# The Full Monty... in Proportion

published Feb 8, 2013**Numb:**1/5 is the same multiple of 1/4 as 3/5 is of 3/4.

**Number:**The ratio of 1/5 to 1/4 equals 3/5 to 3/4.

**Num-best:**The probabilities 1/5 and 3/5 are in proportion to 1/4 and 3/4.

The Monty Hall Problem is the all-time classic probability problem that is emblematic of how counter-intuitive counting can sometimes be. This problem has experienced a renaissance of late, appearing in an expanded version in the Macalester Problem of the Week in October 2008, and getting top billing as the lead article, "Monty Hall, Monty Fall, Monty Crawl" by Jeffrey Rosenthal, in the September 2008 *Math Horizons*publication. Monty has long been a favorite of mine, and I included a variant of it in Alfred University's 2008 Nevins Mathematics Exam for high school students. Monty still causes fits for many professional mathematicians (it was voted the most difficult problem on the Nevins exam), and most solutions that are presented are "hand-waving" at best. So when I read the article by Rosenthal which presents an airtight proof of the answer, and does so using the "proportionality principle," I could not resist sharing with the National Numeracy Network.

The problem states that you are on a game show and must choose one door from three, with the chance of winning a car which is behind one of the doors. After you have made your choice, the host, Monty Hall, graciously opens one of the two other doors which does not contain the car and gives you the option of switching your door to the door he did not open. What should you do? Should you trust your first instinct and stay, or should you switch given the new information, or does it just not matter since there are now only 2 doors and it is 50-50 that the car is behind either one? Marilyn vos Savant answered this question in her "Ask Marilyn" column in *Parade Magazine* on September 9, 1990. She reportedly has the world's highest IQ and did not disappoint by correctly saying you should always switch and double your chances of winning from 1/3 to 2/3! This touched off such a firestorm of controversy with statisticians writing in arguing for the 50-50 solution, that she eventually devoted three more columns to this problem.

A nice way to explain the correct solution is to ask what you would do if Monty had offered you **both** other doors instead of the one you had chosen. In this case everyone agrees you should take two doors for the one and get a 2/3 chance of winning the car; but this is exactly what Monty is doing! The fact that he tells you one of the two doors does not have the car does not change the fact that he is giving you both doors. I have been comfortable with using this explanation for my students until I read the Monty Fall problem in Rosenthal's article. In this version Monty accidentally falls after one too many cosmopolitans and opens one of the two doors by mistake. It is possible in this version that Monty might open the door with the car. If the car is not behind this randomly opened door, should you still switch? The correct solution now is that it doesn't matter, the 50-50 solution is correct here!

This blew my mind away (please return if seen) since this seems to be identical to the original problem. The subtle change is in the randomness of his action. His **intention** makes all the difference in the world and seems to place this problem in the paradoxical realm of Schrodinger's cat. The resolution of this conundrum lies in the **proportionality principle**:

If various alternatives are equally likely, and then some event is observed, the updated probabilities for the alternatives are proportional to the probabilities that the observed event would have occurred under those alternatives...

Yikes! This definition is a good example of a logically sound mathematical statement that takes a bit to digest, and is best understood via example. So let's use a simpler situation of two siblings, Alice and Bob, who are **equally likely** to have eaten an apple. We **observe** an apple core on the counter and know that Alice leaves the core on the counter 1/5 of the time while Bob leaves the core on the counter 3/5 of the time. What are the **updated probabilities** for Alice and Bob to have eaten the apple given the **observed event**?

We have the conditional probabilities of the **observed event**, core, given Alice or Bob. (The vertical slash is read, "given," so the first statement below reads: "The probability of finding a core given Alice ate the apple is 1/5."):

P(core|Alice)=1/5 andP(core|Bob)=3/5

The proportionality principle is trying to tell us that these probabilities are in proportion to the reverse conditional probabilities:

P(core|Alice):P(core|Bob) = 1:3 =P(Alice|core):P(Bob|core).

The next key is to realize that although the first pair of probabilities, 1/5 and 3/5, do not have to add to one, the second pair of probabilities must add to one since they represent the only possible alternatives given the observed event of the apple core. Thus we have:

P(Alice|core) +P(Bob|core) = 1

...

P(Alice|core)=1/4 andP(Bob|core)=3/4 .

We can now apply this technique to the Full range of Monty problems. In Monty Hall, the car is **equally likely** to be behind door #1, or #2, or #3. Let's assume without loss of generality that you choose door #1. The **observed event** is Monty now opening another door, let's say #2. So our three alternatives are car #1, car #2, or car #3; and the observed event is open #2. Once again we have the conditional probabilities for the observed event given where the car might be:

P(open#2|car#1)=1/2,P(open#2|car#2)=0,P(open#2|car#3)=1.

The first probability is 1/2 because if the car is behind door #1 (which you chose) Monty may open door #2 or door #3; while if the car is behind door #3, then Monty must open door #2 (he cannot open the door you chose)! The two non-zero probabilities corresponding to doors #1 and #3 are now in the ratio of 1 : 2, which means the similar reverse probabilities must be in the same proportion:

P(car#1|open#2):P(car#3|open#3) = 1:2.

And since these two probabilities must add to one we have that the probability the car is behind your choice of door #1 given door #2 open is 1/3 and the probability the car is behind door #3 given door #2 open is 2/3. Thus you should switch to door #3!

The Monty Fall problem can be treated identically but now the randomness of Monty opening either door means that they are equally likely to be opened and hence are in the proportion of 1:1, not 1:2, and thus the updated probabilities must both be ½! It is crucial in this version that we are assuming he does not open the door with the car, a point we will return to post haste.

Strictly speaking the proportionality principle is nothing more than the definition of conditional probability applied twice in succession:

Where the A_{i}'s are the equally likely events and B is the observed event (and hence has non-zero probability). Saying these probabilities are "proportional" is referring to the fact that every conditional probability is the same constant multiple of its reverse. This is exactly the same as when two triangles are similar; meaning the ratio of any two associated sides is constant. Thus we could say the two sets of conditional probabilities are similar.

The solution technique above does not require us to actually solve for this constant multiple, we used the additional fact that the sum of the P(A_{i}|B)^{'s} is one. Hence two probabilities in the ratio 1:2 which add to one must be 1/3 and 2/3. There seems to be some magical action-at-a-distance property involved in computing these conditional probabilities by first finding the reverse and then invoking similarity. Conditional probabilities are well-known for their confusing properties and have been popularized in the false-positive medical diagnosis problems. The proportionality principle provides a new twist on these tricky ratios.

One final note for those still not convinced. Marilyn vos Savant admonished her readers who did not believe to play the game with cups and pennies and count the number of times you win by switching. I have my students play a 21^{st} century version of this game by programming Excel to randomly pick a door for the car, and randomly pick a door for the contestant. Now since Monty will open another door without the car, you will win by switching if your door does not equal the car door. Thus you can have Excel count the number of times you win by switching, by simply checking when your door is not equal to the car door! Run this simulation a thousand times and you win on average 2/3 of the time. Interestingly, when students are forced to think of the problem this way they can see why the solution is 2/3, and do not need to run the simulation to count the number of wins.

The Monty Fall problem is also easy to program since the opened door is random (just not equal to the chosen door). Remember that in this version you can now also win if Monty opens the door with the car. I ran this simulation (pennies and cups work here as well) and was expecting to win 50% of the time. Excel however is not interested in the nuances of the proportionality principle, it is a counting machine, and gave me the correct answer. I won't spoil your surprise; that would be letting the cat out of the box!

**Sapere Aude!**

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*The Full Monty... in Proportion*