The Monty Hall problem is a counter-intuitive statistics puzzle:

- There are 3 doors, behind which are two goats and a car.
- You pick a door (call it door A). You’re hoping for the car of course.
- Monty Hall, the game show host, examines the other doors (B & C) and always opens one of them with a goat (Both doors might have goats; he’ll randomly pick one to open)

Here’s the game: Do you stick with door A (original guess) or switch to the other unopened door? Does it matter?

Surprisingly, the odds aren’t 50-50. If you switch doors you’ll win 2/3 of the time!

Today let’s get an intuition for *why* a simple game could be so baffling. The game is really about re-evaluating your decisions as new information emerges.

## Play the game

You’re probably muttering that two doors mean it’s a 50-50 chance. Ok bub, let’s play the game:

Try playing the game 50 times, using a “pick and hold” strategy. Just pick door 1 (or 2, or 3) and keep clicking. Click click click. Look at your percent win rate. You’ll see it settle around 1/3.

Now reset and play it 20 times, using a “pick and switch” approach. Pick a door, Monty reveals a goat (grey door), and you switch to the other. Look at your win rate. Is it above 50% Is it closer to 60%? To 66%?

There’s a chance the stay-and-hold strategy does decent on a small number of trials (under 20 or so). If you had a coin, how many flips would you need to convince yourself it was fair? You might get 2 heads in a row and think it was rigged. Just play the game a few dozen times to even it out and reduce the noise.

## Understanding Why Switching Works

That’s the hard (but convincing) way of realizing switching works. Here’s an easier way:

**If I pick a door and hold, I have a 1/3 chance of winning.**

My first guess is 1 in 3 — there are 3 random options, right?

If I rigidly stick with my first choice no matter what, I can’t improve my chances. Monty could add 50 doors, blow the other ones up, do a voodoo rain dance — it doesn’t matter. The best I can do with my original choice is 1 in 3. The other door must have the rest of the chances, or 2/3.

The explanation may make sense, but doesn’t explain *why* the odds “get better” on the other side. (Several readers have left their own explanations in the comments — try them out if the 1/3 stay vs 2/3 switch doesn’t click).

## Understanding The Game Filter

Let’s see why removing doors makes switching attractive. Instead of the regular game, imagine this variant:

- There are 100 doors to pick from in the beginning
- You pick one door
- Monty looks at the 99 others, finds the goats, and opens all but 1

Do you stick with your original door (1/100), or the other door, which was filtered from 99? (Try this in the simulator game; use 10 doors instead of 100).

It’s a bit clearer: Monty is taking a set of 99 choices and *improving* them by removing 98 goats. When he’s done, he has the top door out of 99 for you to pick.

Your decision: Do you want a *random* door out of 100 (initial guess) or the *best* door out of 99? Said another way, do you want 1 random chance or the best of 99 random chances?

We’re starting to see why Monty’s actions help us. He’s letting us choose between a generic, random choice and a *curated, filtered* choice. Filtered is better.

But… but… shouldn’t two choices mean a 50-50 chance?

## Overcoming Our Misconceptions

Assuming that “two choices means 50-50 chances” is our biggest hurdle.

Yes, two choices are equally likely when you know *nothing* about either choice. If I picked two random Japanese pitchers and asked “Who is ranked higher?” you’d have no guess. You pick the name that sounds cooler, and 50-50 is the best you can do. You know nothing about the situation.

Now, let’s say Pitcher A is a rookie, never been tested, and Pitcher B won the “Most Valuable Player” award the last 10 years in a row. Would this change your guess? Sure thing: you’ll pick Pitcher B (with near-certainty). Your uninformed friend would still call it a 50-50 situation.

Information matters.

## The more you know…

Here’s the general idea: **The more you know, the better your decision.**

With the Japanese baseball players, you know more than your friend and have better chances. Yes, yes, there’s a *chance* the new rookie is the best player in the league, but we’re talking *probabilities* here. The more you test the old standard, the less likely the new choice beats it.

This is what happens with the 100 door game. Your first pick is a random door (1/100) and your other choice is the champion that beat out 99 other doors (aka the MVP of the league). The odds are the champ is better than the new door, too.

## Visualizing the probability cloud

Here’s how I visualize the filtering process. At the start, every door has an equal chance — I imagine a pale green cloud, evenly distributed among all the doors.

As Monty starts removing the bad candidates (in the 99 you didn’t pick), he “pushes” the cloud away from the bad doors to the good ones on that side. On and on it goes — and the remaining doors get a brighter green cloud.

After all the filtering, there’s your original door (still with a pale green cloud) and the “Champ Door” glowing nuclear green, containing the probabilities of the 98 doors.

Here’s the key: Monty does not try to improve your door!

He is purposefully *not* examining your door and trying to get rid of the goats there. No, he is only “pulling the weeds” out of the neighbor’s lawn, not yours.

## Generalizing the game

The general principle is to re-evaluate probabilities as new information is added. For example:

A Bayesian Filter improves as it gets more information about whether messages are spam or not. You don’t want to stay static with your initial training set of data.

Evaluating theories. Without any evidence, two theories are equally likely. As you gather additional evidence (and run more trials) you can increase your confidence interval that theory A or B is correct. One aspect of statistics is determining “how much” information is needed to have confiidence in a theory.

These are general cases, but the message is clear: more information means you re-evaluate your choices. The fatal flaw of the Monty Hall paradox is *not taking Monty’s filtering into account*, thinking the chances are the same before and after he filters the other doors.

## Summary

Here’s the key points to understanding the Monty Hall puzzle:

- Two choices are 50-50 when you know nothing about them
- Monty helps us by “filtering” the bad choices on the other side. It’s a choice of a random guess and the “Champ door” that’s the best on the other side.
- In general, more information means you re-evaluate your choices.

The fatal flaw in the Monty Hall paradox is not taking Monty’s filtering into account, thinking the chances are the same before and after. But the goal isn’t to understand this puzzle — it’s to realize how subsequent actions & information challenge previous decisions. Happy math.

## Appendix

Let’s think about other scenarios to cement our understanding:

**Your buddy makes a guess**

Suppose your friend walks into the game after you’ve picked a door and Monty has revealed a goat — but he *doesn’t* know the reasoning that Monty used.

He sees two doors and is told to pick one: he has a 50-50 chance! He doesn’t know why one door or the other should be better (but you do). The main confusion is that we think we’re like our buddy — we forget (or don’t realize) the impact of Monty’s filtering.

**Monty goes wild**

Monty reveals the goat, and then has a seizure. He closes the door and mixes all the prizes, including your door. Does switching help?

No. Monty started to filter but never completed it — you have 3 random choices, just like in the beginning.

**Multiple Monty**

Monty gives you 6 doors: you pick 1, and he divides the 5 others into a group of 2 and 3. He then removes goats until each group has 1 door remaining. What do you switch to?

The group that originally had 3. It has 3 doors “collapsed” into 1, for 3/6 = 50% chance. Your original guess has 1/6 (16%), and the group that had 2 has a 2/6 = 33% of being right.

## Leave a Reply

1149 Comments on "Understanding the Monty Hall Problem"

You could explain it like this too:

If you stay with the door you picked initially you succeed if the initial door has a car, which has a chance of 1/3. If you’re strategy is to switch then you succeed if your initial pick is a goat, which has a chance of 2/3.

You have a 2/3 chance of initially picking a goat. Since it is more likely you initially picked a goat it makes sense to switch.

This is all I needed to hear for it to make sense. Thanks!

Fantastic summation. This explanation helped me immensely.

All that I just read and although interesting still left me puzzled, this explanation was a lot clearer and removed the confusion, thanks.

These two sentences were much more clear than pages upon pages of other verbose and intricate explanations, thank you. E.B. White would be proud.

The combinations of your first choice are:

Car

Goat(1)

Goat(2)

That’s a 1/3 chance of winning the Car – as you would expect.

After you make your first choice the combinations behind the remaining two doors are:

Goat(1) & Goat(2)

Car & Goat(2)

Car & Goat(1)

If Monty Always picks a Goat then the remaining door contains:

Goat (1 or 2)

Car

Car

Hence if you switch there is a 2/3 chance of picking the Car.

Best concise and clearly laid out explanation. Better than that 10 pages-long contrived and non-convincing explanation above.

I never understood this before, but as I read your explanations, the following came to mind:

The doors can be divided into two groups; the one you picked, and the ones you didn’t pick. When you pick one out of 100 doors, chances are very high that the prize is in the group you didn’t pick, you just don’t know which of those 99 doors has the car.

Luckily for you, Monty narrows it down for you by opening all but one door. The odds of the selected door having the prize hasn’t changed at all. You just know more a lot about the doors you didn’t pick.

This is how I thought about it: imagine that door A is the one you pick. There are 3 equally likely scenarios. 1: Door A is correct. In this case, you would be better off staying. 2: Door B is correct. Monty would open C and you would be better off switching. 3: Door C is correct, Monty would open B and you would be better off switching. Two out of three times, you are better off switching.

Well, that’s correct up until he opens one of the doors. Then that choice goes out the window and we are left with only 2 choices

The problem with your explanation is this: the car was originally randomly placed into one door of the three doors; it wasn’t randomly placed into one of the two remaining doors (one of which being your pick).

So, your choice after Monty eliminates a door is between a 1/3 chance and a 2/3 chance.

Initially we make a random choice with a 33% probability that the car is behind the door. The other two doors contain 66% of the probability. Once the goat is revealed the 66% probability is condensed into a single door. No extra probability is handed to the first door until a new random choice is made. Thus if no new random choice is made the original choice maintains its 33% probability but the other door now has all of the 66% probability of its original set. That’s why switching to the other door wins 66% of the time. If a new random choice is made of those two remaining doors, then it’s truly a 50% chance.

I think this is overly wordy. What helped me is:

if you picked the door with a goat, the other doors have a car and a goat. Monty then CAN’T OPEN the door with the car because that would ruin the game.

the probability that you picked the right door out of 3 is 1/3. so you have a 2/3 of having picked the WRONG door, at which juncture, Monty opens the door with the goat. that means the other door has a 2/3 chance, and your door has a 1/3 chance.

I had to read ovet 20 comments to understand it, the second sentence of your explination made it click, thanks.

Ridiculously complicated. The probability the car is not behind ‘my’ door is 2/3. And now there’s only one door to pick. Change doors!

The way I visualize this problem is with an adversarial scenario. First I imagine it’s 100 doors, then I think of it this way:

Imagine you’re playing AGAINST Monty Hall for a car. You each must pick a door. You go first. You are forced to pick your door at random.

Monty Hall, however, has the luxury of looking at every door other than yours, and not only can but MUST pick the car door if he finds it.

He has a wide grin on his face (99 out of 100 times) and is already celebrating his new acquisition when all of a sudden you are offered to trade places with Monty and get to pick his door! Do you switch then? I think so!

I tend to a view that a two-in-three approach is clearest, combined with the fact that the host KNOWS which is the prize window, but will never reveal it.

If there are two windows NOT the prize, but there is only one window THE PRIZE, then the first guess is two-in-three likely NOT to be the prize.

If the first guess is NOT the prize, then the host, who knows which IS the prize, but will never reveal it, has only one option of window to reveal, the second window ALSO not the prize.

If you switch, you win.

That is two-in-three likely.

On one occasion in three, the first guess will BE the prize, and then switching will lose.

You are twice as likely, but not guaranteed, to win if you switch.

If you stick, you have straight one-in-three chance of landing the prize.

That doesn’t make any sense. If that is a part of the game, the host revealing one of them for you, how is this exactly improving your chances by switching on over?

If you have a 2/3 chance of choosing the wrong door, than you probably will, when he gets rid of one wrong door you still most likely have a wrong door, so if there are no more wrong doors to pick than you would switch to the probably right one.

Yes. This is the explanation that I understood best.

so if two people play the game at the same time – one starts with door number 1 the other guy with door number 3. Monty opens door number 2 -> Goat. Both guys now go on and swap their picks – both have a higher chance to win now? The odds change on paper cause one variable has been taken out of the equation in reality it changes nothing.

you’re assuming randomness. It’s not a random selection. If it were then Monty would randomly remove a door (but that means he only has the same 1/3 chance of removing a goat)

If he randomly removed a goat then its is 1/2 for the car, cause the odds changed for both players. He’d have a 50/50 for a car in the remaining door as, as do you.

Monty does however, already own the other 2 doors. His chances were always 2/3 because its not random and he was allowed to have both doors (opting to remove the goat)

So the puzzle is saying, hey when you started, you got 1/3 and Monty gets 2/3. His odds were always better. So you’d always take his chances over yours- which is why you switch

Try working it backwards: first, actually play the game and log what happens. Do it until you’re convinced it actually works, i.e. that if you stick, you’ll only win 1/3 of the time, but if you switch, you’ll win 2/3 of the time. Once you see that that’s fact (it is), then try to understand why. You can use everybody’s explanations but you can also make yourself understand it in whichever way works for you.

The thing that’s certain is that it’s actually true that switching will double your odds of winning (from 1-in-3 to 2-in-3)… anyone contesting that simply doesn’t want to be convinced, because you can prove it to yourself simply by playing the game a bunch of times and logging the results, as I said.

You should read the problem carefully before providing such flawed opinions. The original problem clearly states that only after you choose a door does Monty examine the two remaining doors knowing he is sure to find at least one goat for him to reveal to you. From your statement, if two people play at the same time and both of them choose different doors, Monty could not examine but one door, making it possible for him to find a car in there and not be able to open the door, letting you both know where the car is. So no, your example is way off base.

You are confusing results with probability. Yes they both would be increasing their odds of winning but you don’t think they are since one person will still get the goat. Think of your same problem with a 100 doors. They each pick one and then the other 97 doors are opened. Should they both switch? Yes they should as it increases theirs odds.

You are also leaving out the situation where each person picks a door with the goat and he can’t open the door. If they both realize this then they would both win. which shows the situation where both peoples odds were increased and both win.

@lewikee: Awesome, I like that formulation! Monty is basically picking the best door he can out of the remaining choices. Very nice ;).

Actually Monty does you a confusing (to most people) disservice by throwing one away early. If he left the unpicked two doors closed and said you could open them simultaneously, and if you see a car … you win. Almost all would then trade for the two doors.

Thank you Don, now it’s finally clicked. When initially trying to understand this problem, I was so focused on the red herring of Monty throwing away door that I failed to see the basic logic as you so clearly explained.

This article was so long and drawn out and seemed to apply a somewhat unrelated rationale for conclusion (more information allows you to make a more informed choice, hence increasing probability of decision). Whilst that is true, I don’t think it’s necessary the key to understanding this particular problem.

Best explanation yet!

Brilliant. That, right there, says it all.

I’ve read a bunch of articles about this, but the multiple doors made it click and made me think of it in a way no one has mentioned.

The reality is that 66.7% of the time, I have picked a goat. This means that 66.7% of the time, Monty still has the car and the other goat. So, 66.7% of the time, Monty is revealing the 2nd goat I did not pick.

Another way that came to me is that there are only 2 versions of this game.

1) I have picked the car and Monty is showing me one of the two goats.

How to win: stay with my first choice because it is the car.

2) I have picked a goat and Monty is showing me the other goat that I did not pick.

How to win: change my choice because the car is behind the 3rd door.

I don’t know which version of the game I’m playing, but I DO know that “version 1” only happens 33.3% of the time, while “version 2” happens 66.7% of the time. Since this is true, I am going to use the winning strategy that applies to 66.7% of the games I play. I will win about 66.7% of the time.

Welcome back!

I actually blogged about this a while back:

http://blog.amhill.net/2009/07/24/from-the-archives-monty-hall/

And one of my uncles was dead-sure that I was wrong about it, so I wrote up an application (source code included) to show the percentage differences between switching and not switching. It’s linked to from that blag post above.

I like your flash version though!

You should stick to the history of American Civil War and probably leave mathematical puzzles alone.

There are many explanations as to why switching doors increases your chances of winning to 2/3 – most of them are presented within the comments here (which either you didn’t read or didn’t understand) Perhaps the simplest is this:

“IF your first pick is a goat (which you know is a 2/3 chance) THEN you are guaranteed to win the car by switching doors”.

Everything hinges on making a random choice. Those that argue it is a 50% chance problem after the goat is revealed fail to see where the original probability of 33% came from. It came from making a random choice. If there was no random choice initially then there’d be no 33% chance. If the contestant was told what to pick, they’d claim their chances were not even. So after the goat is revealed, the only way to have a 50% chance is to flip a coin and make another random from the two remaining doors. But since 66% chance of a win was held in the other two doors and condensed from those 2 doors into 1, it makes sense to just pick that other door that has the 66% chance and NOT make any more random choices.

The way I explain to friends is this:

Let’s make it a iron-clad rule that I switch, after Monty opened another door to reveal a Goat. And of course, the initial choice gives me a 1/3 chance of picking the Car, and a 2/3 chance of picking a Goat.

This means if I picked the Car at first, I will invariably lose it, as the switched choice will invariably be a Goat. And the opposite is true: if I picked a Goat at first, I will invariably switch to the winning door with the Car behind it. In other words, if I picked right at the beginning I lose, and vice versa.

Don’t forget: the odds of the initial choice remains the same, but would would just reverse your outcome. So, at first I had a 1/3 chance of picking the Car but I am bound to lose it, but at the same time, I had a 2/3 chance of picking a Goat which means I will get the Car.

If I stay with the initial choice, Monty opening another door would not change anything: I would still be stuck with an unfavorable odds of 1/3 of picking the car.

Great article – I especially like the visualization part with the green clouds. Sometimes I think the hardest part with mathematical concepts is being able to visualize them.

Yes, great post. After I had time to think about this problem a bit, I thought about it similar to Aaron/uwe. But I use slightly different wording: 1st, I pick a door (1 out of 3). Then, I’m given the opportunity to either 1) keep my pick or 2) choose the other 2 doors (i.e., knowing that Monty would take one away for me later so to speak). Thanks for the articles, they are generally my favorite in the blogosphere.

I understand the problem perfectly. Thinking of a pack of cards, there’s a one in fifty two chance that you have picked the Ace of Spades. It’s a much higher probability that the Ace of Spades will be in the remaining fifty one cards, whether or not fifty of those cards are revealed to you before you choose to swap. Although in swapping, you are not guarantee a win, statistically, it will yield better results over repeated tries. So it is with three cards, or three doors.

It took me so long to shed the 50/50 notion though… My simply refused to see that swapping would be better!

Typo fix: 2nd to last para. above: First part should read “1/3 divided by 2/3 is 1/2. The choice is constrained to only two doors before you’ve even started, because “door number three” isn’t the door LABELLED “Door 3.” It’s whichever door Monty wants it to be..which is one of the two with the goat…”

Point being: If Monty is always going to remove from the “universe” 1 of the 2 goats, then that itself defines the condition under which the game is played. That’s your sandbox and you can’t get out of it.

@ZenasDad: If the chance were 50/50 for a new player with your same card choices… The chance *must* be 50/50 for you as well.

faulty logic. here’s how it works for the 3 doors:

player 1 picks door 1. 1/3rd chance of being correct. there is a 2/3 chance that the car is actually behind EITHER door 2 or door 3. monty then eliminates the least-valuable door from 2 and 3. call that remaining door “monty’s door”. “Monty’s door” has the car in every game where the player’s door does not – 2/3 chances. That’s because player 1 just has “door 1” whereas monty has “whichever door from 2 or 3 holds the car” – the statement “door 2 or door 3 hold the car” is true in 2/3 games.

Ok, now consider a “player 2” who just came in. he chooses at random as he doesn’t know which door player 1 picked, and which door Monty left. let’s assume the 1/3 vs 2/3 chance is correct:

1/3 chance door “A” is the winner

2/3 chance door “B” is the winner.

if player 2 picks door “A” he wins 1/3 * 1/2 = 1/6 games

if player 2 picks door “B” he wins 2/3 * 1/2 = 2/6 games

1/6 + 2/6 = 1/2 chance of winning.

Let’s do it for a 1/10 to 9/10 chance for the two doors:

1/10 chance door “A” is the winner

9/10 chance door “B” is the winner.

if player 2 picks door “A” he wins 1/10 * 1/2 = 1/20 games

if player 2 picks door “B” he wins 9/10 * 1/2 = 9/20 games

1/20 + 9/20 = 1/2 chance of winning.

So, you see the relative chance of the car being behind either door doesn’t benefit someone who doesn’t know which door is more likely. But if you DO know the odds per door, you can win that often.

in the classic Monty Hall problem the player knows which door he chose first, thus he knows that that specific door had a 1/3 chance since he picked it at random from a group of 3. Since Monty always leaves a winning door unopened, the door he didn’t open must have a 1 – 1/3 = 2/3 chance of being the correct one, since probabilities always add up to 1.

But, someone who doesn’t know which is monty’s door vs the player’s door has no better than blind luck in choosing.

Monty is making you a good offer to switch. He confuses the issue by opening a door. There will always be a goat behind at least one of the two doors you did not choose. You should know that and not be bothered when Monty shows you.

Your original choice is a set of one door. Switching allows you to chose a set of two doors. It is a set of two doors even if when you know what is behind one. It is a better choice because it is two doors.

A good deal is staring you in the face.

902. “As I have noted before to no response, in the simulation, if the computer opens the right-most of the two doors there is a 100% chance that switching will win …”

Yes, there is a shortcoming in the simulation if as you say the computer prefers the leftmost door. The MHP contestant has the information in the MHP question to go on, and no more, which gives them no indication of the host’s preference or whether he has any.

While I do not believe you have, with respect, explained the point as plainly as might be possible (“tags”?), you’re on to something Pat, as I have noted several times above, most recently at 902.

To be clear, the location of goats is determined when all doors are closed.

The anomaly I have observed as I describe it is that

(a) if you pick the correct door on first guess;

(b) the “Host” always opens the left-most of the two remaining doors.

This means that whenever the Host opens the right-most of the two remaining door, a switch will be successful 100% of the time.

This is caused by the Host’s undiagnosed OCD IRT the order in which doors should be opened.

The treatment is some random-number-generating code that varies the door to be selected when the contestant has picked the car on first selection, n’est ce pas?

905/6 The simulation does demonstrate 2/3 by switching, however it’s a case of ‘right answer, wrong working’.

The simulation violates the assumption that the host has no ulterior motive, and thus opens a random goat when there are two, though does not necessarily violate the condition that the contestant is unaware of any such motive. The instructions partly replicate that condition by suggesting play-and-hold fifty times and then play-and-switch twenty times (which prevents the contestant selectively switching), then to vary the strategy over a few dozen more (which, if they spot the pattern, does not). As stated, the simulation reveal should really have two goats randomised not leftmost.

However, as pointed out, in the simulation as it is switching still achieves 2/3, never more, as the opportunity for 100% (rightmost goat) occurs 1/3 of the time and is balanced by the greater frequency of the 50% chance (leftmost goat) 2/3 of the time.

This is a case of too much information.

Let’s say you pick a door – you now have a 1 in 3 chance to win.

Let’s now say that Monty got drunk, forgot what to do and instead tells you you can trade your 1 door for both of the other 2.

Would you do it? . . . Of, course you would. Because your chance to get the car would double.

Monty Hall opening a door only confuses a very simple math problem – because it’s irrelevant.

2>1 is all you need to know.

@Lump

“Let’s now say that Monty got drunk, forgot what to do and instead tells you you can trade your 1 door for both of the other 2.”

Except of course that doesn’t happen, a useful analogy maybe, an accurate description of the problem, no..

Let’s now say Monty got drunk and forgot which door hid the car, picked one of the 2 doors at random , and luckily a goat was revealed.

Would you switch? The answer is it makes no difference whether you do or not, you’re equally likely to win the car by staying

It’s possible to invent all sorts of imaginary scenarios that don’t reflect what actually happens. I don’t see the point

No, sorry. You are wrong this time although you were correct in an earlier comment some time ago. In effect you are being asked if you want to trade your original 1 door in exchange for the other 2 doors -which of course you would do. You know as a certainty that one of those 2 doors is a goat so Monty actually showing you a goat in a door gives you no new information whatsoever. Whatever Monty does is of no relevance and you should always switch.

@BobUSA

I find it odd that so many people choose to interpret the question “Do you stick with door 1 or switch to door 3?” as “Do you stick with door 1 or switch to both door 2 and door 3?” just because both questions give the same answer.

The statement “Monty actually showing you a goat in a door gives you no new information whatsoever.” is incorrect. If Monty didn’t open door 2 there would be no advantage in switching to door 3

Here is a short explanation:

Each door has 1/3 chance of success.

You pick a random one…with 1/3 chance of success.

The other two COMBINED AS A GROUP hold 2/3 chance of containing a success.

Removing one unsuccessful door from that group leaves one door that STILL has 2/3 chance of success.

Obviously it makes sense to switch to that door, instead of sticking with the first.

@Joe: I was thinking in your track as well. I will try to describe my understanding as follow:

Say, Finding a car behind door 1 is event (1), behind door 2 is event (2) and behind door 3 is event (3).

Following statements describe the steps.

(A) Chance to find a car behind ANY ONE of the THREE doors is P(1)=P(2) =P(3) =1/3.

Then from (A) it follows that the chance of finding the car behind ANY TWO of the THREE doors is (1 minus 1/3) =2/3

(B) The player chooses door 1, not knowing what’s behind. Chance for 1 is still 1/3.

(C) Since finding the car behind door 2 or 3 are mutually exclusive, the possibility that the car is EITHER behind door (2) or (3) is P[(2) OR (3)], in other words,

P[(2)U(3)] =2/3. = P(2) + P(3)……………..(Equation X) by using reasoning from statement (A).

where, OR =U =symbol for Union Set.

(D) For our example, assume Monty opened door 2 to reveal the goat.

(E) So, form “D” we know that P(2) =0

(F) Inputting value of P(0) in Equation 1, we get P[(2)U(3)]=0+P(3) =2/3.

So, P(3) =2/3.

(G) P(1)<P(3). So, switching makes sense.

Correction: In statement (A) I meant to say “Chance of finding the car behind ONE of THE ANY TWO doors from a SET OF THREE doors” is (1 minus 1/3) =2/3