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Field and Swamp: Animals and Their Habitats


The Monty Hall Problem:  A Notoriously Misunderstood Probability Problem

This is an old problem that reputedly first appeared in Martin Gardner's column in the Scientific American, but is memorialized as the Monty Hall Problem.  Monty Hall was the host of a game show called Let's Make a Deal that stopped running over 30 years ago.  Marilyn vos Savant made the problem famous in her column "Ask Marilyn" in 1990.  It got yet more attention when Leonard Mlodinow published a book on the public's statistical innumeracy called The Drunkard's Walk: How Randomness Rules Our Lives.  In spite of everything, it still arouses fiery controversy.  So I thought I'd try my hand at explaining it.

One thing I've learned over the years is that a proof is not a formula or an algorithm, but rather a form of persuasion.   It will only convince someone who is assuming the same things you are, asking the questions you can think of and answer.   In the professional mathematics world, there are certain conventions that proofs follow designed to make communication among such professionals easier.  But presenting a persuasive argument supporting the truth of a mathematical assertion to people from a variety of backgrounds is harder, and the stormy history of this problem and its proof bears this out.  So here I will present several alternative arguments, each aimed at a different mindset.

Below is a statement of the problem, in the setting of a game show episode in which you, the guest, hope to make some money by working smart, not hard.  NOTE: this is about making the choice with the better odds, not necessarily the right one.  ASSUMPTION:  In a series of these trials, you make your choice of Door #1 completely blindly and randomly, OR the changing arrangements do not follow any detectable pattern, i.e., at the very least, the proportion of times the car was hidden behind each door was 1/3.

Given:  Three closed doors, one hiding a car (the prize) behind it and hiding two goats (which you presumably can't take home even if you wanted to)

Procedure:   You pick a door (Door #1), hoping it has the car behind it.  The game show host opens another door (Door #2) which s/he knows has a goat behind it, and gives you the option of changing your choice to the remaining door.  You choose the door with the better odds of having the prize behind it, either 1) the one you picked first or 2) the other closed door (Door #3).

Issue:  Many people believe that there is an equal chance of the car prize behind either of the doors the host hasn't opened.  

What's the right choice and why?

A Table of possibilities and their probabilities (1=the car    0=a goat).  X, Y and Z are possible placements for the car and the goats, with equal probabilities for each, given the procedure described above.

Possible arrangements
Door #1  has prize? Door #2 has prize? Door #3  has prize? Odds out of 3 for each possibility There are three doors, with a 1 in 3 chance for the prize being behind each door. 
X 1 0 0 1 X: If you pick correctly the first time (1 in 3 chance), you will lose if you change your door selection.
Y 0 0 1 1 Y or Z: If you pick incorrectly the first time (2 in 3 chance),
Z 0 0 1 1 you will win if you change your door selection.
odds of winning (out of 3) à 1 0 2 3 Changing gives you twice the odds of winning.

Basic Explanation of the Above Table:

There are three doors, therefore three choices with equal probability of being right (A, B and C).  The 1 in 3 probability of guessing the right door in one try is represented by the three white-background cells in the "Door #1 has prize" column?  The white-background cells in the "Door #2 has prize?" column are all zeroes because the host must pick a non-winning door.  This leaves the third column, which contains the values that give row totals of 1 ("Odds out of 3 for each possibility.")  The bottom row gives the probabilities of each choice (out of 3) yielding the prize (in red) and their total (3, in orange).  Therefore, picking Door #3, i.e., switching, gives you a 2 in 3 probability of getting the prize.

It seems obvious that your initial guess has a 1 in 3 probability of being right.  However, where things seem less clear is when the host presents information about which of the other two doors hide a goat.  Does this change the probability of the first guess being right?  After all, if the host opened the other two doors, thereby revealing the answer, yes: the probability of each door hiding the prize at that moment would certainly be different!   You would have a 1 in 1 chance of guessing which door was right.   Does that mean the chance of your initial guess being right if the host opened one "wrong" door would be somewhere between 1 in 3 and 1 in 1?

 Addressing the "1 in 2" Issue:

Let's consider the case where the host might open a "wrong" door before you even made a guess: your chances of guessing right would be 1 in 2.  So what is different if you make a guess first?  In the latter case, the host faces only two closed doors (since you've already picked one) instead of three.   Two-thirds of the time, the host will be limited to opening one door (if you had first picked the one with the goat); otherwise (if you picked the car) s/he will have two choices.   There's something else you'll notice about the set-up when the host can open only one door (remember, that was two-thirds of the time!) and you made the wrong initial choice: those are the times when you should switch.  Conclusion: two-thirds of the time, you'll win by switching.

Another angle: Combinations vs. Permutations:

Another confusing factor is the difference between combinations and permutationsContent alone defines combinations whereas order also defines permutations.  The set of numbers 1 and 0 would have only one combination, i.e., {1,0}, while their two permutations would be {1,0} and {0,1}.   As you discover by flipping two coins at a time, you'll discover that, given enough trials, you'll get two heads 25% of the time, tails 25% also -- and a head and a tail 50% of the time.  The rule here is: each permutation has an equal probability of occurring, while some combinations sometimes occur more often than others.  This rule applies because you have equal chances of getting heads or tails for an individual toss.

How does this apply to the Monty Hall problem?  Well, when you think of having a 1 in 2 choice after the host picks, you're mistaking a permutation for a combination.  The combinations: 1) Door A (which you initially picked) and 2) Door B (the door neither you nor the host picked).   The permutations:  Door A (which you initially picked) and 2) Door B (since the host picked C) or 3) Door C (since the host picked Door B).  In other words, A or B vs.  A and (B or C) or A and (C or B).  Those two identical rows in the table (except for "B" and "C" in the leftmost column) represent different permutations but the same combinations.

Brute Force:  All the possibilities laid out, without abstraction

These are all the permutations for your choices and their outcomes according to the procedure given above.  Your choices, the host's choices, the outcomes.  The yellow rows show those permutations for which you lose if you switch, i.e., where the black numbers are zeroes.

Door A Door B Door C
1 0 0
0 1 0
0 0 1
1 0 0
0 1 0
0 0 1
1 0 0
0 1 0
0 0 1

A Third Angle:  Applying Bayesian Statistical Inference

Another way to model this is to apply Bayesian statistics, which calculates probability in situations where partial knowledge of the facts is known.  In essence, if you know the "truth value" of a logical statement describing certain probabilities (in the form of "probability that A implies B") you can find the probability of the converse of that statement (in the form of "probability that B implies A") by using it and another two pieces of information ("probability of A" and "probability of B.") 

Although this would seem to be a natural way of doing this, it's actually quite complex, especially if you try to model it using tables.  In essence, IMHO, if you can understand someone else's explanation of how to apply it, you can figure out an easier way to do the proof on your own.


Martin Gardner, Mathematical Games, Scientific American, 201 (1959), October 180-182, November 188








Copyright © 2008 by Dorothy E. Pugh