox using decision trees, there’s also a way to prove our point on the MCQ problem using decision trees. While I study to polish my understanding of decision trees, building them for either of these problems will be a work in progress. And when I’ve figured it all out, I’ll put them up here. A decision tree for the Monty Hall Paradox can be accessed here.

To end this post, I’m going to complicate our main question a little bit and leave it out in the void. What if on your initial attempt you have no idea which of the answers is correct or incorrect but on your second attempt, your mind suddenly focuses on a structure flaw in one or more of the options? Assuming that an option with a structure flaw can’t be correct, wouldn’t this be akin to Monty showing the goat? One possible structure flaw, could be an option that doesn’t make grammatical sense when combined with the stem of the question. Does that mean you should switch? Leave your comments below!

Hope you’ve found this post interesting. Adios for now!

Flesch reading ease score:  72.4
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SMOG index: 10.7 Intuitive Biostatistics, by Harvey Motulsky The Monty Hall Problem: The Remarkable Story Of Math’s Most Contentious Brain Teaser, by Jason Rosenhouse

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Written by Firas MR

August 11, 2009 at 2:50 pm

8 Responses

1. Salaam

Um your analysis is incorrect. Your expected winning isn’t \$800 if you switch, because you can still only choose one of the remaining four. If you could choose all four and win if one of the four was correct, then it would indeed be \$800. So your chances if you switch are still \$1000 * (4/5)(1/4) = \$200 (chances of your current answer NOT being the correct one, multiplied by the chance of the new answer being correct).

The reason this is different from the Monty Hall problem is because you’re given new information after you choose. You’re shown the empty door and that changes around the probabilities. Suddenly the combined chances of two doors collapses into one door because one is shown to be empty.

Be careful with the math 🙂 BrownSandokan

August 11, 2009 at 4:42 pm

• W’salam,

Interesting 🙂 . I’d actually thought about this. But I think it makes sense if you think of the two pathways as two independent branches. In fact this is how the problem is approached in the book. Your decision to pick an answer is not necessarily correlated or dependent upon your earlier choice. Every answer is an equal candidate of being the right option, as long as you can’t establish that for a fact – that is your outcomes are truly random.

And when you want to figure out the probabilities for each outcome – \$0 or \$1000, I think it makes sense to multiply each outcome with its own probability. To calculate the combined probability of the two mutually exclusive outcomes just find the sum.

Thus, if your calculation can be written as [(4/5 * 0) + (1/4 * 1000)], you end up with \$250, and you’re still better off switching 🙂 .

I agree, the Monty Hall problem is a little different. I’m going to make a mention of that in the post. Geez, I’m so awful at arithmetic! I might have to completely rewrite this if I’m wrong! Firas MR

August 11, 2009 at 5:15 pm

• Nope .. absent any new information, you can’t get any more out of the system than an expected winning of \$200.

I don’t see how you got \$250. What are you multiplying? You have to multiply the 1/4 with 4/5 also because both events must happen. BrownSandokan

August 11, 2009 at 5:22 pm

• @BrownSandokan
Yes, I understand where you’re coming from and your point makes sense. Absent any new information, how can you get anything more than \$200? Valid question. Tell you what, I’m going to send you the excerpts on decision trees from his book and maybe you can help me learn something! And I’ll keep updating this post as I go along. Maybe I should turn this article in to a wiki 🙂 . Firas MR

August 11, 2009 at 5:54 pm

• Just as a quick refresher to readers, there are three different kinds of probabilities:

Independent :- Two events x and y are considered independent if the occurrence of x doesn’t preclude the occurrence of y and vice versa. Not only that, the occurrence of x doesn’t tell you anything about the occurrence of y. To find the combined probability of two independent events x AND y multiply like so

Pr(x AND y) = Pr(x) * Pr(y)

Example: What is the combined probability that a patient has a foot infection and is bald?

Pr(foot infection AND baldness) = Pr(foot infection) * Pr(baldness)

Conditional: Two events x and y are said to be conditional if giving the probability of one tells you something about the other. Not only that, the occurrence of x doesn’t preclude the occurrence of y or vice versa. To find the combined probability of two conditional event x and y, where y is conditional on x:

Pr(x AND y; if y given x) = Pr(x) * Pr(y given x)

Example: What is the combined probability that a person has diabetes AND atherosclerosis? We all know atherosclerosis and diabetes are usually seen together and having one predisposes to the other.

Pr(atherosclerosis AND diabetes, if diabetes given atherosclerosis) = Pr(atherosclerosis) * Pr(diabetes given atherosclerosis)

Mutually exclusive: Two events x and y are mutually exclusive if the occurrence of x precludes the occurrence of y or vice versa. To find the combined probability of two mutually exclusive events x and y, find the sum like so:

Pr(x OR y) = Pr(x) + Pr(y)

Example: What is the combined probability that patient has hyperacidity OR hypoacidity?

Pr(hyperacidity OR hypoacidity) = Pr(hyperacidity) + Pr(hypoacidity) Firas MR

August 11, 2009 at 6:13 pm

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