Who would’ve thought that an old TV game show could inspire a statistical problem that has tripped up mathematicians and statisticians with Ph.Ds? The Monty Hall problem has confused people for decades. In the game show, Let’s Make a Deal, Monty Hall asks you to guess which closed door a prize is behind. The answer is so puzzling that people often refuse to accept it! The problem occurs because our statistical assumptions are incorrect.

The Monty Hall problem’s baffling solution reminds me of optical illusions where you find it hard to disbelieve your eyes. For the Monty Hall problem, it’s hard to disbelieve your common sense solution even though it is incorrect!

The comparison to optical illusions is apt. Even though I accept that square A and square B are the same color, it just doesn’t seem to be true. Optical illusions remain deceiving even after you understand the truth because your brain’s assessment of the visual data is operating under a false assumption about the image.

I consider the Monty Hall problem to be a statistical illusion. This statistical illusion occurs because your brain’s process for evaluating probabilities in the Monty Hall problem is based on a false assumption. Similar to optical illusions, the illusion can seem more real than the actual answer.

To see through this statistical illusion, we need to carefully break down the Monty Hall problem and identify where we’re making incorrect assumptions. This process emphasizes how crucial it is to check that you’re satisfying the assumptions of a statistical analysis before trusting the results.

## What is the Monty Hall Problem?

Monty Hall asks you to choose one of three doors. One of the doors hides a prize and the other two doors have no prize. You state out loud which door you pick, but you don’t open it right away.

Monty opens one of the other two doors, and there is no prize behind it.

At this moment, there are two closed doors, one of which you picked.

The prize is behind one of the closed doors, but you don’t know which one.

Monty asks you, “Do you want to switch doors?”

The majority of people assume that both doors are equally like to have the prize. It appears like the door you chose has a 50/50 chance. Because there is no perceived reason to change, most stick with their initial choice.

Time to shatter this illusion with the truth! If you switch doors, you double your probability of winning!

What!?

## How to Solve the Monty Hall problem

When Marilyn vos Savant was asked this question in her *Parade* magazine column, she gave the correct answer that you should switch doors to have a 66% chance of winning. Her answer was so unbelievable that she received thousands of incredulous letters from readers, many with Ph.D.s! Paul Erdős, a noted mathematician, was swayed only after observing a computer simulation.

It’ll probably be hard for me to illustrate the truth of this solution, right? That turns out to be the easy part. I can show you in the short table below. You just need to be able to count to 6!

It turns out that there are only nine different combinations of choices and outcomes. Therefore, I can just show them all to you and we calculate the percentage for each outcome.

You Pick | Prize Door | Don’t Switch | Switch |

1 | 1 | Win | Lose |

1 | 2 | Lose | Win |

1 | 3 | Lose | Win |

2 | 1 | Lose | Win |

2 | 2 | Win | Lose |

2 | 3 | Lose | Win |

3 | 1 | Lose | Win |

3 | 2 | Lose | Win |

3 | 3 | Win | Lose |

3 Wins (33%) | 6 Wins (66%) |

Here’s how you read the table of outcomes for the Monty Hall problem. Each row shows a different combination of initial door choice, where the prize is located, and the outcomes for when you “Don’t Switch” and “Switch.” Keep in mind that if your initial choice is incorrect, Monty will open the remaining door that does not have the prize.

The first row shows the scenario where you pick door 1 initially and the prize is behind door 1. Because neither closed door has the prize, Monty is free to open either and the result is the same. For this scenario, if you switch you lose; or, if you stick with your original choice, you win.

For the second row, you pick door 1 and the prize is behind door 2. Monty can only open door 3 because otherwise he reveals the prize behind door 2. If you switch from door 1 to door 2, you win. If you stay with door 1, you lose.

The table shows all of the potential situations. We just need to count up the number of wins for each door strategy. The final row shows the total wins and it confirms that you win twice as often when you take up Monty on his offer to switch doors.

## Why the Monty Hall Solution Hurts Your Brain

I hope this empirical illustration convinces you that the probability of winning doubles when you switch doors. The tough part is to understand *why* this happens!

To understand the solution, you first need to understand why your brain is screaming the incorrect solution that it is 50/50. Our brains are using incorrect statistical assumptions for this problem and that’s why we can’t trust our answer.

Typically, we think of probabilities for independent, random events. Flipping a coin is a good example. The probability of a heads is 0.5 and we obtain that simply by dividing the specific outcome by the total number of outcomes. That’s why it *feels* so right that the final two doors each have a probability of 0.5.

However, for this method to produce the correct answer, the process you are studying must be random and have probabilities that do not change. Unfortunately, the Monty Hall problem does not satisfy either requirement.

**Related post**: How Probability Theory Can Help You Find More Four-Leaf Clovers

## How the Monty Hall Problem Violates the Randomness Assumption

The only random portion of the process is your first choice. When you pick one of the three doors, you truly have a 0.33 probability of picking the correct door. The “Don’t Switch” column in the table verifies this by showing you’ll win 33% of the time if you stick with your initial random choice.

The process stops being random when Monty Hall uses his insider knowledge about the prize’s location. It’s easiest to understand if you think about it from Monty’s point-of-view. When it’s time for him to open a door, there are two doors he can open. If he chose the door using a random process, he’d do something like flip a coin.

However, Monty is constrained because he doesn’t want to reveal the prize. Monty very carefully opens only a door that does not contain the prize. The end result is that the door he doesn’t show you, and lets you switch to, has a higher probability of containing the prize. That’s how the process is neither random nor has constant probabilities.

Here’s how it works.

The probability that your initial door choice is wrong is 0.66. The following sequence is totally deterministic when you choose the wrong door. Therefore, it happens 66% of the time:

- You pick the incorrect door by random chance. The prize is behind one of the other two doors.
- Monty knows the prize location. He opens the only door available to him that does not have the prize.
- By the process of elimination, the prize must be behind the door that he does not open.

Because this process occurs 66% of the time and because it always ends with the prize behind the door that Monty allows you to switch to, the “Switch To” door *must* have the prize 66% of the time. That matches the table!

**Related post**: Luck and Statistics: Do You Feel Lucky, Punk?

## If Your Assumptions Aren’t Correct, You Can’t Trust the Results

The solution to Monty Hall problem seems weird because our mental assumptions for solving the problem do not match the actual process. Our mental assumptions were based on independent, random events. However, Monty knows the prize location and uses this knowledge to affect the outcomes in a non-random fashion. Once you understand how Monty uses his knowledge to pick a door, the results make sense.

Ensuring that your assumptions are correct is a common task in statistical analyses. If you don’t meet the required assumptions, you can’t trust the results. This includes things like checking the residual plots in regression analysis, assessing the distribution of your data, and even how you collected your data.

As for the Monty Hall problem, don’t fret, even expert mathematicians fell victim to this statistical illusion!

Brent says

June 14, 2019 at 11:55 amIs it not, then, equally likely that you choose doors “2 or 3” when the prize is behind door one? Once more, the outcome would be the same if you switch no matter which door you chose, and if we use the incorrect doors as a unit in one instance, we’re forced to do so in any instance. How can it be that listing two separate outcomes is only artificially inflating the outcome when it benefits your hypothesis to call it so? Citing established belief as the only “correct” solution is a poor precedent to be setting when that belief relies on biased calculation

Jim Frost says

June 14, 2019 at 2:37 pmHi Brent,

If choose door 2 or door 3 and the prize is behind door 1, that’s covered by different rows in the table.

The table lists multiple doors under the Monty Opens column only when your initial choice is the door that prize is behind. From Monty’s perspective, he will reveal one of the two choices. It doesn’t matter which one because it won’t change the outcomes or probabilities.

The logic behind this problem is that your initial choice is most likely to be wrong. There’s two-thirds chance that you’ll pick the wrong door. That’s why it’s beneficial to switch because your initial choice is probably wrong. Monty helps by removing one of the incorrect doors from the set of doors you didn’t select.

The reason this hurts your brain so much, and the point behind the blog post, is that most people go into this problem with the wrong assumptions. We’re thinking independent, random probabilities. In reality, it’s a mixture of random probabilities when you first choose but then Monty acts with intention to modify the probabilities going forward.

No one is saying that this is true because it’s “established belief.” It’s true because it’s been proven mathematically and it’s been proven using computer simulations that run through it thousands of times. Further, this is an experiment you and a friend can do on your own where one of you is the contestant and the other is Monty. Just run through this scenario a number of times, record the results, and see how you fare when you switch and don’t switch. The Mythbusters did this and verified it empirically.

Yeah, I know, it hurts the brain, but it’s true!

Frank Goring says

April 9, 2019 at 3:08 amHi Jim,

Another way to look at it:

1. The probability the contestant chooses the door with the car behind it is 1/3

2.The probability Monty chooses the door with the car behind it is 0

3. The probability the unchosen door hides the car is therefore 2/3

And the contestant can deduce this before he even makes his choice. ðŸ™‚

Jim Frost says

April 9, 2019 at 10:19 amHi Frank,

That’s a great way to look at it too! It’s interesting how there are common sense ways of seeing the problem like that, yet the solution caused such a commotion originally!

Zach Dorman-Jones says

March 27, 2019 at 4:58 pmJim,

I do understand the difficulty of presenting the information in an intuitive way, considering how subjective intuitiveness is to begin with. I guess what I would do is to accompany the table with a probability tree, which is a good way to visually illustrate the conditional probabilities.

By the way, I love the optical illusion in the beginning. I needed to use a color picker to convince myself that the colors of squares A and B are (nearly) identical. Even knowing that, I still “see” the difference between them, and always will. Mind-blowing!

To the folks asserting that Jim has cheated by clumping outcomes together, consider this somewhat simpler scenario. I plan to flip a fair coin. If I get heads, I will drink coffee and eat chocolate cake. If I roll heads, then I will flip the coin again. If I gets heads that time, I will eat ice cream. Tails, and I will eat pecan pie.

Here is a table of possible outcomes:

| First action | Second action | Third action

—————————————————————–

| Roll heads | Drink coffee | Eat chocolate cake

| Roll tails | Roll heads | Eat ice cream

| Roll tails | Roll tails | Eat pecan pie

The way I have presented this makes it look like these are three equally likely outcomes. They are not! The probability that I end up eating chocolate cake is 50%; ice cream and pecan pie are 25% each.

Presenting the table shown earlier with separate rows for each of Monty’s possible choices introduces a similar catch. Those rows would not be equally probable as the ones where Monty has only one choice. Individually, they would be half as probable, for the same reason that eating ice cream in my example is half as probable as eating chocolate cake: conditional probability.

Zach Dorman-Jones says

March 26, 2019 at 12:21 pmThat is a cogent observation, and it seems you are on the right track to understanding the solution with a little more consideration. No offense to Jim Frost, but I think that the table is a little bit misleading, and somewhat obscures the correct solution. The key thing to understand about the table is that the “Monty Opens” column is extraneous; it’s just a function of the first two columns. I think the example would be clearer with the “Monty Opens” column removed entirely.

Jim Frost says

March 26, 2019 at 4:45 pmHi Zach,

I sort of see what you’re saying. The “Monty opens” column might add a bit of confusion. And winning is a function of the first two columns. However, I’m trying to clarify the process–specifically Monty’s non-random decision about which door to open. You pick one door. Monty opens one of the remaining doors based on his knowledge, which connects to what is ultimately behind the door you switch to.

I’ll have to think if there is a clearer way to present that process information.

Rachel Williams says

March 19, 2019 at 1:39 amthat is still incorrect. You are deliberately ignoring different possibilities. Your percentages represent of percentages of total possibilities and you are deliberately two separate outcomes as one. Your very first column of your chart, for example, has “2 or 3” as the door that Monty opens. If you choose the correct door you cannot lump together whether or not he chooses one incorrect door or another incorrect door as a single possibility. They are two separate outcomes and when calculating a percentage of total outcomes, you have to be included for the sake of correct mathematics. I do not know how to express my thoughts in terms of probabilities but based on your explanation I would expect that the probability calculations are doing something similar and do not account for different outcomes when choosing the correct door, instead and correctly lumping them together because it is “irrelevant” which it most certainly is not.

Jim Frost says

March 19, 2019 at 10:50 amHi Rachel, I know the answer doesn’t appear to make sense at first glance, but it’s been the recognized solution for decades now.

As for the table, yes, Monty can do one or the other but not both. And, both actions produce the same outcome. So, the table is correct and consistent with the recognized solution.

Matthew Allen says

March 17, 2019 at 6:53 pmI think the key is that Monty’s turn disregards the door you have already chosen. This means that the probability of your choice being incorrect remains as 2/3. Monty’s turn doesn’t change this because he knows where one of the goats are and your chosen door wasn’t part of his selection.

To put it another way, if you lump together the two unselected doors (before Monty opens one), they have a combined probability of 2/3. The chances of of these two cards containing the prize are 2/3 regardless of Montys selection. This doesnt change because your selected door isnt included in his selection so there is still a 2/3 chance that the remaining door contains the prize.

Jim Frost says

March 17, 2019 at 10:07 pmHi Matthew,

That’s a great way to explain it!

Billy says

October 30, 2018 at 3:57 pmThat is just wrong. You are listing one outcome for 2 different outcomes. If he opens 2, that is not the same as opening 3. You are grouping them together, they are separate outcomes. The chance when having 2 doors is 50/50, and you cannot count the odds of the door already eliminated. This isn’t very hard.

Jim Frost says

October 30, 2018 at 9:01 pmHi Billy,

Back in the 1980s, there was some debate about the correct answer. However, over the intervening decades, the consensus has converged on 67/33 split as the correct answer. If you search around, you’ll see that there is no longer any debate about the correct answer. People have run computers simulations and have gotten this result. So, there’s no doubt that you have twice the chance of winning if you switch doors.

As I explain in the post, it’s easy to get caught up in the illusion that it’s 50/50 because you have two doors. However, there are underlying assumptions behind that outlook which are not being met. Namely, you’re assuming random, constant probabilities. Instead, Monty acts based on inside knowledge which changes the probabilities in a non-random manner. And, that’s why the results aren’t what you expect.

As for the cells in the table that have two outcomes. As I explain in the post, Monty can open either one and it does NOT change the outcome.

I will update this post at some point to include a computer simulation where I can run this experiment many thousands of times.

Dan Holgate says

August 22, 2018 at 5:09 pmOne way to grasp the concept of how Monty’s knowledge has affected the final choice is to multiply the number of doors. If there were 100 doors, and then Monty eliminated 98, you can see that unless your initial choice was the 1 out of 100 right door, then Monty has shown you where the prize is located.

Joe McCollum says

March 5, 2018 at 8:52 amI would say this problem is a mischaracterization of what Monty actually did. For the final deal, he always showed the least valuable prize first even if a contestant picked that door. Then he would show the middle prize, and finally the most valuable prize, even if nobody picked it.

For the regular part of the game, he would offer what the contestant turned down to other contestants. Then he would reveal the zonk – so the contestant could wind up with a zonk on the first selection.

Jim Frost says

March 5, 2018 at 10:02 amHi Joe, that may very well be the case. I did catch a few episodes of the show way back when. However, when people talk about the Monty Hall Problem, they’re historically referring to the scenario as described in this post. This scenario might be slightly different than the various ways he presented the information in the game show. It’s apparent you know how he worked better than I! Thanks for the information.

Jack2 says

January 23, 2018 at 8:27 pmYour table is missing lines. the “Pick door 1” section should read

1 1 2

1 1 3

1 2 3

1 3 2

There are 2 doors Monty can open if you happen to choose the right door the first time. Don’t know why you lump them together.

He can only choose 1 door if you choose the wrong door the first time

Jim Frost says

January 23, 2018 at 8:36 pmHi Jack, I explain the reason for this in the post. When you pick door 1 and the prize is behind door 1, Monty can pick either door 2 or 3. However, the outcome is the same. For example, if you don’t switch, you win. However, if I list both options on separate lines, that artificially inflates that outcome because it is listed twice. It’s the same one outcome for one scenario, so it’s listed once. You can see this for yourself. If you fill in the table as you show it, the probabilities don’t work out correctly.

Jack says

April 19, 2017 at 5:58 amMonty knows which door to open. So the choice of which door he opens is not probabilistic. That changes the probabilities. If he were to open a door at random, then your analysis would hold true, but he doesn’t, he removes a known empty door from the game.