The "Monty Hall Problem" in statistics comes from the old TV game show Let’s Make a Deal. In the game, Monty asks you to guess which door a prize is hidden behind. This problem is interesting because the solution is so counterintuitive that it's tripped up some of the world’s leading mathematicians! In fact, this problem's mindbending quality reminds me of the nature of optical illusions.
I’ve always been fascinated by optical illusions because some of them remain very compelling even after you know the truth about them. For example, even though I know square A is the same shade of grey as square B in the picture below, that still doesn't appear to be the case!
Optical illusions occur when the default algorithms your brain uses to process visual data produce results that don’t match reality. To see through the illusion, you have to carefully deconstruct the image so you can bypass your mental defaults.
The Monty Hall Problem holds the same fascination for me. It’s a statistical illusion that occurs because your default mental algorithms for assessing probability produce results that don’t match reality. As with an optical illusion, this illusory perception can feel more real than the true solution. In fact, many people want to reject the true answer to this problem when they first learn it.
Let's explore the Monty Hall Problem and deconstruct it like an optical illusion in order to reveal the truth. We'll also learn an important statistical lesson, too, because resolving this problem highlights the importance of verifying the assumptions behind any analysis!
The Monty Hall Problem
Monty Hall asks you to pick one of three doors. A prize is behind one door and no prize is behind the other two. You choose a door, but you can’t open it yet.
Monty then opens one of the remaining two doors, and there is no prize behind it.
At this point, two unopened doors are in play: the door you picked and the remaining door you didn’t pick. The prize is behind one of those two doors, but you don’t know which one.
Monty asks you, “Do you want to switch doors?”
The Decision: What Door Should You Choose?
If you’re like most people (including the noted mathematicians), you’ll assume that the prize is equally likely to be behind either door. Since it appears to be a 50/50 choice, most people stick with their original pick. They see no reason to change.
But now the counterintuitive nature of the problem raises it head! It turns out that the prize is not equally likely to be behind either of the remaining doors. In fact, if you switch, you’re twice as likely to win!
Huh?
The Solution
Marilyn vos Savant was asked this question in her “Ask Marilyn” column in Parade magazine. She answered that the player should switch in order to have a 2in3 chance of winning. The answer was so unbelievable that she received 10,000 letters saying that it was impossible! In fact, 1,000 letters came from readers with Ph.D.s! One of them was Paul Erdős, a noted mathematician who was convinced only after seeing the results of a computer simulation.
So, are you thinking that it’ll be difficult for me to show you the solution? Not really, I can do it in the small table below. The only math skill that you’ll need is the ability to count to 6!
There are not too many possible scenarios for all of the outcomes, so I can show you empirically what the Ph.D.s had a hard time figuring out theoretically! The table below shows every combination of choices and the outcomes for switching and not switching:
You Pick 
Prize Door 
Monty Opens 
Don’t Switch 
Switch 
1 
1 
2 or 3 
Win 
Lose 
1 
2 
3 
Lose 
Win 
1 
3 
2 
Lose 
Win 
2 
1 
3 
Lose 
Win 
2 
2 
1 or 3 
Win 
Lose 
2 
3 
1 
Lose 
Win 
3 
1 
2 
Lose 
Win 
3 
2 
1 
Lose 
Win 
3 
3 
1 or 2 
Win 
Lose 



3 Wins (33%) 
6 Wins (66%) 
In the table, the first row indicates that you picked door 1 and the prize is actually behind door one. Because the two remaining doors don’t have a prize, Monty can show you either of those two and the outcome is the same. If you don’t switch, you win. If you switch, you lose.
In the second row, you again pick door 1, but the prize is actually behind door 2. Monty can only show you door 3. Of the two doors that you didn’t pick, door 3 is the only one that doesn’t have the prize. If you don’t switch to door 2, you lose. If you switch, you win.
And so on; the table shows all of the possible scenarios. All we need to do is count up the number of wins for switching and for not switching. As you can see in the tally, you’re twice as likely to win if you switch!
The Reason Your Brain Hurts
After this empirical demonstration, I hope you’re on board with the idea that you are twice as likely to win if you switch doors. But why does it work out like this? The key to understanding this solution lies in understanding why our intuition is screaming—incorrectly—that the probability should be 50/50! We need to see through the statistical illusion.
We are used to probabilities based on independent, random events, such as flipping coins or rolling a die. In those cases, we simply divide the specific outcome by the total number of outcomes. For example, the probability of getting a head on the coin is 1/2, or 0.5. The probability of rolling a 1 on a die is 1/6, or 0.166. So, it seems logical that when you are down to two doors you have a 50/50 chance, rather than the true 33/66.
However, you must satisfy certain assumptions in order to use this procedure to calculate the probabilities. Specifically, the process must be random and the probabilities must not change.
The Monty Hall Problem violates both assumptions.
The random part
Your initial choice is the only random part of this process. You pick one of three doors, and you have a 0.33 probability of choosing the correct one. Look at the “Don’t Switch” column in the table and you’ll see that the table confirms this with wins 33% of the time. Random probability only applies up to this point.
The nonrandom part
However, once Monty starts using his insider knowledge of the prize's location, the process is no longer random and the probabilities are no longer constant. Monty knows where the prize is and, very selectively, he only opens 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. Here’s how it works.
Twothirds of the time, your initial choice is incorrect and the prize is behind one of the two doors you didn’t pick. Therefore, the probability of the following sequence is 0.66:
 By random chance, you pick the wrong door and the prize is behind one of the other two doors
 Of those two other doors, Monty is completely constrained in which door he can open because only one doesn’t have the prize
 He opens the door that doesn’t have the prize
 By the process of elimination, the prize is behind the door that he doesn’t show you!
Notice how the process is completely deterministic after you make your initial random choice? If you chose the incorrect door, the prize must be behind the unopened door that Monty offers you.
Because you’ll pick the wrong door twothirds of the time, the prize must be behind the door that Monty offers you twothirds of the time. This is borne out empirically by the “Switch” column in the table where the winning percentage is 66%!
Closing Thoughts
I hope the rationale for switching doors is more apparent after this explanation. The counterintuitive outcome happens because Monty uses his insider information to affect the results in a nonrandom manner.
And as promised, there is also a larger statistical lesson here. In the field of statistics, we often remind you to check the assumptions for the specific test or model you are using. This applies to everything from checking the residuals for a regression analysis to checking the distribution of your data. If you don’t satisfy the assumptions, your results may be incorrect.
In this case, our original mental model was one of independent, random events. However, that is an incorrect model for the Monty Hall Problem. Only the initial choice among the three doors follows this model. From that point on, the events are highly dependent and not random. And that’s why our intuition was wrong!
In statistics, checking the assumptions of the analysis is your key to escape the trap of illusory results! But, don’t feel bad...even Ph.D.s fall prey to this potential trap!
Time: Friday, January 11, 2013
this is awesome. i was talking about this same problems with my buddys the other day. I knew there was something more and that the actual probabilities were not 50/50. however, i couldnt remember what they were and how to go about to figure them out. I really like your table, good job!
Time: Saturday, January 12, 2013
Sorry, that table is incorrect in determining all possibilities.
You can't merely use the "2 or 3", "1 or 3", and "1 or 2" lumped together. You have to consider them separately.
For example, when you lump together the scenario where you have chosen the correct door, say door #1, you have to consider Monty's selection of opening door #2 or door #3 as separate situations; lumping them together is what causes a perception that switching increases the odds of winning.
That table correctly formulated for all situations would read:
Your selection...........Correct..........Reveal.............Keep.........Switch
..........1.........................1.....................2.................Win.............Lose
..........1.........................1.....................3.................Win.............Lose
..........1.........................2.....................3.................Lose...........Win
..........1.........................3.....................2.................Lose...........Win
And so on. The odds are still 50/50
Time: Tuesday, January 15, 2013
Hi DM,
In your table, the first 2 rows cover the one scenario where you pick door 1 and the prize is behind door 1. The Reveal column in your table indicates that for this one scenario Monty will open both door 2 and door 3. Clearly, that's not what he is going to do.
For this one scenario, Monty opens one door and there is one set of outcomes. That's why you must have one row per scenario. It's valid to use an OR to indicate this because the choice doesn't change the outcomes listed under Don't Switch and Switch.
Monty's reveal is contigent on your choice, it's not random. He'll never open a door that has the prize. Most of the time, that rule constrains him to one door, but sometimes he can open either. That's why some rows have OR and some don't.
It's a tricky problem. However, my table matches the results of both computer simulations AND theoretical proofs.
Personally, I think the most concrete way to grasp this is to understand the following:
1) If you pick the wrong the door initially, the prize is definitely behind the door that Monty offers you.
2) Because you'll pick the wrong door twothirds of the time, the prize is behind the door that Monty offers you twothirds of the time.
Thanks for reading!
Jim
Time: Tuesday, January 22, 2013
I use this twist on the problem to try and help people work out the right answer for themselves:
You and a friend are both invited to take part in a quiz show. There are three doors, behind one of them is a car. The host of the show asks you to discuss with your friend and agree on a door you want to choose (for completeness, you are trying to win the car). Having agreed between you which door you want, you tell the host your joint choice. In order to increase the excitement the host, who knows which door the car is behind, opens one of the other doors to reveal an empty doorway. He then asks each of you in turn whether you want to stick with the original choice or change your mind and switch to the other remaining door...at this point you and your friend can each make different decisions.
You get invited to play the game 100 times. In each game, you always decide to stick to the original door jointly selected, whereas your friend always decides to switch to the other door. On average, how many times would each of you win the car?
Time: Friday, February 15, 2013
Hi DM,
Your table is tempting, but there is an obvious flaw. It predicts that the contestant has a 50% chance of choosing the winning door, because your table lists them choosing door 1 twice. Surely it's got to be 1/3, if they have only one free choice out of 3 doors.
Best wishes, Brian
Time: Tuesday, February 19, 2013
I have explained this by imagining a scenario with 100 doors. You pick a door and then the host opens 98 more and then asks if you want to switch to the other one still closed. Nearly everyone recognizes in this scenario that the odds are not 5050, but heavily weighted in favor of the other door (you chances of picking the right door to start with were 1 in 100). Once you see this it is a natural extension to understand how the noswitch strategy would lead to a one in three chance when there are only three doors.
Time: Tuesday, February 19, 2013
I do think it's important to specify that Monty will *always* offer you the choice to switch doors after he reveals one of them to be empty. In the original show (which I am old enough to remember), he did not always offer that choice.
Time: Wednesday, February 20, 2013
I've heard the same thing about the game show Deal or No Deal. When you pick your original suitcase (out of 26) and slowly eliminate the others one by one, at the very end when there are just two, you are given the option to switch. You should, as the article explained, because if you switch the odds are 50/50. If not, the odds are slightly less, I don't know the probability. Minitab problem anyone?
Time: Thursday, February 20, 2014
In Deal or No Deal it makes no difference whether you stay or switch when you're down to 2 cases  it's 50/50 for each case.
One point regarding the article: the initial choice isn't the only random part of the game. Given a choice of 2 goat doors to open, in order for the conditional probabilities for the 2 remaining doors to be 1/3 and 2/3 the host must pick a door at random
Time: Thursday, February 20, 2014
RE: Palmer Eldritch  nice! Always happy to see a PKD reference!
Time: Thursday, February 20, 2014
Hi PalmerEldritch,
Thanks for writing! This is a tricky puzzle!
If you initially choose a nonprize door, the host has no choice in which door he can choose. He must choose the one nonprize door to open.
If you initially choose the prize door, the host can randomly choose either of the 2 remaining doors. However, neither door has the prize so his random choice doesn't change the outcome.
Remember, it's the host's knowledge of where the prize is located, and how that knowledge affects his door choice in a nonrandom manner, that ultimately skews the probabilities.
Jim
Time: Thursday, April 3, 2014
Hi Jim,
Thanks for your response. As you acknowledge, if you initially picked the door, the host makes a random choice between the 2 remaining doors.
If the host doesn't choose at random (when you initially pick the car) then you can get some interesting results.
Say the host will always open the highest number door that contains a goat. In this scenario if you pick Door1 and the host opens Doors3, then it's 50/50 whether the car is behind Door1 or Door2 (of course if the host opened Door2 instead, then it's 100% certain the car is behind Door3).
Overall, over multiple trials, you'll still win 2 times out of 3 on average, but in any individual game the probabilities aren't 1/3 and 2/3 for the 2 remaining doors.
Time: Thursday, April 3, 2014
Hi,
I understand what you're saying. It's true that the probabilities vary depending on your initial choice.
So, let's figure out the probabilities assuming that you will take Monty up on his offer to switch.
If your initial door choice is the one with the prize, the probability of you winning is zero.
If your initial door choice is not the one with prize, your probability of winning is 100%.
The reason it works out to an average probability of 2/3 is because twothirds of the time your initial door choice won't be the one with prize. So, twothirds of the time your probability is 100% and onethird of the time it is zero. Hence, the overall probability of winning is 66.6%!
Thanks for taking the time write!
Jim