**Update:** This article was originally posted in May 2010. We’ve just learned that Canadian-American game show host Monty Hall died last Saturday at the age of 96. Read more about the puzzle that was named after him below.

It’s a fact: a lot of people get confused over probability problems. For example, most geek reading this site will probably realize that if a coin is tossed ninety-nine times and comes up heads every time, then in probability terms, a hundredth toss still has an equal shot of coming up heads or tails. Those who think otherwise are either misreading the question (looking at the probability of the hundred-head streak, not the individual toss) or are sensible enough to take a practical approach (if a real coin comes up heads ninety-nine straight times, it’s probably weighted.)

But other probability issues cause much more headscratching. One of the most famous is the so-called Monty Hall problem which is written in several different ways but follows these principles:

- There are three doors with a car and two goats placed behind them at random. The game show host knows which is placed where.
- You must start off by choosing a door.
- The game show host opens one of the two doors which you did not choose, revealing a goat. (He or she will always open a door that will reveal a goat. He will never open a door which will reveal the car.)
- The host then offers you the chance to change your original pick.

The question is whether it is better to change or stick with your original choice. The answer — which can be and regularly has been demonstrated by running the scenario over and over — is that you are more likely to win if you change. But many, if not most people simply can’t process this and insist that it cannot make any difference whether or not you switch and that your chances of winning are the same either way.

What fascinates me is that the real puzzle isn’t what the correct answer is, but rather how to explain it in a way that people can understand why their initial belief is incorrect. As an exercise in both linguistic and numerical persuasion, I suspect it may be that different explanations will be more effective for different people. Here’s a few of my attempts:

**Explanation one:**

(This is the most basic explanation, but even when people read and accept it, they sometimes still can’t overcome their original “logic.”)

There are three equal possibilities. For example, if you originally pick A:

1) It’s in A (sticking wins)

2) It’s in B, so C is revealed (switching wins)

3) It’s in C, so B is revealed (switching wins)

So switching is more likely to win.

**Explanation two:**

One way to understand it is to realize that because the host opens one of the doors you don’t pick, you can treat both of the two doors you don’t pick as effectively being one singular option.

So say that the doors are A, B and C and you pick A. There’s a 1 in 3 chance that the car’s behind door A and a 2/3 chance that it’s behind either door B or C.

What you then have to take into account is that if the car isn’t behind A, it doesn’t matter whether it’s behind B or C: If it’s behind B, the host reveals the goat behind C. If it’s behind C, the host reveals the goat behind B. Either way, it’s always been more likely that the car is behind B/C than A. Your first guess will always probably be wrong, which is why you are better off switching once you no longer need to worry about having to choose between B and C.

**Explanation three:**

Even though it’s a mathematical problem, it can be better understood by looking at the situation in a completely literal and practical manner. What is physically behind the doors never changes. That’s why you can’t apply mathematical “logic” after the reveal and call it a 50-50 chance. The prize goes behind one door at the start. Either it’s behind the door you choose first, or it isn’t. What happens with the reveal doesn’t physically change that by making it more or less likely.

Thinking it’s a 50-50 chance is taking a logical approach to a *different* situation. If you were to put a prize behind one of two doors, it would indeed give a 50-50 chance of being behind either door. But that hasn’t happened here. The prize is already in position *before* you come to the point where two boxes are left.

To say the same thing a different way: Probability relates to random events, not to states. The random event in this situation is the placing of the car and goats. Selecting a door to open, whether that be by the contestant or the host, has no bearing on this event.

**Explanation four:**

Regardless of the wording he or she uses, the two questions the host asks you are actually very different. The original question is to guess the correct door, which is difficult (1 in 3 chance). The question after the reveal shouldn’t be thought of as a repetition “What door is correct?” but rather “Was your original guess wrong?” And as we’ve just established, the chances are it was.

So, are you any clearer now? Or did you understand it already? And if so, have you got a clearer way to explain it?

I will be the brave soul and admit that I've always had a difficulty with this problem. I've consoled myself by believing that the people trying to explain it are misstating the situation or question. With issues such as this, the precise language is very important.

You've done the best job I've seen yet. Explanation two makes the most sense to me.

What I've found most difficult to accept is the idea that a series of subsequent actions that should be meaningless—I mean, no matter what, they happen in the same way—will always result in you changing your choice.

When you make your initial choice, you already know for a certainty that at least one of the other doors will have a goat, and you know that Monty will show it to you. You already have that information at the start. How can his follow-through affect the assessment of your original decision? Further, if you had chosen B, Monty might still have shown you C, and then we'd say that you should switch to A??? It almost seems like when you go up on stage and you're presented with three doors, each one has a 50/50 chance of being the right one. So not that sticking with your first choice gives you better odds, but neither should switching.

"When you make your initial choice, you already know for a certainty that at least one of the other doors will have a goat, and you know that Monty will show it to you. You already have that information at the start. How can his follow-through affect the assessment of your original decision?"

The point is that his follow-through CAN'T affect the assessment of your original decision. The probability that your original choice was right STAYS at 1/3. Having chosen A, what you don't know is whether the host will open door B or door C. That new information (which door is opened) changes the probabilities for those door (the one opened goes from 1/3 to 0 and the other goes from 1/3 to 2/3).

If I may add in regards to Explanation One…

Aren't there are more than three equal possibilities? If you choose A:

1. It's in A and Monty shows you B, sticking is correct.

2. It's in A and Monty shows you C, sticking is correct.

3. It's in B and Monty shows you C, switching is correct.

4. It's in C and Monty shows you B, switching is correct.

We take just winning scenarios, so if you've already chosen right door, host's revealing doesn't influence anything

Your 4 possibilities are not equally likely. The prize was placed randomly, so the probability that it is in A is 1/3. If the host picks randomly when he has the choice of 2 goats, your first 2 possibilities have each a probability 1/6. So, 1 & 2 each are 1/6 and 3 & 4 are each 1/3.

If this argument worked, then the following situation would be in your favor…

You flip a coin. If you flip heads, I do nothing. If you flip tails, I will roll a 6-sided die. If I roll anything other than the number 6, I'll give you a cookie. So, before we begin, what are the odds you'll get a cookie? Well, we could just enumerate the possibilities:

1. You flip tails and I roll a 1. COOKIE!

2. You flip tails and I roll a 2. COOKIE!

3. You flip tails and I roll a 3. COOKIE!

4. You flip tails and I roll a 4. COOKIE!

5. You flip tails and I roll a 5. COOKIE!

6. You flip tails and I roll a 6. NO COOKIE.

7. You flip heads. (I don't roll a die.) NO COOKIE.

Well, it looks like five of the seven possibilities end with you getting a cookie, so your chances must be 5/7, or about 71% right? But if that were true, it would mean that the fact that I'm ABOUT to roll a die somehow makes you much likelier to flip tails (it happened in 6/7 of those possibilities), which makes no sense. The actual probability is instead the PRODUCT of the two events, namely 5/6 times 1/2 = 5/12, or about 42%.

Likewise, this fact — "If and only if you choose the car, Monty will pick a random goat" — does NOT make you likelier to choose the car in the first place. The actual probability of the outcome "I pick the car and Monty reveals Goat B" (for example) is 1/6, not 1/4. If it were 1/4, then the example with the coin and the die would work the same way, which would be a rather strange universe indeed.

Will the host open B or C if the correct answer is A? In other words will he always ask if you want to change your mind or will he automatically show you that you are right when you pick the right door without showing the other two?

Will the host open B or C if the correct answer is A? In other words will he always ask if you want to change your mind or will he automatically show you that you are right when you pick the right door without showing the other two?

As I was taught the problem, the host MUST always show you a door and offer you the chance to change.

If the host is following a script – that is, if the host is completely random – then the proper explanation is:

opening the goat door lowers the pool size to 2/3 the original size. You are still choosing randomly one possibility from the pool (which is now 2).

Your chance of winning has gone up from 33% to 50%.

Ran the statistics, and, indeed, your explanation is true.

I see the problem: the second choice is not the 50/50 random (which it would be if you flipped a coin between the two choices, the diminished pool choice). The choice is the split pool you describe.

If the host is following a script – that is, if the host is completely random – then the proper explanation is:

opening the goat door lowers the pool size to 2/3 the original size. You are still choosing randomly one possibility from the pool (which is now 2).

Your chance of winning has gone up from 33% to 50%.

Ran the statistics, and, indeed, your explanation is true.

I see the problem: the second choice is not the 50/50 random (which it would be if you flipped a coin between the two choices, the diminished pool choice). The choice is the split pool you describe.

I look at problems like this completely different than most people. If you pick, say, Door #1 first, then you are either right or you are wrong. That is two options (right or wrong). Two options divided by one whole gives you a 50% chance of being right and a 50% chance of being wrong. Given that logic, it doesn't matter if there are 2 doors or 200, no matter what door you pick, you always have a 50% chance of being right. Anything else is just extra data meant to confuse you.

That doesn't tell you which door you should choose once the host reveals the goat.

Just because there are only two outcomes doesn't mean they are equally likely. Babies can either be born alive or be stillborn, yet there are far more successful births than stillbirths. The odds aren't 50/50 just because there's two choices.

If you have 200 doors and only one has a car, and you are asked to pick the one that has a car, then you only have a 1 in 200 chance of getting it right. After the host reveals 198 of the 199 goats, the odds that you picked the right door originally is STILL only 1 in 200. The question now becomes "Is the car behind the door you picked… or is it behind one of the other 199 doors?"

The common, but faulty, thinking behind ignoring the goat when it's revealed is what made this game such a winning proposition for Let's Make a Deal!

Well said anon.

You are incorrect. Before the reveal of the 198 "bad" doors in this scenario, there is 1/200 chance that you have the "good" door. After the reveal, you are left with a second choice of the two remaining doors, one of which has the car, one of which doesn't. Each with an equal chance of being the car. What your scenario is proposing is that when given the choice of a or b, b somehow has twice the odds of having a favorable outcome. That flies in the face of logic

The odds don't change after the reveal! Even if you want them to.

Let's say I'm holding a deck of cards face down and ask you to pick a card without looking at it. I take my deck of 51 cards and say – "Okay. I'll give you a hundred bucks if you found the Ace of Spades. Do you think it's the card you took, or is it somewhere in the rest of the deck?"

At this point it should be pretty obvious that the chance that you actually hit the Ace of Spades on your first try is 1/52, and the odds that it was in the rest of the deck are 51/52.

Now let's say I go looking through my deck for the Six of Clubs. I take it out, show it to you, and then put it aside, keeping it out of the deck. Does this change the fact that when you originally picked your card, the odds that it were the Ace of Spades were 1 in 52? No. Does it change the fact that the deck still has a 51 in 52 chance of containing the Ace? Again, no. Revealing the six doesn't change any of the previous odds (except for the six itself, which we KNOW isn't the Ace!)

But we could take this further. I could pull out the rest of the clubs, or all the red cards, or, in fact, I could take out 50 cards that aren't the Ace of Spades. The odds that you found the Ace of Spades on your first go is only 1 in 52, and showing you extra cards doesn't make you any luckier there. The odds that it was somewhere in the deck were 51 and 52… but now that I've showed you 50 bunk cards, the odds that the single card I'm holding, my one-card "deck", are the Ace, is 51 in 52. For a single card! If you want those dead presidents you sure as heck better point at my hand if I ask you where the Ace is.

If when we started the game, we had a three-card deck (say, the Ace and the red deuces), then the same idea still holds, but the odds are 1/3 for your first pick and 2/3 for the rest of the "deck". And this is just a restatement.

The reason that revealing useless cards, or goat doors, doesn't help the odds of your original choice, is because you're not making any decisions there. Your first guess doesn't get any luckier just because I showed you a goat… it only helps the pool of the options you DIDN'T pick.

If I'm still not making any sense and you're not simply trying to troll, you might find the Wikipedia article to be enlightening.

I look at problems like this completely different than most people. If you pick, say, Door #1 first, then you are either right or you are wrong. That is two options (right or wrong). Two options divided by one whole gives you a 50% chance of being right and a 50% chance of being wrong. Given that logic, it doesn’t matter if there are 2 doors or 200, no matter what door you pick, you always have a 50% chance of being right. Anything else is just extra data meant to confuse you.

That doesn’t tell you which door you should choose once the host reveals the goat.

Just because there are only two outcomes doesn’t mean they are equally likely. Babies can either be born alive or be stillborn, yet there are far more successful births than stillbirths. The odds aren’t 50/50 just because there’s two choices.

If you have 200 doors and only one has a car, and you are asked to pick the one that has a car, then you only have a 1 in 200 chance of getting it right. After the host reveals 198 of the 199 goats, the odds that you picked the right door originally is STILL only 1 in 200. The question now becomes “Is the car behind the door you picked… or is it behind one of the other 199 doors?”

The common, but faulty, thinking behind ignoring the goat when it’s revealed is what made this game such a winning proposition for Let’s Make a Deal!

Well said anon.

You are incorrect. Before the reveal of the 198 “bad” doors in this scenario, there is 1/200 chance that you have the “good” door. After the reveal, you are left with a second choice of the two remaining doors, one of which has the car, one of which doesn’t. Each with an equal chance of being the car. What your scenario is proposing is that when given the choice of a or b, b somehow has twice the odds of having a favorable outcome. That flies in the face of logic

The odds don’t change after the reveal! Even if you want them to.

Let’s say I’m holding a deck of cards face down and ask you to pick a card without looking at it. I take my deck of 51 cards and say – “Okay. I’ll give you a hundred bucks if you found the Ace of Spades. Do you think it’s the card you took, or is it somewhere in the rest of the deck?”

At this point it should be pretty obvious that the chance that you actually hit the Ace of Spades on your first try is 1/52, and the odds that it was in the rest of the deck are 51/52.

Now let’s say I go looking through my deck for the Six of Clubs. I take it out, show it to you, and then put it aside, keeping it out of the deck. Does this change the fact that when you originally picked your card, the odds that it were the Ace of Spades were 1 in 52? No. Does it change the fact that the deck still has a 51 in 52 chance of containing the Ace? Again, no. Revealing the six doesn’t change any of the previous odds (except for the six itself, which we KNOW isn’t the Ace!)

But we could take this further. I could pull out the rest of the clubs, or all the red cards, or, in fact, I could take out 50 cards that aren’t the Ace of Spades. The odds that you found the Ace of Spades on your first go is only 1 in 52, and showing you extra cards doesn’t make you any luckier there. The odds that it was somewhere in the deck were 51 and 52… but now that I’ve showed you 50 bunk cards, the odds that the single card I’m holding, my one-card “deck”, are the Ace, is 51 in 52. For a single card! If you want those dead presidents you sure as heck better point at my hand if I ask you where the Ace is.

If when we started the game, we had a three-card deck (say, the Ace and the red deuces), then the same idea still holds, but the odds are 1/3 for your first pick and 2/3 for the rest of the “deck”. And this is just a restatement.

The reason that revealing useless cards, or goat doors, doesn’t help the odds of your original choice, is because you’re not making any decisions there. Your first guess doesn’t get any luckier just because I showed you a goat… it only helps the pool of the options you DIDN’T pick.

If I’m still not making any sense and you’re not simply trying to troll, you might find the Wikipedia article to be enlightening.

It seem to me that if the host will ALWAYS show you a goat, then you can discard one door entirely, because the REAL decision only occurs when you have two doors and a goat to choose from.

It seem to me that if the host will ALWAYS show you a goat, then you can discard one door entirely, because the REAL decision only occurs when you have two doors and a goat to choose from.

'The question is whether it is better to change or stick with your original choice. The answer — which can be and regularly has been demonstrated by running the scenario over and over — is that you are more likely to win if you change.'

'The question after the reveal shouldn’t be thought of as a repetition “What door is correct?” but rather “Was your original guess wrong?” And as we’ve just established, the chances are it was.'

My problem is that this article makes it out to be mathematical fact that your first guess is ALWAYS more likely to be incorrect than it is to be correct.

Pick 1 door out of A, B, C. You pick A. You have a 1 in 3 chance of being correct and a 2 in 3 chance of being incorrect. In this case, YES, it is more likely that A is the incorrect choice than it is that A is the correct choice. It does not, however, make it more likely that B is correct, or that C is correct. They are each individually weighted with exactly the same odds as A.

C is revealed to be a goat, i.e. 1 of the incorrect options is removed from the equation. You are left with 2 doors, 1 with the car, 1 empty. 50-50 odds.

Do you see what I'm saying? Odds only dictate you more likely to lose if you don't change BEFORE one of the dud doors is revealed. Once one of the non-car doors is opened, your first choice suddenly becomes 16.666 (recurring) % more likely to be a winner than it was when you had 3 doors to choose from.

‘The question is whether it is better to change or stick with your original choice. The answer — which can be and regularly has been demonstrated by running the scenario over and over — is that you are more likely to win if you change.’

‘The question after the reveal shouldn’t be thought of as a repetition “What door is correct?” but rather “Was your original guess wrong?” And as we’ve just established, the chances are it was.’

My problem is that this article makes it out to be mathematical fact that your first guess is ALWAYS more likely to be incorrect than it is to be correct.

Pick 1 door out of A, B, C. You pick A. You have a 1 in 3 chance of being correct and a 2 in 3 chance of being incorrect. In this case, YES, it is more likely that A is the incorrect choice than it is that A is the correct choice. It does not, however, make it more likely that B is correct, or that C is correct. They are each individually weighted with exactly the same odds as A.

C is revealed to be a goat, i.e. 1 of the incorrect options is removed from the equation. You are left with 2 doors, 1 with the car, 1 empty. 50-50 odds.

Do you see what I’m saying? Odds only dictate you more likely to lose if you don’t change BEFORE one of the dud doors is revealed. Once one of the non-car doors is opened, your first choice suddenly becomes 16.666 (recurring) % more likely to be a winner than it was when you had 3 doors to choose from.

You're still seeing doors b and c as separate choices, when in fact, they are only one.

This is how I look at it:

No matter how many choices you have, there are only two "pools." The pool that is incorrect, and the pool that is correct.

If you have three choices, and you pick A, then A is one pool, which has a 1 in 3 chance of being correct. That means B and C are the second pool, which after the reveal, has a 2 in 3 chance of being correct.

The reason the second pool is better is because if there is a goat behind A, a goat behind B, and a car behind C, you've already eliminated one goat by picking A. Then the host is forced to show you the second goat, which is behind B, so by switching to C, you've hit on the car.

I will be the brave soul and admit that I’ve always had a difficulty with this problem. I’ve consoled myself by believing that the people trying to explain it are misstating the situation or question. With issues such as this, the precise language is very important.

You’ve done the best job I’ve seen yet. Explanation two makes the most sense to me.

What I’ve found most difficult to accept is the idea that a series of subsequent actions that should be meaningless—I mean, no matter what, they happen in the same way—will always result in you changing your choice.

When you make your initial choice, you already know for a certainty that at least one of the other doors will have a goat, and you know that Monty will show it to you. You already have that information at the start. How can his follow-through affect the assessment of your original decision? Further, if you had chosen B, Monty might still have shown you C, and then we’d say that you should switch to A??? It almost seems like when you go up on stage and you’re presented with three doors, each one has a 50/50 chance of being the right one. So not that sticking with your first choice gives you better odds, but neither should switching.

If I may add in regards to Explanation One…

Aren’t there are more than three equal possibilities? If you choose A:

1. It’s in A and Monty shows you B, sticking is correct.

2. It’s in A and Monty shows you C, sticking is correct.

3. It’s in B and Monty shows you C, switching is correct.

4. It’s in C and Monty shows you B, switching is correct.

We take just winning scenarios, so if you’ve already chosen right door, host’s revealing doesn’t influence anything

Your explanation doesn't really make sense.

Doors A, B, C, if I pick A and B is revealed to have a goat, you say I should then change my pick to C. if I picked C originally, and they revealed B to have a goat, then I should pick A if I understand what you are saying correctly.

In actuality you have a 1/3 shot when first picking, once one of the doors are out of it you then have a 1/2 shot, so just by not doing anything, your chances are increased that the door you picked originally is the correct door.

"In actuality you have a 1/3 shot when first picking, once one of the doors are out of it you then have a 1/2 shot, so just by not doing anything, your chances are increased that the door you picked originally is the correct door."

By what form of time-travel do the laws of causality change the probability of your initial guess after the choice has been made?

Your explanation doesn’t really make sense.

Doors A, B, C, if I pick A and B is revealed to have a goat, you say I should then change my pick to C. if I picked C originally, and they revealed B to have a goat, then I should pick A if I understand what you are saying correctly.

In actuality you have a 1/3 shot when first picking, once one of the doors are out of it you then have a 1/2 shot, so just by not doing anything, your chances are increased that the door you picked originally is the correct door.

“In actuality you have a 1/3 shot when first picking, once one of the doors are out of it you then have a 1/2 shot, so just by not doing anything, your chances are increased that the door you picked originally is the correct door.”

By what form of time-travel do the laws of causality change the probability of your initial guess after the choice has been made?

“In actuality you have a 1/3 shot when first picking, once one of the doors are out of it you then have a 1/2 shot, so just by not doing anything, your chances are increased that the door you picked originally is the correct door.”

By what form of time-travel do the laws of causality change the probability of your initial guess after the choice has been made?

The wikipedia section on why its NOT 50/50 is what proved this to me.

The wikipedia section on why its NOT 50/50 is what proved this to me.

But what kind of car is it??? I might rather have a 100% chance of getting a goat!

But what kind of car is it??? I might rather have a 100% chance of getting a goat!

It's true… the real puzzle is trying to explain it. When i tried to explain it to my class mates i had to use the "infinite" resource. I told them to imagine 10000 doors. One has a car. Choose one. What are the chances to pick the right one??!! your chances are 1/10000. Now i open the rest of doors except for one. Do you want to change now?… it's pretty obvious in that way. All of them wanted to change the original pick.

This is the greatest idea to explain it. Even after reading the article my mind was still puzzled (even though i knew logically he is right) but after your explanation with the ""infinite" resource" mi mind now knows it too.

No it's not obvious. In your scenario, you start with 10,000 doors and are asked to pick one. The door you picked has a 1/10,000 chance of having the car. Now remove all the doors except for the one you picked and one other, leaving only 2 doors. Both of these doors had an initial 1/10,000 chance of having the car, with your first choice. However, since all other possibilities are removed, we now know that one of these two doors does have the car. So now, each door has a 1/2 chance of being the winning door. I see nothing in your math or logic that would indicate that switching from my original choice to the other one would increase the odds. Both doors have a 50/50 shot since we are now no longer dealing with a choice of 10,000 options, but only 2.

Let's go with your 10000 door example.

If you choose a door, you have a .0001 probability of getting it right on your first guess. This means that .9999 of the time, you will pick the wrong door at first.

When the host reveals all but one of the remaining doors, .9999 of the time he is forced by the rules of the game to offer you the correct door. (He is required to eliminate the other 9998 false choices since the one you initially chose was wrong).

It is true that a third person choosing randomly between the two doors would have a 50/50 chance, but he never had to consider the other 9998 losing doors.

You are right… i made a mistake trying to explain it, at the end you had to leave at least 2 more options.

but the idea is the same …if you have only 3 options to choose one is not pretty obvious. But if you had 10000 doors, then choose one, and then they open the rest except for two, you'd wanted to change the original pick.

Aww, i didnt read well… what you say is that theres nothing in my sample indicating that changing the original pick is the best option. Wich means my point failed, cause im not arguing if this is the best option since that is clear for most of us. My point is to make you understand it in the easiest way posible. And because you couldnt understand it i failed.

I can help you saying this, open your mind and forget the 50/50 illusion. If you choose a door among 10billion is almost imposible to choose the one with the car… but if AFTER you choose it, they tell you they are going to open all the doors except for one and let you choose again it is pretty obvius you need to change your pick to win the car.

It’s true… the real puzzle is trying to explain it. When i tried to explain it to my class mates i had to use the “infinite” resource. I told them to imagine 10000 doors. One has a car. Choose one. What are the chances to pick the right one??!! your chances are 1/10000. Now i open the rest of doors except for one. Do you want to change now?… it’s pretty obvious in that way. All of them wanted to change the original pick.

This is the greatest idea to explain it. Even after reading the article my mind was still puzzled (even though i knew logically he is right) but after your explanation with the “”infinite” resource” mi mind now knows it too.

No it’s not obvious. In your scenario, you start with 10,000 doors and are asked to pick one. The door you picked has a 1/10,000 chance of having the car. Now remove all the doors except for the one you picked and one other, leaving only 2 doors. Both of these doors had an initial 1/10,000 chance of having the car, with your first choice. However, since all other possibilities are removed, we now know that one of these two doors does have the car. So now, each door has a 1/2 chance of being the winning door. I see nothing in your math or logic that would indicate that switching from my original choice to the other one would increase the odds. Both doors have a 50/50 shot since we are now no longer dealing with a choice of 10,000 options, but only 2.

Let’s go with your 10000 door example.

If you choose a door, you have a .0001 probability of getting it right on your first guess. This means that .9999 of the time, you will pick the wrong door at first.

When the host reveals all but one of the remaining doors, .9999 of the time he is forced by the rules of the game to offer you the correct door. (He is required to eliminate the other 9998 false choices since the one you initially chose was wrong).

It is true that a third person choosing randomly between the two doors would have a 50/50 chance, but he never had to consider the other 9998 losing doors.

You are right… i made a mistake trying to explain it, at the end you had to leave at least 2 more options.

but the idea is the same …if you have only 3 options to choose one is not pretty obvious. But if you had 10000 doors, then choose one, and then they open the rest except for two, you’d wanted to change the original pick.

Aww, i didnt read well… what you say is that theres nothing in my sample indicating that changing the original pick is the best option. Wich means my point failed, cause im not arguing if this is the best option since that is clear for most of us. My point is to make you understand it in the easiest way posible. And because you couldnt understand it i failed.

I can help you saying this, open your mind and forget the 50/50 illusion. If you choose a door among 10billion is almost imposible to choose the one with the car… but if AFTER you choose it, they tell you they are going to open all the doors except for one and let you choose again it is pretty obvius you need to change your pick to win the car.

Two different events change the odds.

The first is a one out of three chance of selecting correctly. Next the host reveals one(for example door: "C").

Full stop. Forget what happened before.

Now host says: (begin 2nd scenario)

Car is behind door A or B. Even before host offers something else the odds are still 1 out of 2. You only have two doors and one definitely has and one definitely has not.

Now hose says Do you want the cash I am offering you instead of your pick? 100% chance of cash.

Finally: Your choice is a 50% chance of car or a 100% choice of cash.

(The odds will again chance if they offer a "curtain" instead of cash because it becomes yet another 50% chance of cool prize or 50% chance of crap….versus your 50/50 chance of car. You will then have 2/4 chances of getting crap..1/4 chances of car and 1/4 chance to get some other cool thing.)

Two different events change the odds.

The first is a one out of three chance of selecting correctly. Next the host reveals one(for example door: “C”).

Full stop. Forget what happened before.

Now host says: (begin 2nd scenario)

Car is behind door A or B. Even before host offers something else the odds are still 1 out of 2. You only have two doors and one definitely has and one definitely has not.

Now hose says Do you want the cash I am offering you instead of your pick? 100% chance of cash.

Finally: Your choice is a 50% chance of car or a 100% choice of cash.

(The odds will again chance if they offer a “curtain” instead of cash because it becomes yet another 50% chance of cool prize or 50% chance of crap….versus your 50/50 chance of car. You will then have 2/4 chances of getting crap..1/4 chances of car and 1/4 chance to get some other cool thing.)

The way I've best explained it kind of removes the bad door the host gets rid of. Once you choose your first door, we will call this Option A, there is a 1/3 chance this option contains the prize and a 2/3 chance that Option B (the other two doors) contains the prize. So when he eliminates the bad door and gives you the option to change, he is effectively (mathematically) asking if you want to stick with Option A, which will have the prize 1/3 of the time, or go with Option B (switch), which will have the prize 2/3 of the time. Him eliminating the bad door does not change that Option B will have the prize 2/3 of the time. This works with any number of doors. Say 10,000 and you choose one door, which is option A. Option A has 1/10000 of a chance of containing the prize than Option B (all the other doors).

You can also think of this scenario in the following way that is mathematically the same situation. In this case he doesn't show you the goat(s)/eliminate a door(s) after you pick the first door. First, he lets you pick a door, which becomes option A. Option B becomes ALL other doors. So, after picking the door that becomes Option A, he then gives you the choice of staying with A or switching to Option B, where you only need the prize to be in one door if you choose B. So, by switching you have a pool of n-1 doors to win from, rather than 1 door to win from. Probability of winning on switching (n-1)/n. Don't switch: 1/n. Where n represents the number of doors.

The way I’ve best explained it kind of removes the bad door the host gets rid of. Once you choose your first door, we will call this Option A, there is a 1/3 chance this option contains the prize and a 2/3 chance that Option B (the other two doors) contains the prize. So when he eliminates the bad door and gives you the option to change, he is effectively (mathematically) asking if you want to stick with Option A, which will have the prize 1/3 of the time, or go with Option B (switch), which will have the prize 2/3 of the time. Him eliminating the bad door does not change that Option B will have the prize 2/3 of the time. This works with any number of doors. Say 10,000 and you choose one door, which is option A. Option A has 1/10000 of a chance of containing the prize than Option B (all the other doors).

You can also think of this scenario in the following way that is mathematically the same situation. In this case he doesn’t show you the goat(s)/eliminate a door(s) after you pick the first door. First, he lets you pick a door, which becomes option A. Option B becomes ALL other doors. So, after picking the door that becomes Option A, he then gives you the choice of staying with A or switching to Option B, where you only need the prize to be in one door if you choose B. So, by switching you have a pool of n-1 doors to win from, rather than 1 door to win from. Probability of winning on switching (n-1)/n. Don’t switch: 1/n. Where n represents the number of doors.

I too have had thoughts on the simplest possible explanation.

I think rather than saying, "what are your chances of initially choosing the right door?",

you should ask,

"What are your chances of initially choosing the wrong door?"

You have a 66% chance of picking the wrong door.

In this scenario, only one door can be revealed by the host. In other words, there's a 66% chance that the TV host will have no choice but to reveal the wrong door.

I too have had thoughts on the simplest possible explanation.

I think rather than saying, “what are your chances of initially choosing the right door?”,

you should ask,

“What are your chances of initially choosing the wrong door?”

You have a 66% chance of picking the wrong door.

In this scenario, only one door can be revealed by the host. In other words, there’s a 66% chance that the TV host will have no choice but to reveal the wrong door.

What about explaining it like this:

Imagine that the host didn't open any doors. So you get to choose the first door you picked, or you get to choose two doors at the same time, and if the car is between either one of them, you win it.

So you have a 1/3 chance by sticking with the first door, and a 2/3 chance by switching to the other two doors.

In effect, this is exactly what's happening – except that the host happens to open one of the two doors for you. This in itself does not change the overall odds.

What about explaining it like this:

Imagine that the host didn’t open any doors. So you get to choose the first door you picked, or you get to choose two doors at the same time, and if the car is between either one of them, you win it.

So you have a 1/3 chance by sticking with the first door, and a 2/3 chance by switching to the other two doors.

In effect, this is exactly what’s happening – except that the host happens to open one of the two doors for you. This in itself does not change the overall odds.

What about explaining it like this:

Imagine that the host didn’t open any doors. So you get to choose the first door you picked, or you get to choose two doors at the same time, and if the car is between either one of them, you win it.

So you have a 1/3 chance by sticking with the first door, and a 2/3 chance by switching to the other two doors.

In effect, this is exactly what’s happening – except that the host happens to open one of the two doors for you. This in itself does not change the overall odds.

I wrote up something similar a long time ago:

http://www.panix.com/~mshaw/3door.txt

I wrote up something similar a long time ago:

http://www.panix.com/~mshaw/3door.txt

MAKE A CHOICE:

|—–| |—–| |—–|

| A | | B | | C |

| o| | o| | o|

| | | | | |

|_____| |_____| |_____|

|———-|———————————–|

| | YOU CHOOSE |

| |———————————–|

|———-| A | B | C |

|R s |—-|———–|———–|———–|

| | A | stay | switch | switch |

|E p |—-|———–|———–|———–|

| | B | switch | stay | switch |

|A o |—-|———–|———–|———–|

| | C | switch | switch | stay |

|L t |—-|———–|———–|———–|

|———————————————-|

Only one door has a car, but two doors have a goat. You are more likely to pick a goat. The host then shows you the goat you didn't pick. Now, because you have good evidence to believe you just picked a goat, and the host showed one goat, it means that the door you didn't pick is more likely a car.

In short, you should switch, because there is a 66% that your first guess was WRONG.

MAKE A CHOICE:

|—–| |—–| |—–|

| A | | B | | C |

| o| | o| | o|

| | | | | |

|_____| |_____| |_____|

|———-|———————————–|

| | YOU CHOOSE |

| |———————————–|

|———-| A | B | C |

|R s |—-|———–|———–|———–|

| | A | stay | switch | switch |

|E p |—-|———–|———–|———–|

| | B | switch | stay | switch |

|A o |—-|———–|———–|———–|

| | C | switch | switch | stay |

|L t |—-|———–|———–|———–|

|———————————————-|

Only one door has a car, but two doors have a goat. You are more likely to pick a goat. The host then shows you the goat you didn’t pick. Now, because you have good evidence to believe you just picked a goat, and the host showed one goat, it means that the door you didn’t pick is more likely a car.

In short, you should switch, because there is a 66% that your first guess was WRONG.

That is really upsetting. I just spent five minutes making a detailed table, and this stupid thing deletes spaces.

That is really upsetting. I just spent five minutes making a detailed table, and this stupid thing deletes spaces.

Some number problems become more obvious when taken to extremes.

– There are 1000 doors, and only one of them is the winning door.

– You have to guess which door out of 1000 is the winner. (you will most likely choose wrong)

– After you select a door, all the doors except the one you chose and one other door are revealed to be losing doors.

There are two possible scenarios here.

Either:

A) You correctly guessed 1 in 1000, and the other door was randomly chosen from the remainder; or

B) You guessed wrong, and the other door is the winner.

Which is more likely?

The idea is the same with any number of doors. Fewer doors makes it less certain that you chose wrong the first time, but it's still most likely that you did.

i don't understand.

just pretend that the host didn't and won't reveal the goat. initially, if you choose A, you got 1/3 chance of getting it right. By switching means B and C is considered as ONE choice, but they hold 2/3 chance. So its better. It doesn't matter if the goat is in B or C. so by switching means you are choosing B and C, pretending the host didn't and won't reveal the goat.

An interesting point about this problem is that it only holds true IF the host is actually aware of which door holds the car. The basis for the logic behind this is that the host always reveals a goat. But if he sometimes reveals the car during the first round then there is no logical difference between switching and staying, because the chance is still equal.

I think that's the part that trips some people up. They don't realize that the fact that a goat having to be revealed is part of the logic. They assume that the game is being played fairly, and thus assume that the probability is equal the whole way through.

Its quite simple really. when you make the first choice you have a 1/3 chance of making the right choice and a 2/3 chance of getting it wrong. meaning your more likely at choosing the wrong door. so when your asked a second time and the odds are 1/2 (50%) at choosing the right door, you are more likely to have chosen the wrong door originaly and it is a statisticaly safer option to swap

OMG! Imagine that you pick one door, and you have a friend with you who automatically gets the other two doors you didn't pick. Who is more likely to win the car?! YOUR FRIEND. By switching after the reveal you get the same effect: You can stick with Door 1, or switch and get the BEST of Doors 2 AND 3. By sticking you are limited to one door and 33.33% success; by switching you double your doors and your success to 66.666%. AND IT DOESN'T MATTER that Monty revealed a goat behind one of the other doors–one of them HAD to have a goat! Geez. To all those who have developed new statistical theorems whereby having 3 Doors and 1 pick somehow gives you a 50/50 chance of having initially picked correctly, I salute you.

One more, let's take it to extremes: suppose that there are 100,000 doors, you pick door 1 and your friend gets Doors 2 through 100,000. Who is more likely to have the car?? Your one stinking door, or your friend's 99,999 Doors? Then the host opens 99,998 of your friend's doors to reveal goats, so he has one door unopened and your door is unopened. If you believe that it's a 50/50 chance that your door has the car, then there's no hope for you.

The user Anon said it in a matter that made me understand :

“The odds don’t change after the reveal! Even if you want them to.

Let’s say I’m holding a deck of cards face down and ask you to pick a card without looking at it. I take my deck of 51 cards and say – “Okay. I’ll give you a hundred bucks if you found the Ace of Spades. Do you think it’s the card you took, or is it somewhere in the rest of the deck?”

At this point it should be pretty obvious that the chance that you actually hit the Ace of Spades on your first try is 1/52, and the odds that it was in the rest of the deck are 51/52.

Now let’s say I go looking through my deck for the Six of Clubs. I take it out, show it to you, and then put it aside, keeping it out of the deck. Does this change the fact that when you originally picked your card, the odds that it were the Ace of Spades were 1 in 52? No. Does it change the fact that the deck still has a 51 in 52 chance of containing the Ace? Again, no. Revealing the six doesn’t change any of the previous odds (except for the six itself, which we KNOW isn’t the Ace!)

But we could take this further. I could pull out the rest of the clubs, or all the red cards, or, in fact, I could take out 50 cards that aren’t the Ace of Spades. The odds that you found the Ace of Spades on your first go is only 1 in 52, and showing you extra cards doesn’t make you any luckier there. The odds that it was somewhere in the deck were 51 and 52… but now that I’ve showed you 50 bunk cards, the odds that the single card I’m holding, my one-card “deck”, are the Ace, is 51 in 52. For a single card! If you want those dead presidents you sure as heck better point at my hand if I ask you where the Ace is.

If when we started the game, we had a three-card deck (say, the Ace and the red deuces), then the same idea still holds, but the odds are 1/3 for your first pick and 2/3 for the rest of the “deck”. And this is just a restatement.

The reason that revealing useless cards, or goat doors, doesn’t help the odds of your original choice, is because you’re not making any decisions there. Your first guess doesn’t get any luckier just because I showed you a goat… it only helps the pool of the options you DIDN’T pick.”