It always surprises me when I talk about the Monty Hall problem, how many people argue with me about what the “right” answer is. It goes against most everyone’s ideas of what is right. Here’s the problem:

## The Three Doors

You are presented with three doors, and are told that behind ONE of them, there is a shiny, new car. Behind the other two doors, however, there are goats. The object of this game, obviously, is to choose the door that has the car.

So make your choice. Do you choose Door #1, Door #2, or Door #3?

## The Reveal

Once you have chosen a door, your gracious host, Monty Hall, chooses to open one of the doors you did not select. He will ALWAYS open a door that contains a goat. For the purposes of illustration, let’s assume you chose door #2 initially. In this case, then, Monty knows that there is a goat behind Door #1. So he opens it.

## The Choice

This is when the Monty Hall problem rears its ugly head. Once that door is revealed, he offers you a choice. You can keep the door you initially selected, or you can switch to the other unopened door. It’s entirely your choice. What would YOU do? Would you stay with your original door? Would you choose the other one? **(Leave a comment on this post with your choice right now…before you continue reading…)** In our example, again, we are going to stay with our initial choice.

## The Result

After you’ve made your choice, Monty will open the door you DIDN’T select. Here’s a picture of how our scenario played out:

So the point of this whole post comes down to probability:

## Why You Should Always Switch Doors

This is where the controversy comes in. Many people (probably including you) disagree with this. The common thought is that no matter what you do, your initial choice of the three doors can’t actually be improved. I’m writing this to tell you that you’re wrong.

In fact, you can DOUBLE your odds of winning by choosing the other unopened door. Let me illustrate the flow of choices you get to make, and how they improve with a “switch.” In this example, we assume that the “contestant” always chooses Door #1.

Recall back to how the problem works: You pick one of three doors. So your odds are 33% (1/3) that you’re going to pick the car. Said another way, there’s a 66% (2/3) chance that you’re going to pick a goat.

Here’s what that means, as you look at the image above: 2 out of 3 choices result in you choosing a door with a goat. Switching, when you have selected a goat initially, results in getting the car 100% of the time. So if you ALWAYS switch, you’ve got a 66% chance of winning the car. If you NEVER switch, which means you stick with your initial choice, you are forcing yourself to take a 33% chance of the car being behind your door.

Yes, that does still mean that 1/3 of the time you will still lose by switching, but I’ll take that over losing 2/3 of the time by NOT switching.

In the end, you DOUBLE your odds of winning the car by switching every time.

## Let’s Discuss This Further

I’d like to think that this topic will generate some conversation. You probably have stared at the diagram, and yet, inside, you still disagree. You might say that switching is a 50/50 shot. That’s where I disagree. If you’re simply playing the percentages, switching will always give you the best results.

Leave a comment. Leave several. Let’s have a lively discussion on this one.

UPDATED: Marvin S. Schwartz, PhD. has written a simple C# application that lets you play this game. Here’s a link to his Monty Hall source code.

@Mel Grubb,To win by staying, you have to make the correct (1/3) choice on your first shot.OR make an INcorrect (2/3) choice on your first shot, and then have a 50% chance on your second shot.50% of 2/3 is 1/3.Total = 1/3 (first shot) + 1/3 (second shot) = 2/3 by staying.YMMV.Eunoia,PS: They told me I was gullible and I believed them 😉

Just imagine there being 100 doors instead of 3 and you pick 1 at random, then the host opens the other 98 of them revealing goats, leaving just 2 doors, yours and the other one. Switching your choice at that point becomes a no brainer.

Initially:

p(1) + p(2) + p(3) +…p(100)= 1

You pick p(1)

Then

p(3) + …p(100) = 0 (are eliminated, leaving you with 1 and 2)

So

p(1) + p(2) + 0 =1

Therefore

p(1) = p(2)

And i don’t even need to specify p values for the initial situation

@Cosmin Visan

p(1) + p(2) = 1

Means that they sum to 1. but that doesn’t mean they’re equal.

Say you *always pick door 1 out of 100. By your reasoning, then door 1 has a 50% chance of being right, and the other remaining door (call it door X) also has a 50% chance of being right. Right? So, that means the car always has a 50% chance of being behind whichever door you picked first. Picking a random door out of 100, this is an obvious logical absurdity.

… think about it this way, say that no swap is ever offered, and you pick a random door out of 100. With 100 doors and 1 prize a random choice will only get your the prize 1% of the time.

Monty can waggle all the other doors opened and closed in any order you like, that won’t chance the odds that you picked right in the first place – 1 game per 100 tries.

Mel Grubb’s explanation is the best I’ve heard.

I don’t think you can double your odds by switching. The fallacy of the argument is that the host’s choice has a bearing on the outcome and must be considered statistically. In the first case on the diagram the host has two choices making for

fourpossible scenarios, yet it is calculated statistically as if it were one in three. Of the four possible cases you lose two if you switch and lose two if you stay. Looking at it another way, since the host willalwayschoose a goat, thereby removing it statistically, there are really onlytwoactual cases. The constestant picked either a door with a goat or the car. Switching or not switching, it’s 50/50.incorrect. there are 3 possible permutations with 2 goats, 1 car, and 3 doors. odds are 1/3 if you stay, 2/3 if you switch. Think of it like this. there are 2 sets.. set#1 is your door… set#2 is the other two doors combined. Odds of the car being in set#2 = 2/3. By eliminating a door from set#2, you get that entire set for the price of 1 door.

Yeah, the “4” choices are only 3 equally-likely scenarios, but in one of those scenarios, Monty can choose two different things. That doesn’t mean “4 equal scenarios”. It means two scenarios which happen 1/3rd of the time each, and a remaining scenario that has two equal outcomes of 1/6th each.

An analogy would be this game:

Flip a coin. If it’s heads, I buy you an ice-cream. If it’s tails, flip again: If heads I buy you a soda, if tails, I buy you a hamburger.

There are “three” outcomes here, you get ice-cream, a sode or a hamburger. But it would be silly to say each one was “equally likely”. Clearly the Ice-cream will happen 50% of the time and the other two 25% of the time. It’s exactly the same for the 1/3 scenario where Monty has to “coin flip” which door he opens.

Go with your gut! If it’s rigged, then it’s rigged…nothing you can do about that!

The plain-language explanation goes like this: Decide your strategy beforehand. Are you going to stay or switch.To win by staying, you have to make the correct (1/3) choice on your first shot.To win by switching, you have to make an INcorrect (2/3) choice on your first shot, because Monty will then eliminate the other incorrect choice FOR you.

I would switch doors but that is because I have seen this problem before. Originally I wanted to stay with the first choice. My question to you: What scenario in real life patterns this problem? (other than the game show)

I often hear this referred to as the Monty Hall “paradox”, because of how unintuitive it is at first.

More specifically, branch prediction.

Because Monty Hall ALWAYS opens a losing door, isn’t that door really just irrelavant to the problem? To make this truly a problem of x/6 then Monty Hall should choose a door randomly as well. The true problem is whether or not you choose the correct door from one of the two.

No, because you picked first, hence it’s 1/3. If you are not offered a swap, or you decline the swap, it doesn’t matter what order Monty opens doors AFTER you pick. You’re 1/3 likely to have picked the correct door.

Consider this scenario: you pick a door, then Monty open a Goat door. But instead of offering a “swap” Monty says: “one of these two remaining doors is the car. But I’m claiming the remaining door, and I win if you lose”. Since you still have your 1/3 chance of having picked the correct door, that means Monty has a 2/3 chance of keeping the car for himself. Clearly allowing a “swap” of the prizes or the doors they were behind would have helped you.

I’m going with switching doors. My calculations are that it would about double my chances of being correct. I love these type of problems 🙂

That’s a fun problem that requires you to put aside your feelings and put trust in statistics. I thought it had become somewhat of a household knowledge from the movie 21 last year, where it was featured. Maybe the movie wasn’t that popular. The main character looked like a rockstar, knowing the answer.

I remember arguing about this for a long time with the other nerds at math camp when i was in high school… Good times.

How soon hath Time, the subtle thief of youth, Stolen on his wing my three-and-twentieth year

Love and you shall be loved.

Elijo la puerta 2

best explanation ever

I think that it should be taken as a new problem. 2 Doors. 50% and 50%.

Nope, because you initially picked “your” door out of 3. So never switchng wins the exact chance of getting it right at first – 1/3. Whereas Monty’s remaining door will always be a winner if yours was not – 2/3rd chance.

Consider this example with cards, with the goal of getting the Ace of Spades:

1. you pick a card at random. 1/52 chance of getting the Ace of Spades. You place that face down in front of you, unseen.

2. The dealer knows where the Ace of Spade is, so he reveals 50 “losing” cards ans places one last card face down in front of himself.

3. Now, what are the chances that you have the Ace of Spades, and what are the chances that the dealer has it? And would you swap to his card?

Only if you ignore the information you already have. The player chose 1 door from 3 at random. That’s a 1/3 chance of success if you STICK with your door. The other constraint is that either sticking or switching MUST win. Since “sticking” wins 1/3 then switching wins all other times, or 2/3.

look at this scenario run infinite times up until this point. 66.67% of people will have chosen a door with a goat behind it, and 33.33% of people will have chosen a door with the car. If everyone switched, 66.67% of the people who entered the competition will be winning a car. There’s no new problem

Swithching doors only works if there are 3 doors….

no no no.. the odds depend on (1) number of doors and (2) number of doors monty reveals.

At first, I was among the many who said, “No, way! It’s 50/50.” I even diagramed all possible outcomes. But what I failed to do was recognize that some of the outcomes I diagramed were not possible outcomes due to the fact that when Monty chooses a door, he does NOT do so at random. Otherwise, Monty would sometimes open a door with a car behind it. Since those outcomes have been eliminated as possible, the net 50/50 figure is no longer applicable.

Let me put it to you very simply. Let’s say you choose to stay with your first choice 100% of the time. Statistically, you will get the car 1/3 times no matter what happens after that, period.

No matter what door you choose, there is a 100% possiblity that at least one losing door remains, and it (or one of the two) will be opened 100% of the time. This still has no bearing at all on the fact that you will only choose the winning door 1/3 of the time initially.

An analogy I can make is – Image there are two convenience stores each of which sells lottery tickets. You know that convenience store A sold one ticket, and convenience store B sold two tickets. There is one (only one) winning ticket 100% of the time. If you were asked which store sold the winning ticket, there would be a 1/3 chance that store A sold it, and a 2/3 chance that store B sold it. Since you KNOW that store A sold only one ticket, that has a bearing on probability and it would therefore always be beneficial to guess store B sold the winning ticket. If you did not know which store sold one ticket and which sold two, the ticket must be in one and only one store and from your point of view now, you have a 50/50 chance of guessing the store that sold the winning ticket. I actually tested this empircally, and when I knew which store sold only one ticket, I guessed it correctly about 2/3 of the time, by guessing the store I knew had sold two tickets. When I diliberately stayed ignorant of which store sold one and which store sold two tickets, I was forced to essentially guess randomly, and sure enough I was correct about 50% of the time. There is NO mathematical paradox here, but rather, it simply had to do with how much you know.

Back to the Monty problem. When you guess your first door, you essentially created something similar to the store that sold only one ticket while the other “store” would be the remaining two doors. This is true no matter which door you start off with. You always separate into something similar to the two stores and it is always in your best bet to guess the “store with two tickets.” The fact that a losing ticket is revealed has no bearing on anything. It is simply stating that it exists, which is a meaningless statement.

Finally, I tested three scenarios empircally. First by always sticking with my first door, where I got it right 1/3 of the time. Second, by always switching, where I got it right about 2/3 of the time. Lastly, I randomly chose whethere or not I would switch doors, and guess what, I got it right about 50% of the time. (Do the tables on that and you will see why). The logic error that most people make when first confronted with this problem is that they think as if they had the same information as the person who randomly chose whether or not to switch, and that would give you the 50/50 so many insist must be there, but when you have more information (and dismiss irrellevant information such as the opening of a losing door) you can make a more informed choice.

It is just like the situation where one person knows something and therefore chooses the right course of action, but another person has no idea which is right and which is not and has to take his best guess. Of course the person who knows more will get it right more often. The “trick” to the whole problem, is that you know more than you initially realize!!!

Long winded, but I hope it makes sense.

Grammatical correction – At some point I said “image” when I should have said “imagine.” Sorry…

I am going against the flow — it is 50/50. Please relax and think about the rules of the game, and you will see it is very simple.

1. 3 choices will be presented with 1 correct choice

2. You “choose” one door.

3. 1 incorrect choice is removed.

4. Your choice in #2 is now irrelevant given #3 is happening 100% and we get to repick. Even when you pick the car correctly, Monty still removes a goat. So no matter what we are always playing a game between 2 choices with 1 car and 1 goat. Monty taking an incorrect choice away doesn’t help us at all.

Now think of another scenario–

1. 3 choices will be presented with 1 correct choice

2. 1 incorrect choice is removed.

3. You pick a between the two choices, which clearly have identical probabilities.

What you guys are proposing is that *the act of choosing* affects probabilities when it does not. You use the logic “we had 1/3 chance of being wrong originally” but actually that is not true. You had a 0% chance of being wrong, since the choice was not a final choice, but merely a placeholder since one wrong choice is always being removed. There is no 1/3 chance, there never was a choice between 3.

“There is no 1/3 chance, there never was a choice between 3.” I think you might want to reconsider that statement. 3 doors, 1 car. Pick a door and you think there’s a 50% chance of getting the car. You should go to Vegas and play the roulette tables, you’d make a fortune.

I go with you. Everyone is fooled in believing the initial 1/3 chance of picking the car door is unchanged by subsequent events.

Consider the 8 toss coin question. The chance of tossing 8 heads in a row is 1/256. However if you have actually tossed 7 heads in a row, the chance of the 8th toss being a head is no longer 1/256 but 50/50.

Subsequent revelations change odds. Ask a poker player.

Completely different situations: the coin tosses are independent events, you picking a door and Monty then opening a goat door are dependent events. Your initial odds of 1/3 of picking the car remain unchanged.

It is a well proven mathematical puzzle, the only people fooled are those who believe it’s 50/50 when there’s 2 doors left.

Think of it this way. Imagine you have not three, but a thousand doors. In one of them is a prize. You pick one, and the host opens 998 doors, giving you the chance to switch. Would you?

inb4doesntmatterbecause50%chance

Very well put Brian, and I agree with you.

My simple way of looking at it is too that there has never been a 3 door option. If, for example I refuse to choose, or Monty doesn’t know which door I chose, he will still reveal a door and reduce it to a 1/2 game.

Just because it can be demonstrated with an equasion doesn’t mean it is true.

I’ll prove that 0.999… = 1 with an equasion:

x = 0.999…

10x = 9.999…

10x – x = 9.999… – 0.999…

9x = 9

x = 1

Just because you can use maths to ‘demonstrate’ something doesn’t necessarily mean it is right.

Well you’re mistaken too then. Your simple way of looking at it is the wrong way, if you refuse to choose then you’re playing a different game – surely you can see that?

And btw, 0.9999….. does actually = 1, so you got that wrong as well LOL

The Monty Hall question a broken question really. It makes two very clear conditions that contradict each other. One condition is that there are three entities each with a probability of being a car. the other is that the host will select an entity that is not a car, when by the first condition he can only select one that has a probability of not being a car.

The answer can depend on which condition you disregard as being a mistake. If the first condition is wrong and the car is behind a predetermined door, then it is a better chance to switch, if the second condition is wrong and it is a matter of hypothetical probability, it is a 1/3rd chance of winning no matter if you stay or switch.

For some reason though most people ignore the first condition rather than the second, I can only attribute this to people mis-quoting the puzzle, as the blog writer does in adding the predetermined door to the question, or to people trying to test it with real world scenarios like playing cards, or cups with balls, forgetting that this also makes the same assumption that isn’t in the original question.

The scenario you described however does not ignore either condition, it does however add a third to fix the broken puzzle, which is that the act of choosing doesn’t affect the puzzle.

As for me, I think the best answer is 1/3rd chance either way, however, as to which condition you ignore or which condition you have to add in order to make the puzzle work, this is just a matter of opinion on which is best.

Though according to what I’ve read a lot of math geniuses and PHD’s say its 50% too, so that opinion probably has more weight than the others.

Someone explain to me how switching is a 1/3 chance of winning when there are only two cards to choose between at that point, the third “choice” is gone, so you have TWO choices each time, adding up to 3/6. How are you guys chosing something that can’t be chosen?

Because you picked from 3 doors at first, if you do nothing else, then you will win 1/3rd of all game. The order Monty opens doors in doesn’t change where the car is.

Think of this: 100 doors and you pick 1 at first. 1/100 chance of being correct, right?

Monty then opens other losing doors 1 at a time. Each time he say you can switch to any other unopened door. But you do not. Eventually, it’s down to 2 doors. Do you switch then? Or is your first pick now magically right 50% of the time because it’s “down to 2 doors”?

Crikey. There is no 50/50 chance. You will win 2/3 of the time if you switch doors. One more time: imagine that you have a friend with you. You pick one door and your friend gets the other two. Who will most often win the car? Your friend, of course–he will win the car 2/3 of the time and you will win 1/3 of the time. You win if the car is behind door 1, and he wins if the car is behind either door 2 or 3. Switching doors has the exact same effect. If you stay with door 1, you will win 1/3 of the time. but switching allows you to win when the car is behind either of the other two doors. Switching lets you win when the car is behind Door 2 OR Door 3! 2/3 of the time.

You deniers are way over-thinking this.

There are only 3 possible outcomes behind Doors 1,2, and 3: Car, goat, goat–Goat, Car, goat–or, Goat, goat, Car. If you don’t switch and always choose Door 1, you win 1/3 of the time. But if you switch, you will always win when the car is behind EITHER Door 2 OR Door 3. You double your chances of winning.

When Monty opens one of the remaining doors to reveal a goat, that does NOT magically enhance your odds to 50/50 by staying. There were 3 doors when you started, you had a 1/3 chance of picking the correct door. However, by switching you are in effect getting to pick TWO doors–2 and 3–and this WILL increase your odds of winning.

What if there were 1000 doors? You have a tiny 1/1000 chance of picking the correct door! And there is a 999/1000 chance that the car is behind ANY ONE of the other 999 doors. If 998 of the remaining 999 doors are opened to reveal goats—and you know that the car was MOST LIKELY behind one of those 999 doors–would you switch? Of course, yes. Because you had an astronomical 1/1000 chance of picking the correct door–and the car was likely behind one of the 999 you didn’t pick. But if you’re allowed to switch you’re essentially being allowed to choose all Doors 2 through Door 1000. Those are good odds–so you switch.

An excellent synopsis, Laurence; however there IS a 50/50 chance which lies at the seat of the “paradox”: A person walking in to the scenario after the goat-door has opened and who does not have knowledge of previous events has a 1/2 chance of randomly picking the first-selected door (which has a 1/3 chance of the car), and a 1/2 chance of randomly picking the other unopened door (which has a 2/3 chance of a car).

(1/2 x 1/3) + (1/2 x 2/3) = 1/2 chance of winning the car.

This is RANDOM choice is the 50/50 chance which confuses everyone who doubts that the odds can indeed be increased by switching.

This becomes intuitively easier to understand if the number of doors is scaled up. Imagine that there are 1000 doors and only one car. After you pick a door, the host will let out goats from 998 doors. Two doors remain: the one you chose and the one he didn’t open. But isn’t it obvious they don’t each have a 50% chance of hiding a car?

I will forever disagree with this. I’ve considerd the problem with a million doors, I still disagree. You pick door #1, the host opens door #3. Door #3 is a goat, so why are you still considering it in your statistics? Door #3 is irrelevant, it no longer exists. You now have a choice between 2 doors, not 3. I disagree, because every argument I’ve seen still considers door #3 a choice. Door #3 is not a choice, you have 2 choices, so how can you possibly have a 2/3 chance if there are only two choices?

I understand your problem. I had this trouble, because simple maths says 1 prize behind 2 doors gives a 1/2 probability for each door, because the probabilities add up to 1. But this is only true if we assume the probabilities are equal for each door, to begin with. If we can show the probabilities are not equal, then the simple maths is not valid. Obviously, at the start of a horse race, 10 horses running does not give each horse a 1/10 chance of winning. Each horse has different probabilities. The maths involves adding the horses probabilities in such a way that the total =1.

This is the simplest Monty Hall explanation I know. If the contestant chooses door 1, he has a 1/3 probability of winning. And there is a 2/3 probability of the prize being behind one of doors 2 or 3. It doesn’t matter which door is left, as long as the loosing door is revealed. And the door left has the 2/3 probability of winning, double that of door 1. In fact, it doesn’t even matter if the loosing door was chosen by luck randomly, or was chosen deliberately. A bit of prior knowledge on what has happened can improve your chances. Same as betting on horses! Good luck.

If Monty doesn’t know what’s behind each of the doors and opens a losing door by luck then it makes no difference whether you stay or switch, as you’ve got a 50% chance of having picked the car in the 1st place.

show host revels 1 goat in 1 of the doors, that leaves us with 2 doors!

any way you look at its 100% pure bullshit!

designed to test gullibility level in people facing with “statistics”.

Well it’s 100% certain that you don’t understand basic probability or simple logic.

If you 1st choose a goat then switching GUARANTEES you win the car (as Monty must reveal the other goat). The probability of this happening is 2/3 (1st pick a goat) * 1 (Monty reveals a goat) = 2/3

The only way to win by staying is if you 1st choose a car.The probability of this happening is 1/3 (1st pick a car) * 1 (Monty reveals a goat) = 1/3,

So by switching you’re twice as likely to win by switching than by staying.

It’s not that hard to understand.

Because Monty always reveals a goat, that simply isn’t considered a possibility anymore, leaving only 2 possibilities and a 50/50 chance. By saying 1/3 and 2/3, you are implying that there are 3 choices because the denominator is the possible outcomes. You are suggesting that the already revealed goat could still be a possible car.

All the scheamatic examples in favour of switching assume six possible outcomes. But if you initially choose the car, Monty has to make a choice between which of two goats to show. Schematically, the problem then has 8 outcomes. With an even chance of you getting a goat or a car. See below.

You are asked to choose a door

A B C

You choose You choose You choose

goat A goat B car C

Monty eliminates B Monty eliminates A Monty eliminates A Monty eliminates B

Stick Switch Stick Switch Stick Switch Stick Switch

GoatA Car GoatB Car Car Goat B Car GoatA

1 2 3 4 5 6 7 8

I’ve done this againt because my previous post is illegible. I hope the dashes will force it to line up correctly.

—————————–You are asked to choose a door——————————–

——–A———————-B————————————C——————————-

—-You choose———You choose————————You choose —————————goat A—————goat B——————————— car—————————

Monty eliminates B–Monty eliminates A–Monty eliminates A–Monty eliminates B

–Stick—–Switch——-Stick—-Switch——Stick—— Switch——-Stick—–Switch–

–GoatA—–Car——–GoatB—-Car———-Car——Goat B——–Car——GoatA–

I’ll try this one last time. Up to you moderator if it doesn’t work.

I’ve done this againt because my previous post is illegible. I hope the dashes will force it to line up correctly.

—————————–You are asked to choose a door——————————–

——–A———————-B———————————————C——————————-

—-You choose———You choose————————You choose ———————

——goat A—————goat B——————————— car—————————

Monty eliminates B–Monty eliminates A–Monty eliminates A–Monty eliminates B

–Stick—–Switch——-Stick—-Switch——Stick—— Switch——-Stick—–Switch–

–GoatA—–Car——-–GoatB—-Car———-Car—-—Goat B——–Car——GoatA–

Although there are 8 outcomes they are not all equally likely. If you pick a car (1/3 chance) 50% of the time the host reveals Goat A and 50% of the time Goat B, so the probabilities are respectively 1/6 and 1/6 giving a total probability of 1/3 where switching loses. In the other 2/3 cases where you pick a goat switching wins the car (because the host is forced to reveal the other goat)

I understand the 2/3 vs 1/3 argument. But I am still having a hard time accepting it. Someone please help me with this. If someone walk in after the host revealed the 3rd door with the goat, and all they say where 2 doors, but the host tells him that there were 3 doors but the 3rd had a goat….or he could tell me he eliminated 1000 goat doors. Whatever the path that got me there, I am looking at 2 doors…..in the original scenario if I was going to stay with door #1 but suddenly have a seizure and forget what my first choice was, and pick door 2, do I have a 66% or 50% change of getting the car.It is almost like people are saying the idea of the door in my head is changing the odds?

If someone walks in after the door(s) has been opened and simply sees 2 closed doors then they have no information on which to base their decision and it becomes a simple coin flip (50/50). Similarly if you as the contestant forget which door you picked 1st then you’re in the same position – a 50/50 choice.

But the rules of the problem mean that you have a 2/3 chance of winning if you switch because you have some information. Monty has complete information and has 100% chance of picking the car (he knows where it is)

Your chances of winning depends on your knowledge of what protocol the host uses for opening the doors.

Consider the following 3 protocols:

i) The host has decided that whatever door you choose he will open a different door with a goat, and if he has a choice between two goats he will choose one of them with equal probability.

– This is the setting for the problem described in this post, and as several people have mentioned switching would give you a 2/3 chance of winning.

ii) The host has decided that whatever door you choose he will open a different door, and will do so at random.

– Since you were never told he would show you a goat, the chances are now 1/2 for you to win by switching (this is under the given circumstance that he picked a door with a goat and now you are presented with a choice between two doors)

iii) The host has decided that whatever door you choose he will open a different door with a goat. However if he has a choice between two goats, call them Alan and Bill, then he chooses Bill with probability b. If the host shows you Bill then your probability of winning by switching is 1/(1+b) (this is a nice exercise!)

When I first learned about the Monty Hall problem, I understood it. But later, I still knew the results but I forgot why they were what they were. Rewatching the video that first helped me learn about it, I understood once again why switching doubles your chances of winning. Lets see if I can explain it simply.

I’m sure everyone agrees that, originally, the probability of choosing the one car out of the three doors is 1/3, and choosing a goat is 2/3. If you stay with your original choice those are the final probabilities. But if you switch, they reverse.

If you chose the car (1/3), the host opens a door to one of the two goats, and you switch to the other door, you will get a goat. This means that you get a goat 1/3 of the time by switching because originally you picked a car (1/3) then switched.

On the other hand if you originally picked a goat (2/3), the host opens the only other goat door leaving you with only the car if you switch. This makes the probability of getting a car 2/3 when switching. This is because you have 2/3 probability of choosing a goat, then when switching will always get the car.

If you still don’t understand, you can watch the video I watched to help me understand. It’s on YouTube called “The Monty Hall Problem” uploaded by niansenx.

Ahi les va la ñonga (explicación muy clara de lo que está pasando):

Debemos tener en cuenta que el problema se divide en dos etapas (se le llama probabilidad condicionada): cuando se elige una de las tres puertas (probabilidad 1/3) y la otra cuando se destapó una puerta y se puede elegir entre sólo dos puertas.

Cuando sucede la primer etapa, podemos imaginar que tenemos muchas más puertas, digamos 1 millón, y una sola de ellas con el coche (las demás con cabras).

Elegimos una de ellas.

Por lo tanto lo más probable es que se haya elegido una cabra en este primera etapa. Es decir, es casi seguro que tengo una maldita cabra detrás de la puerta elegida.

Entonces el presentador elimina todas las puertas dejando sólo la escogida y una más.

Este es el truco del problema.

El presentador no está dejando al azar estas dos puertas. Lo hace sabiendo que quitó 999998 cabras, dejando sólo una cabra y un coche.

En este momento es cuando tomando en cuenta que hay una primer etapa previa que implicó que mi puerta elegida muy seguramente tiene ya detrás una cabra, al optar por el switcheo (donde ya sólo hay una puerta más y que lo más probable es que sea el coche) me dará casi la total certidumbre de acertar.

Si el problema se aisla eliminando la primer etapa, entonces daría lo mismo el switchear o no. El “truco” es el tomar en cuenta lo que sucedió en la primera etapa y que nos arrojó con una enorme probabilidad de traer una cabra en la puerta elegida.

Gracias a exagerar (1 millón) las opciones es que se puede mejor vislumbrar el origen del problema.

Lo puedo escribir en inglés pero me da gueva y tengo ya prisa por írmela a jalar un rato, jeje.

Saludos.

ESFM, IPN, Méx.

Yet another strange way to understand this:

1st Choice:

There is a p of 2/3 that you do not pick car.

2nd Choice:

There is then a p of 1/2 that you do not pick the car after switching.

Multiply the two probabilities to get p of 1/3 that you will not pick the car at the end.

A key to making the math work is to remember that not switching doors is a nonevent and as such will not change your original 1/3 probability of picking the car with the first door.

It’s obviously 50/50. Since Monty always removes a wrong door, that door can simply be removed from the equation, so it is no longer a 1/3 choice because that one obviously isn’t a possibility anymore. Therefore this only leaves 2 doors and a 50/50 chance of winning the car.

If you still disagree with that, imagine this: Monty reveals the losing door BEFORE you choose a door. Then it becomes more obvious that it’s a 50/50 chance. It is no different than if you choose a door before he reveals the losing one. The sequence of events in no way affects the probability.

“Monty’s Door” isn’t really Monty’s Choice. This is an important part of why it works. Given 1 car and, 2 goats and a player choosing at random. If the player picks a goat (2 out of 3 times), then Monty’s choice is constrained to the specific other door which holds goat#2. If the player picks the car at first (1/3 chance) then Monty can remove a door of “his” choice, but in all cases he’s removed an identical door: a goat door when you had the car.

The rules constrain Monty to opening a specific door, or flipping a coin between two identical choices.

The player is making real choices that affect the outcome, Monty never is.

Solution here:

Call the contestant’s choice Door 1, Monty’s goat Door 2 and the remaining ‘switch’ option Door 3.

The chance that before the game a car was put behind door 1 is 1/3. The chance that a car was put behind any other door is 2/3. When you see Monty’s goat, it’s a clue as to which other door that chance went to …

Door 1 was filled from a ratio of 1 car to 2 goats. Door 3 was filled from a ratio of 2 cars to 1 goat. It’s not random (50/50) because Monty peeps.

What everyone is missing, according to me, is that if Monty Hall chooses the door (say #1) with the goat, that should increase the chances that the car is behind #2 AND #3. Thus increasing chances that you chose correctly in the first time as well as increasing the chances of choosing correctly by switching.

The problem that both sides of the spectrum are having is consolidating two different factors into one problem.

Thus we have people on both sides using two completely different processes to explain something where the problem lies in the ability of the problem itself to align fractional mathematics and boolean logic.

To illustrate that both sets of people are both wrong and right in different ways:

The explanation of the issue is that the contestant is more likely to have chosen something “other than a car” originally. This is true. However this statement concludes that the real loss is in the initial choice. This is true.

What is lost in translation however is that you were never under any circumstances going to be allowed to pick “Monty’s” door. The problem itself has to follow its own rules which state that “Monty” is entitled to a door which is never going to be a door that you pick.

Mathematicians are using a choice which exists only under the assumption of the contestant assuming he can pick each door.

The contestant can pick (1) goat door – it does not matter which, because “Monty” is already assigned to the other (1) variable.

The contestant has thus the choice of “goat” or “car” because “Monty” will always be a placeholder for one of those goat doors.

You can never choose “Monty’s Door” and therein lies the problem with the fraction being represented as n/3 wherein n = 1 or 2.

To label this:

Monty’s Door = can only be assigned to him

Initial Choice = contestant choice

Switch Choice = contestant choice

How many choices do we really have available? It appears to be 3 if we include “Monty’s Door” however that is a constant not a variable. Monty will always own (1) goat, with the door being the “placeholder” for that goat. The goat he owns is independent of the door. He will . Always have a goat behind his door, rendering it impossible for you to “choose” that particular goat.

The problem here is simple:

With (3) doors and (1) prize you will always have a 1/3 chance of car and 2/3 chance of a goat.

Monty will always have a goat door so that door is not available for choice. Monty existing and owning a door in this entire problem has changed the denominator. The denominator has the illusion of being 3 so mathematicians are using the right operations and operands but the wrong denomination, due to the assumption that “Monty’s Door” was ever to be considered in the pool of options.

You only ever had two choices:

Stay or Switch

One will always be a goat and one will always be a car.

From those two axioms, the conclusion should be evident.

The choice was always:

Car on initial

Car on switch

Monty’s door always holds one goat. Its a choice you can never make.

Further:

The mental confusion arises when assigning fixed identifiers to rotating objects. There is never a door 1, 2, or 3 or A, B, or C. We assigned them identifiers to fix their positions, however “Monty” will never be assigned to A, B, or C because he has to go where a goat goes, not with a specific door.

Using fixed identifiers changes the way we view the problem. A variable cannot be a constant and a constant cannot be a variable. The two are boolean opposites.

It is an information problem, not a “mathematics” problem.

The problem that both sides of the spectrum are having is consolidating two different factors into one problem.

Thus we have people on both sides using two completely different processes to explain something where the problem lies in the ability of the problem itself to align fractional mathematics and boolean logic.

To illustrate that both sets of people are both wrong and right in different ways:

The explanation of the issue is that the contestant is more likely to have chosen something “other than a car” originally. This is true. However this statement concludes that the real loss is in the initial choice. This is true.

What is lost in translation however is that you were never under any circumstances going to be allowed to pick “Monty’s” door. The problem itself has to follow its own rules which state that “Monty” is entitled to a door which is never going to be a door that you pick.

Mathematicians are using a choice which exists only under the assumption of the contestant assuming he can pick each door.

The contestant can pick (1) goat door – it does not matter which, because “Monty” is already assigned to the other (1) variable.

The contestant has thus the choice of “goat” or “car” because “Monty” will always be a placeholder for one of those goat doors.

You can never choose “Monty’s Door” and therein lies the problem with the fraction being represented as n/3 wherein n = 1 or 2.

To label this:

Monty’s Door = can only be assigned to him

Initial Choice = contestant choice

Switch Choice = contestant choice

How many choices do we really have available? It appears to be 3 if we include “Monty’s Door” however that is a constant not a variable. Monty will always own (1) goat, with the door being the “placeholder” for that goat. The goat he owns is independent of the door. He will . Always have a goat behind his door, rendering it impossible for you to “choose” that particular goat.

The problem here is simple:

With (3) doors and (1) prize you will always have a 1/3 chance of car and 2/3 chance of a goat.

Monty will always have a goat door so that door is not available for choice. Monty existing and owning a door in this entire problem has changed the denominator. The denominator has the illusion of being 3 so mathematicians are using the right operations and operands but the wrong denomination, due to the assumption that “Monty’s Door” was ever to be considered in the pool of options.

You only ever had two choices:

Stay or Switch

One will always be a goat and one will always be a car.

From those two axioms, the conclusion should be evident.

The choice was always:

Car on initial

Car on switch

Monty’s door always holds one goat. Its a choice you can never make.

Further:

The mental confusion arises when assigning fixed identifiers to rotating objects. There is never a door 1, 2, or 3 or A, B, or C. We assigned them identifiers to fix their positions, however “Monty” will never be assigned to A, B, or C because he has to go where a goat goes, not with a specific door.

Using fixed identifiers changes the way we view the problem. A variable cannot be a constant and a constant cannot be a variable. The two are boolean opposites.

It is an information problem, not a mathematics problem.

zionsreveille

Posting the same comment twice doesn’t improve your argument – you’re still wrong,, Switching doors is a 2/3 chance. It’s mathematically proven and ALL the statistical data supports that proof.

If you pick a door at random from the 2 remaining you have a 50% chance of winning, but the question posed is “Is it advantageous to switch doors”, and the answer is unequivocably YES. It’s simple enough to prove it to yourself with 3 playing cards if you don’t understand the logic behind the answer.

I’m aware. That comment was made in December when I was still having trouble visualizing the problem. I ran some simulations and realized exactly how it works and why its 2/3… Hence, why I haven’t posted in a long while…having resolved that problem already mentally. Thanks for your patience though bearing with me in between, some of us learn differently than others, or need different visual aids.

Yeah William, it’s quite obviously just a silly fallacy invented by some bored statisticians, showing how easily some bad logic can lead to a complete divergence of math and reality.

except that it was indeed a REAL TELEVISION SHOW where people had a real opportunity to win a goat.

I am a dealer and have three cards: 1 king and two queens, your objective is to get the king.

I shuffle the cards and present them to you face down, you take one but are not allowed to look. There is a 1/3 chance you have drawn the king.

Both of us know, without even looking, I hold in my hand at least one queen. But if I gave you the opportunity to trade hands now – my two cards for your 1 card – would you take it? Of course you would, because my hand carries better odds at having the king.

I am required to offer you the chance to trade hands anyway, because that is a rule of the game, so does it make any difference if I peek at my hand of cards before offering you the trade? No because you still don’t know where the king is (even though I now do), and no odds have changed, since you are the one that gets to choose to trade or not.

Now before you make your choice, to demonstrate that indeed I do hold in my hand one queen I take one of the cards in my hand which I know to be a queen, flip it around and show it to you. Does it change the odds you are holding a king? No…because we both knew I was holding at least one queen from the moment you drew your card, I simply showed one to you. You would still want to trade hands, right?

Does it matter if that queen is part of the trade? It shouldn’t since a queen can’t win the game, so let’s burn it.

What has happened during all of this to improve the chance of the card you selected being a king from 1/3, to anything better? Nothing, however by eliminating the queen from my hand for you, I have improved the chance that the single card left in my hand is a king from 1/3 to 2/3.

That queen I burned only ever offered any value to me at the time you made your selection since it reduced your odds of drawing a king into your hand. My hand was always better than your hand and you’d be a fool to not trade.

But….your hand is 2/3, mine is 1/3, and you flamed your queen. Fair enough. Your hand is NO LONGER 2/3. Why? I am now presented with a NEW choice, irrespective of all other choices made in the past. My odds are NOW 50:50. They have changed in light of superior information…AND the choice I am given…Just as the would change for a new guy if he was offered MY card or YOUR card.

A new choice (perhaps a coin flip) should drive selection – not some arbitrary “switch now.” The reason all the simulators say “switch…” is they are erroneously, wrongly, and really rather stupidly, being programmed to state your hand is always 2/3 and stays that way if you burn your queen (with my knowledge that it has been burnt and thus one choice eliminated).

Said another way, my ability to choose again starts the problem over. The simulators are mis-programmed as fakes.

I think the easiest way to understand this is with an example

…. …. … … … … … … … .. … you.. . .. .host.. ..you. .. . .switch

door#1.. door#2.. door#3.. choose.. reveals.. stay… to door?

..car.. .. ..goat… .. .goat.. .. . ..#2.. .. .. .#3.. . lose… …#1-win

.goat.. . .. car.. .. …goat.. .. .. .#2… .. . .#1.. ..win.. .. ..#3-lose

.goat.. .. .goat.. . …car.. .. .. ..#2… .. .. #1.. ..lose.. .. .#3-win

.. … .. … … … … … … .. .. odds of winning.. . .1/3.. .. .. 2/3

or if it helps.. initially.

.. door#1.. odds of winning = 1/3

.. door#2.. odds of winning = 1/3

.. door#3.. odds of winning = 1/3

now we divide into 2 sets

… set#1 = door #1.. … … … .. .. .1/3 chance of winning

.. .set#2 = door #2 + door #3… . 2/3 chance of winning

by revealing door #2 or door #3 to NOT be the winning door, the host is giving you the entire set (door #2 + door #3) for the price of 1 door.

**********

here’s a variation you might like

.. 2 prizes behind 3 doors

.. you pick a door

.. host reveals a prize behind one of the other 2 doors

Should you switch?

what are the odds of winning a prize without switching and with switching?

Yeah, one thing I learned from the Monty Hall discussions is that the “odds” of the doors definitely DO change as you get new information. e.g. a lot of people get stuck on the “each door has a 1 in 3 chance, and opening doors doesn’t change that”. But …no door ever had a 1 in 3 chance. The “odds” are only an estimate based on your current level of information, and this can change individually per door as per the “rules”.

To Monty, one door has a 100% chance, and the other two doors had zero% chance. Monty knows the true odds, everyone else is only estimating.

To “Player 1” choosing at random from all 3 doors, each door has a 33% chance. One a goat is revealed, then that door drops from a 33% chance to zero% chance. After the contestant locks in their first door, its odds are locked in at 33%. After the contestant locks in their first door, the doors are divided into two sets now: doors you picked (33% chance of winning) and doors you didn’t pick (66% chance of winning). Each “set” of doors now has it’s own separate set of odds, which allows the odds per door to vary in unequal ways.

For a hypothetical “Player 2” who didn’t observe the first round, they can only assign 50/50 odds to the two remaining doors: because they lack information about the selection process. So here you had three people who all have different views on the final “odds” of the two remaining doors. For Monty: the doors are 100/0/0, for “Player 1”, they are 66/33/0, and for “Player 2” they are 50/50/0.

Nope. They are 50:50, summing to 1. We can NOT have two observers, irrespective of their pasts, experiencing different a priori odds….effectively, the problem starts over.

Wrong, and utterly so. In effect the game starts over when the choice of switch or stay is offered. The odds are 1/2 per door, period – and…here’s why. If Monty picks a new contestant, offers her the same two doors, pick one (and me the same two, under the theory we can choose the same door and both get cars) our odds ARE 50:50 and must be 50:50 because all previous odds are invalid. Period. Fundamentals state no two observers in the exact same situation, EXACT, may observe different odds for the exact same set of choices.

you should only switch if monty **intentionally** shows you a losing door. this is inconsistent with how the game is in fact played, which is why so many people “get it wrong”. I have watched many episodes and if monty lets you switch, it’s often to get you to **CHOOSE** the zonk. So on paper, as a probility question, strictly formulated, with monty **intentionally** showing you the losing door, yes you should switch. but this is **not** how the game is played