The post discusses the Monty Hall problem, a brain teaser and a classic problem in probability. The following 5 pictures describe the problem. Several simple solutions are presented.

___________________________________________________________________________

**The Problem in Pictures**

___________________________________________________________________________

**Should You Switch?**

Assuming that you, the contestant in the show, wants to win a car and not a goat, which door should you pick? Should you stick with your original pick of Door 1, or should you switch to Door 2? It is clear that had the game show host not intervened by opening another door, your chance of winning a car would be 1/3. Now that one door had been eliminated, you are faced with two doors, one being your original pick and another door. Would choosing one door out of two means your chance of winning the car is now 1/2? If this were the case, it would mean it does not matter whether you switch your choice or not. Enough people were reasoning in this same way that this problem stirred up a huge controversy back in 1990 when the problem was posed on Parade Magazine. More about that later. For now, let’s focus on the solution.

It is to your advantage to switch your choice. With switching, the probability of winning the car is 2/3, while the winning probability is 1/3 if you stick with the original choice. The rest of the post attempts to convince doubters that this is correct.

___________________________________________________________________________

**The Problem to be Solved**

The problem described by the 5 pictures above is based on the assumption that the contestant picks Door 1 and the host picks Door 3. The problem in the pictures is asking for the conditional probability of winning by switching given that the contestant picks Door 1 and the host picks Door 3. We solve a slightly different problem: what is the probability of winning a car if the contestant uses the “switch” strategy, which is the strategy of choosing the alternative door offered by the game show host. At the end, the first problem is also discussed.

The game involves selections of doors by the contestant and the host as well as by the staff who puts the prizes behind the three doors. It is important to point out that the manner in which the doors are chosen is important. Basically the doors have to be selected at random. For example, the staff of the game show selects the door with the car at random, the contestant selects his/her door at random, and the game show host select his/her door at random (in case that the door with car and the contestant’s door are the same). Under these assumptions, the best strategy is to switch. If the doors are not selected at random, the solution may not be the “switch” strategy.

___________________________________________________________________________

**Simulation**

An effective way to demonstrate that switching door is the optimal solution is through playing the game repeatedly. Let’s use a die to simulate 20 plays of the game. The simulation is done in 4 steps. In Steps 1 and 2, roll a fair die to select a door at random. If it turns up 1 or 2, then it is considered Door 1. If it turns up 3 or 4, then Door 2. If it turns up 5 or 6, then Door 3. In Step 3, the host also rolls a die to select a door if the door with car happens to be the same as the contestant’s choice.

In Step 1, the rolling of the die is to determine the prize (i.e. which door has the car). In Step 2, the rolling of the die is to determine the choice of door of the contestant. In Step 3, the host chooses a door based on Step 1 and Step 2. If the door with the car and the door chosen by the contestant are different, the host has only one choice. If the door with the car and the contestant’s door are identical, then the host will roll a die to choose a door. For example, if Door 1 is the door with the car and the door chosen by the contestant, then the host chooses Door 2 and Door 3 with equal probability (e.g. if the roll of the die is 1, 2 or 3, then choose Door 2, otherwise choose Door 3).

*Step 1- Simulate the Doors with Car*

Play | Door with Car |
---|---|

#1 | 1 |

#2 | 3 |

#3 | 3 |

#4 | 2 |

#5 | 2 |

#6 | 1 |

#7 | 1 |

#8 | 2 |

#9 | 1 |

#10 | 1 |

#11 | 2 |

#12 | 3 |

#13 | 1 |

#14 | 2 |

#15 | 2 |

#16 | 2 |

#17 | 3 |

#18 | 3 |

#19 | 1 |

#20 | 1 |

*Step 2- Simulate the Doors Chosen by the Contestant*

Play | Door with Car |
Contestant’s Choice |
---|---|---|

#1 | 1 | 2 |

#2 | 3 | 1 |

#3 | 3 | 1 |

#4 | 2 | 3 |

#5 | 2 | 3 |

#6 | 1 | 2 |

#7 | 1 | 1 |

#8 | 2 | 3 |

#9 | 1 | 1 |

#10 | 1 | 2 |

#11 | 2 | 2 |

#12 | 3 | 2 |

#13 | 1 | 1 |

#14 | 2 | 2 |

#15 | 2 | 1 |

#16 | 2 | 2 |

#17 | 3 | 2 |

#18 | 3 | 3 |

#19 | 1 | 2 |

#20 | 1 | 3 |

*Step 3- Simulate the Doors Chosen by the Host*

Play | Door with Car |
Contestant’s Choice |
Host’s Choice |
---|---|---|---|

#1 | 1 | 2 | 3 |

#2 | 3 | 1 | 2 |

#3 | 3 | 1 | 2 |

#4 | 2 | 3 | 1 |

#5 | 2 | 3 | 1 |

#6 | 1 | 2 | 3 |

#7 | 1 | 1 | 2 |

#8 | 2 | 3 | 1 |

#9 | 1 | 1 | 3 |

#10 | 1 | 2 | 3 |

#11 | 2 | 2 | 1 |

#12 | 3 | 2 | 1 |

#13 | 1 | 1 | 3 |

#14 | 2 | 2 | 3 |

#15 | 2 | 1 | 3 |

#16 | 2 | 2 | 3 |

#17 | 3 | 2 | 1 |

#18 | 3 | 3 | 2 |

#19 | 1 | 2 | 3 |

#20 | 1 | 3 | 2 |

*Step 4 – Contestant Makes the Switch*

Play |
Door with Car |
Contestant’s Choice |
Host’s Choice |
Switched Door |
/GoatCar |
---|---|---|---|---|---|

#1 | 1 | 2 | 3 | 1 | Car |

#2 | 3 | 1 | 2 | 3 | Car |

#3 | 3 | 1 | 2 | 3 | Car |

#4 | 2 | 3 | 1 | 2 | Car |

#5 | 2 | 3 | 1 | 2 | Car |

#6 | 1 | 2 | 3 | 1 | Car |

#7 | 1 | 1 | 2 | 3 | Goat |

#8 | 2 | 3 | 1 | 2 | Car |

#9 | 1 | 1 | 3 | 2 | Goat |

#10 | 1 | 2 | 3 | 1 | Car |

#11 | 2 | 2 | 1 | 3 | Goat |

#12 | 3 | 2 | 1 | 3 | Car |

#13 | 1 | 1 | 3 | 2 | Goat |

#14 | 2 | 2 | 3 | 1 | Goat |

#15 | 2 | 1 | 3 | 2 | Car |

#16 | 2 | 2 | 3 | 1 | Goat |

#17 | 3 | 2 | 1 | 3 | Car |

#18 | 3 | 3 | 2 | 1 | Goat |

#19 | 1 | 2 | 3 | 1 | Car |

#20 | 1 | 3 | 2 | 1 | Car |

There are 13 cars in Step 4. In 20 simulated plays of the game, the contestant wins 13 times using the switching strategy, significantly more than 50% chance of winning. Of course, if another simulation is performed, the results will be different. We perform 10 more simulations in Excel with 10,000 plays of the game in each simulation. The numbers of winnings in these simulations are:

*10 More Simulations*

Simulation | Number of Games | Number of Winning Games |
---|---|---|

#1 | 10,000 | 6,644 |

#2 | 10,000 | 6,714 |

#3 | 10,000 | 6,549 |

#4 | 10,000 | 6,692 |

#5 | 10,000 | 6,665 |

#6 | 10,000 | 6,687 |

#7 | 10,000 | 6,733 |

#8 | 10,000 | 6,738 |

#9 | 10,000 | 6,625 |

#10 | 10,000 | 6,735 |

Total | 100,000 | 66,782 |

In each of the 10 simulations, the chance of winning a car is between 66% and 67% (by switching). In all of the 100,000 simulated games, the chance for winning a car by switching door is 66.782%. More and more simulations showing essentially the same results should give us confidence that the “switch” strategy increases the chance of winning and that the probability of winning is around 2/3.

If anyone who still thinks that switching does not make a difference, perform simulations of your own. It can be done by rolling a die as done here or by using computer generated random numbers. In fact, a computer simulation can produce tens of thousands or more plays. But anyone who does not trust computer simulations can simply generate 100 games by rolling the die. Simulations, when done according to the assumptions behind the game (the assumptions that the doors are selected at random), do not lie. In fact, many of the skeptics in the Monty Hall problem were convinced by simulations.

___________________________________________________________________________

**Listing Out All Cases**

Note that in Step 4 of the simulation, one pattern is clear about switching to the door offered by the host. The contestant will win a goat if his/her original choice of door happens to be the same as the door with the car. On the other hand, the contestant will win a car if his/her original choice happens to be different from the winning door. The observation is actually another explanation of the solution.

Let’s say the contestant picks Door 1 originally. There are three cases to consider since the winning door can be any one of the three door. Let’s look at what happens if the contestant switches door in each case.

*The Three Cases for the Winning Door*

*Door 1 is Contestant’s Choice*The contestant wins 2 of the three cases by switching door. The contestant wins in only one of the cases if he/she were to stick with the original choice of door. The three cases are equally likely since the winning door is supposed to be randomly selected. So in the “switch” strategy, the probability of winning a car is 2/3.

___________________________________________________________________________

**Looking at Two Diagrams**

The solution in the previous section is a simple but correct solution. For anyone who thinks the solution is too simple, the following two diagrams can supplement the reasoning in the above simple solution.

In Figure 6, assume that the contestant picks Door 1, which as a 1/3 chance of winning a car. Then the other two doors as a group has a 2/3 chance of winning the car.

In Figure 7, the game show host opens a door with a goat (Door 3). The two doors of Door 2 and Door 3 as a group still has a 2/3 chance of winning a car. Since Door 3 is eliminated by the host, Door 3 now has zero chance of winning a car. So Door 2 has a 2/3 probability of winning the car. It is then advantageous for the contestant to switch door.

___________________________________________________________________________

**Tree Diagram**

Another way to view the problem is through a tree diagram. The idea is that such a tree diagram would capture the idea that the game show takes places in stages, e.g. the car is randomly placed behind one door, the contestant randomly picks one door, the host picks a door (randomly if required). The following diagram captures the entire random process.

Because the door is chosen at random, the probability at a branch is 1/3 if there are three branches at a decision point and the probability is 1/2 if there are two branches at any given node. Of course, if there is only one choice of a door, then the probability at the point is 1. There are 12 paths in the tree and each path probability is the product of all the individual probabilities in that path. At the right of the diagram, we compare the stay strategy and the switch strategy. With the stay strategy, the contestant wins in 6 of the paths. If using the switch strategy, the contestant wins in the other 6 paths. But the 6 paths with the switch strategy are twice as likely to happen.

___________________________________________________________________________

**The Problem Described in the Pictures**

The several solutions so far solve a general problem – if the contestant picks the door offered by the game show host, what is the probability of winning a car? In the pictures, the problem is: if the contestant picks Door 1 and the host picks Door 3, what is the winning probability for the contestant under the switch strategy? This is a conditional probability. Figure 8 has all the ingredients for the answer to this problem. In Figure 8, there are two paths in which the contestant picks Door 1 and the host picks Door 3. The following diagram shows these two paths with a check mark.

For one of these two paths, the car is behind Door 1 and for the other path, it is behind Door 2. The path with the car behind Door 2 is twice as likely as the one with the car behind Door 1. With the switching strategy, the path with the car behind Door 1 would lead to a goat (losing path) and the path with car behind Door 2 will lead to a car (winning path). The winning path is twice as likely. Thus the winning probability for the switch strategy is 2/3 if the contestant picks Door 1 and the host picks Door 3. Now this answer is the same as the answer for the general problem discussed previously. However, the conditional problem is a different problem.

___________________________________________________________________________

**Remarks**

The Monty Hall problem is loosely based on a real game show called Let’s Make a Deal. At one point in time, it was only a brain teaser as well as an academic problem (being discussed in statistics journals). In 1990, it appeared in a column by Marilyn vos Savant in Parade Magazine. In that column, vos Savant used the simple 3-case solution above to explain that the contestant should switch. The problem generated so many responses from angry readers, some of them had PhDs in math or statistics. Because of the ensuing controversy, the Monty Hall problem became a famous problem the world over. It is covered in many standard probability and statistics books and courses. Eventually some of those angry readers came to understand that the correct answer is the switch strategy. In fact, Paul Erdős, a famous and prolific mathematician in the 20th century, did not believe that switching is the correct answer. He remained unconvinced until someone showed him a computer simulation (see here).

Some psychologists asserted that the Monty Hall problem causes cognitive dissonance. As a result, the problem is confusing and even troubling to some people. See this article for the discussion.

The Monty Hall problem is all over the Internet. The Wikipedia entry on Monty Hall problem has more detailed information mathematically and in many other aspects. This article from Scientific American is also interesting.

___________________________________________________________________________

2017 – Dan Ma

Pingback: Five probability problems to help us think better | Math in the Spotlight

Very interesting problem-paradox, and well presented. You just left a typo: “If it turns up 2 or 4, then Door 2” in the Simulation paragraph.

Typo corrected. Thanks for point it out.

Pingback: More about Monty Hall Problem | A Blog on Probability and Statistics

Pingback: Monty Hall Problem – Daniel Ma

I had recently been discussing this brain teaser with family, and afterwards googled it just to see what came up. Anyway, your post is absolutely incorrect. The way you phrase the problem makes the odds 50/50. The question is properly framed as such. “Monty Hall asks a player to choose one of three doors, behind one of which is a new car. After choosing, Monty WILL open one of the two REMAINING doors, behind which there is nothing. Monty will then ask the player if they want to change their answer. Should they do so?” The key difference is that this description makes clear that Monty can’t open the door the player has chosen, which is the factor that increases the odds to 2/3. This fact isn’t made clear in the original question (or even the description of the brain-teaser) (though it is shown that way in the diagrams of possible scenarios/outcomes). If Monty were forced to flip a coin between the two doors behind which there is no car, and open it, regardless of whether it is the door chosen by the player or not, the odds would then be 50/50 and changing the answer would make no difference.

I wanted to ask: Would it be alright with you if I used the pictures on this blog post for an article that my college’s math committee is writing on the Monty Hall Problem?