Monty Hall Problem

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loveoforganic

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For the love of God, I am so angry. My math has been violated, and I just want to go hit people out of spite toward this problem. If one of you math saavy people can explain how this doesn't violate the rules of the universe without using a convoluted proof that will make me want to choke you, I will forever be in your debt.

http://en.wikipedia.org/wiki/Monty_Hall_problem

If you don't know what this is, don't look. Save yourself the cognitive misery.
 
loveoforganic said:
For the love of God, I am so angry. My math has been violated, and I just want to go hit people out of spite toward this problem. If one of you math saavy people can explain how this doesn't violate the rules of the universe without using a convoluted proof that will make me want to choke you, I will forever be in your debt.

http://en.wikipedia.org/wiki/Monty_Hall_problem

If you don't know what this is, don't look. Save yourself the cognitive misery.
Click to expand...

What exactly is confusing you about it? It seems like that wikipedia page does a pretty good job of explaining it.
 
Intuitively, after the host shows the goat behind the door, the chances of a car being behind either of the two remaining doors is 50/50, but modeling shows this not to be the case. For there to be a 2/3 chance of picking the car by switching doors after the goat is shown, your initial pick must remain a 1/3 chance of having the goat. I don't buy it though. The odds change once one door is withdrawn from the selection.
 
Wish I hadn't read this cause it's going to bother me. How is it 2/3 by switching!? Only skimmed it but it doesn't make sense to me.
 
loveoforganic said:
Intuitively, after the host shows the goat behind the door, the chances of a car being behind either of the two remaining doors is 50/50, but modeling shows this not to be the case. For there to be a 2/3 chance of picking the car by switching doors after the goat is shown, your initial pick must remain a 1/3 chance of having the goat. I don't buy it though. The odds change once one door is withdrawn from the selection.
Click to expand...

Keep in mind though that the door that is withdrawn is not selected at random. It must be both a) not the door you picked and b) a door containing a goat.

In addition, the problem is not redefined after the first door is opened. That door is not 'removed' from the problem, you are just given information about what is behind the door.

I feel like I am not explaining this well...

How about this-since what is behind the doors is set and doesn't change you can do this:

Door 1: Goat
Door 2: Car
Door 3: Goat

If you pick Door 1, then the host will open door three and switching will cause you to win. Same if you pick Door 3. If you pick Door 2, switching will cause you to lose. So switching gets you 2/3 of the possible winning combinations.

Keep in mind too that the question isn't "What is the probability that the car is behind this door?" but rather "What is the probability that switching is the winning strategy?"
 
If you can explain this permutation of the problem, maybe it'll click with me (thank you for bearing with me by the way!).

You have 100 doors, labeled 1-100. You select door 1. The host opens doors 3 through 100, showing you a goat behind every single door. You're now given the option to keep your selection of door 1 or switch your selection to door 2.

Would the simulator spit out that switching to door 2 gives you a 99 percent chance of winning? Do you have only a 1 percent chance of winning in keeping your original selection?

Edit: I just read that this permutation is true. Ugh
 
The simulator would say that switching gives a 99% chance of winning.

Think of it this way. Which is more likely, that you randomly chose the door with the car (1/100 in the original problem) or that you chose incorrectly and the host was forced to leave you with the car behind an unopened door (99/100 times in the original problem). I think where the confusion is coming from is that you assume once all of the doors are opened, they don't matter anymore. They do though. Just because the host opened all of the other doors doesn't make your original guess more accurate (as it would if the probability collapsed to 50/50).

Edit: I should mention that until I read the explanation on Wikipedia, I would have said that the chances were 50/50 after door(s) were opened. I get the solution, but I wouldn't have thought of it myself.
 
It doesn't make your initial probability of guessing correctly any better as doors are eliminated, but it seems to increase the probability that you got "lucky" with that 1 in 100 guess. Is a critical factor in this that when the host eliminates doors 3 through 100, your selection was never in the running for being eliminated by the host, so your door was not in the sample that steadily gets better odds of containing the winning door?

If this gets too circular, you're welcome to call me a dummy and leave 🙂
 
The host's behavior is important, but I don't know how to explain why since the host is an integral part of the game. If the host could open the door you chose, maybe the probability would be different (I'm too tired to think it through) but then there would be no game. All I can say to make it simpler is that you have to look at the whole problem, not just the endpoint.
 
Well, I mapped out the scenarios on paper using a door a, door b, door c scenario, where the contestant swapped every time, which kind of helped. Here it is, in case it helps anyone else

Car Loc-Choice 1-------Eliminated-------Choice 2--Outcome
1) A-------A----------------C---------------B----------L
2) B-------A----------------C---------------B----------W
3) C-------A----------------B---------------C----------W
4) A-------A----------------C---------------B----------L
5) B-------A----------------C---------------B----------W
6) C-------A----------------B---------------C----------W
 
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i will explain it.

imagine you had 100 cards, and there was only ONE ace of spades in them. imagine i make two piles, one with 1 card, and one with 99. now, in the 99 pile, imagine i removed 98 cards that are NOT the ace of spades. now we have 2 piles of 1 card each.

CLEARLY the ace is in the originally larger pile. i only removed WRONG cards, and initially the ace had a 99% chance of being in the larger pile, and it still has a 99% chance of being in the 2nd pile once i remove 98 cards.

the problem is that the host only removes wrong choices. if there were a chance that he could reveal the car on accident, then yes, there would be a 50/50 chance of getting a car.
 
I have three apples. Two tasty. One tasty but poisonous. Never mind; I got hungry and now have two apples. One tasty. One tasty with poison. Which one do you want?

It's a 1:1 scenario, and more math is unlikely to change my mind.
 
I think the neat thing is it goes beyond mental masturbation "every statement I say is a lie"-type paradox - you'd live 50% more often swapping your apple than keeping it, all assumptions being the same.
 
ReptarBar said:
i will explain it.

the problem is that the host only removes wrong choices. if there were a chance that he could reveal the car on accident, then yes, there would be a 50/50 chance of getting a car.
Click to expand...

With respect, I'm not sure that's right. I had a teacher who explained that the odd probability was due to the fact that the host is actually, in a roundabout way, giving you BOTH of the other doors by allowing you to switch. If he had allowed you to pick both of the other doors, and ignore the one that was wrong, then the odds would clearly be 2/3, and that is actually exactly what he's doing.

If it were possible for the host to reveal a prize the odds should be unchanged (after he fails to do so, of course, but before that the chance that he gives away the game actually helps the player).

Say the player chooses door 1 and the door 3 is the winning door. There's a 50/50 chance the host will give away the prize door, and then the player should obviously switch, and a 50/50 chance the puzzle will proceed as normal, with the player having a 2/3 chance by switching. The intentionality of the host, and the previous chance that he would reveal the prize, are now irrelevant.
 
loveoforganic said:
I think the neat thing is it goes beyond mental masturbation "every statement I say is a lie"-type paradox - you'd live 50% more often swapping your apple than keeping it, all assumptions being the same.
Click to expand...

I still don't see how this is different from: Hey, I have 10^23 things; pick one. Haha...just kidding. It's two things. Pick one.

I have a six chambered-revolver. There is one round in it. Give it a spin and fire. No, it only has two chambers. Give it a spin and fire.

You want one of my three wives/husbands? Too bad. I only have two. Pick one.

Heads: I win. Tails: You lose. Side of coin: Had the sides removed.

I only see 1:1 scenarios.
 
Suppose you have a bushel of apples, most poisonous, some not. You take out one at random to eat, then a native, who can actually tell the difference between the good and bad apples, looks at your bushel and throws out a few bad ones.

Do you eat the apple in your hand or pick out a new one?
 
Diallelus said:
Suppose you have a bushel of apples, most poisonous, some not. You take out one at random to eat, then a native, who can actually tell the difference between the good and bad apples, looks at your bushel and throws out a few bad ones.

Do you eat the apple in your hand or pick out a new one?
Click to expand...

So this person leaves me with two apples...one good and one bad.
 
I didn't mean for it to be just like the Monty Hall problem, just a similar idea. The point is that if you take advantage of new information your chances increase.

So let's say you do have two apples, one good and one bad. You can't assume that the odds are 50/50 for each one, not when we have other information available. The presence of two choices doesn't always imply that the odds are even (consider a boxing match between The Rock and Anna Paquin).

In this case one apple is more likely to be good than the other, because one was chosen based on pure chance, and the other based on chance mixed with some pertinent information. The educated guess has a 2/3 chance of being right, and the uneducated guess only a 1/3.
 
Diallelus said:
I didn't mean for it to be just like the Monty Hall problem, just a similar idea. The point is that if you take advantage of new information your chances increase.

So let's say you do have two apples, one good and one bad. You can't assume that the odds are 50/50 for each one, not when we have other information available. The presence of two choices doesn't always imply that the odds are even (consider a boxing match between The Rock and Anna Paquin).

In this case one apple is more likely to be good than the other, because one was chosen based on pure chance, and the other based on chance mixed with some pertinent information. The educated guess has a 2/3 chance of being right, and the uneducated guess only a 1/3.
Click to expand...

This makes perfect sense
 
Diallelus said:
With respect, I'm not sure that's right. I had a teacher who explained that the odd probability was due to the fact that the host is actually, in a roundabout way, giving you BOTH of the other doors by allowing you to switch. If he had allowed you to pick both of the other doors, and ignore the one that was wrong, then the odds would clearly be 2/3, and that is actually exactly what he's doing.

If it were possible for the host to reveal a prize the odds should be unchanged (after he fails to do so, of course, but before that the chance that he gives away the game actually helps the player).

Say the player chooses door 1 and the door 3 is the winning door. There's a 50/50 chance the host will give away the prize door, and then the player should obviously switch, and a 50/50 chance the puzzle will proceed as normal, with the player having a 2/3 chance by switching. The intentionality of the host, and the previous chance that he would reveal the prize, are now irrelevant.
Click to expand...

I don't know how this answer is different from reptarbar's.. but anyhow

The benefit of switching (1/3 probability -> 2/3 probability) comes from the fact that the host will always open the "false" door.

Let's say you pick a door, then it's got 1/3 probability of having the prize. The other two doors have 2/3 probability of having the prize. Among these two doors, the host will always reveal the door without the prize, so whichever door is remaining will be assigned 2/3 probability all to itself.

Hence the benefit of switching. you gain 1/3 probability of winning just by switching.

If the host was unaware of the whereabouts of the prize, there's no benefit of switching.
 
For those who can't accept that when you actually work it out, you win 2/3 of the time if you switch, think of it this way:

When the host asks you, “Do you want to switch?” What he is really asking you is, “do you think you guessed it correctly the first time?” or “Do you think you guessed wrong?”

For all of us, if given three doors to choose, we know our chances of guessing correctly are not that great. We are more likely to guess incorrectly then to guess correctly. When the host removes one of the doors, there are two options left:

1. You guessed right initially
Or
2. You guessed incorrectly initially

There is no arguing that from the start, given three doors to choose from, you are more likely to guess incorrectly than to guess correctly. Thus, when the host asks you if you think your first guess was wrong, you should say yes. That is what he is really asking by saying “Do you want to switch doors?” If you switch doors, you think your first guess was wrong. If you don’t switch doors, you think your first guess was right.
 
I like this thread because I think I finally get it now. Big thanks to Astareal for that! 👍
 
The frustration many have with the problem is that switching your choice will lead to failure sometimes. 1/3 of the times, to be precise.

It's just like putting in a pinch hitter. The manager takes out a scrawny shortstop with a .222 batting average for a power hitter with a .333 batting average. The .333 hitter strikes out. Should the manager have kept the .222 guy in? It is possible that the .222 guy would have gotten a hit.
 
I'm so glad this question got posted! I've seen this so many times, and never really understood the reasoning behind the right answer. Hell, now I think I could teach it!🙂
 
kupa said:
If the host was unaware of the whereabouts of the prize, there's no benefit of switching.
Click to expand...

I'm sorry, but I don't think so. Intentionality isn't relevant to statistics. So long as one door is opened, and that door is a "no prize" door, the standard "make the switch" argument applies.

Suppose the host doesn't know where the prize is, and opens one of the doors not chosen at random. If this door happens to be a no-prize door, then the player's first chosen door has a 1/3 chance of being right (because it was randomly guessed from 3 choices) and the other door has a 2/3 chance of being right because it is the only other choice.

If the host opens a prize door, I don't know what happens next, because the article doesn't allow this possibility. But the important point is that the possibility of this happening doesn't alter the probability in the former case.
 
This is a poor explanation, but here's a diagram they use in attempt to simplify the proof:

197px-Monty_open_door_chances.svg.png


I believe the catch is that the contestant does not change his assumption of equal probabilities to each door even after the first door is opened (which is stupid to me). I agree with OP, that 1/2 is more intuitive.
 
aSagacious said:
I believe the catch is that the contestant does not change his assumption of equal probabilities to each door even after the first door is opened (which is stupid to me). I agree with OP, that 1/2 is more intuitive.
Click to expand...

Suppose you have one genuine 100$ bill in with a pile of 1000 counterfeit 100$ bills. You remove one bill from the pile arbitrarily. A fraud expert comes in and throws away 990 counterfeit bills. Your friend then chooses one of the remaining 10 bills arbitrarily.

Is it intuitive to believe that both bills have an even probability of being genuine? Your friend's bill has survived a rigorous selection process, while your bill only survived a random guess.

One last example: 1000 people enter a wrestling tournament. The judges choose one person randomly to be removed from the normal bracket and set aside. The champion of the tournament then fights this random person. Who is more likely to win? The person who has bested 998 opponents or the one chosen at random?

EDIT: in case it wasn't clear, these examples are relevant because the Monty Hall problem is the same issue, just on a smaller scale. You have 3 items, and you choose between a random item from the pool and an item that withstood an elimination process. (Note that a random elimination process is still a selective process if the results are available to you, as they are here. Even if the host doesn't know which door has the prize, the fact that the door opens to reveal a goat tells you all you need to know: that a no-prize item has been removed from the set "doors I didn't choose.")
 
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Diallelus said:
One last example: 1000 people enter a wrestling tournament. The judges choose one person randomly to be removed from the normal bracket and set aside. The champion of the tournament then fights this random person. Who is more likely to win? The person who has bested 998 opponents or the one chosen at random?
Click to expand...

Is one of them named Steve Jennum?
 
Diallelus said:
Suppose you have one genuine 100$ bill in with a pile of 1000 counterfeit 100$ bills. You remove one bill from the pile arbitrarily. A fraud expert comes in and throws away 990 counterfeit bills. Your friend then chooses one of the remaining 10 bills arbitrarily.

Is it intuitive to believe that both bills have an even probability of being genuine? Your friend's bill has survived a rigorous selection process, while your bill only survived a random guess.
Click to expand...

Regardless of how the first bill was selected, we know it still has 1/10 odds of being legit.

-We know that there is exactly one legit bill in the pile.
-We know that there are 1000 bills
-We know that 990 of the 1000 bills were removed already and identified as counterfeit, leaving only 10 bills that are possibly legit

...therefore, if our bill is not in the pile of already established counterfeits, then we needn't consider the counterfeit pile when considering the remaining candidates.

If the contestant is presented new information, it would absolutely change the probability.
 
Hey all:

Pretty solid with math (up to and some PDE's). It's quite simple when you think about it. If you chose door 1 you have a 33.3% chance of getting it correct. Then the host shows door 3 as the wrong choice. Now, if you switch and LOSE, you know your initial guess was right with a 33.3% chance of winning, thus your losing strategy had to have 1 - 33.3% = 66.7% (2/3). Therefore, switching does, in fact, give you a 2/3 chance of winning. Looks like the wikipedia site goes through the proof with Bayes' theorem (a very pwerful relationship, I might add) for those of you more statistically inclined.
 
Navier said:
If you chose door 1 you have a 33.3% chance of getting it correct.
Click to expand...
Initial conditions

Navier said:
Then the host shows door 3 as the wrong choice.
Click to expand...
New information is presented. Thus, reevaluate probability (which is now 1/2)

Navier said:
Now, if you switch and LOSE, you know your initial guess was right with a 33.3% chance of winning, thus your losing strategy had to have 1 - 33.3% = 66.7% (2/3).
Click to expand...
Initial conditions are no longer relevant
 
loveoforganic said:
Intuitively, after the host shows the goat behind the door, the chances of a car being behind either of the two remaining doors is 50/50, but modeling shows this not to be the case. For there to be a 2/3 chance of picking the car by switching doors after the goat is shown, your initial pick must remain a 1/3 chance of having the goat. I don't buy it though. The odds change once one door is withdrawn from the selection.
Click to expand...
I thought 50/50 as well. I wish I wouldn't have read it lol
 
aSagacious said:
That's what I'm saying, which is why the probability should be 1/2 instead of 1/3 or 2/3.
Click to expand...

But new information does not retroactively change probability, it only changes your current choices. If you choose 1 bill out of a thousand, then that bill has a 1/1000 chance to be correct. What happens to the remaining bills isn't relevant to this probability. When your friend chooses his bill he has a 1/10 chance, because he's acting on new information. He, in essence, gets to choose 991 bills from the 1000, because the examiner removing 990 bills is statistically equivalent to your friend guessing 990 times, and getting it wrong each time.

Similarly, the host is giving you the option of choosing both of the remaining doors. Revealing 1 wrong door is equivalent to letting you guess one of the two other doors, revealing it to be wrong, and then letting you pick again. Your odds are 2/3 because the situation is equivalent to opening both other doors.

EDIT: "New information is presented. Thus, reevaluate probability (which is now 1/2)"
The problem with this is that you can't assume 50/50 odds based only on 2 possible outcomes. 50/50 odds are default only in the absence of information. That is, the odds of a coin flip being heads are 50/50 because there are so many unknown variables in the coin's flight path. The more information we have about the coin (for example, the weight on each side) the more we can say about the probability. In this case, the relevant information is that one door was chosen at random, and the other after other choices were eliminated. You cannot ignore this relevant information.
 
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Diallelus said:
...Similarly, the host is giving you the option of choosing both of the remaining doors. Revealing 1 wrong door is equivalent to letting you guess one of the two other doors, revealing it to be wrong, and then letting you pick again. Your odds are 2/3 because the situation is equivalent to opening both other doors.
Click to expand...

So, once the host opens the door and reveals that it is a dud, he gives you another chance to pick a door. This now becomes an entirely different scenario that needs to be treated as such (therefore disregard the prior scenario completely)

The 1/2 probability that I am describing only pertains to the new choice (not the initial one). In the moment that I am selecting between doors after the goat has been revealed I am now only concerning myself with the remaining two doors. Why do you suggest that the host is 'letting you choose both of the remaining doors' (bolded above)?
 
Diallelus said:
In this case, the relevant information is that one door was chosen at random, and the other after other choices were eliminated. You cannot ignore this relevant information.
Click to expand...

Why does this matter? After a goat is revealed, all that tells you is that the revealed door is no longer valid, nothing else.

You are then left with a new choice: between the two unopened doors (that you still know nothing about) what is the chance that either one of them have the car? The fact that the host removed one door or a thousand doors is irrelevant, so long as you have a final number of possible doors coming into your final choice.
 
aSagacious said:
So, once the host opens the door and reveals that it is a dud, he gives you another chance to pick a door. This now becomes an entirely different scenario that needs to be treated as such (therefore disregard the prior scenario completely)

The 1/2 probability that I am describing only pertains to the new choice (not the initial one). In the moment that I am selecting between doors after the goat has been revealed I am now only concerning myself with the remaining two doors. Why do you suggest that the host is 'letting you choose both of the remaining doors' (bolded above)?
Click to expand...
\

How about this...

After your decision (choice A), there is a 2/3 chance it is behind the other two doors. The host only has two choices to reveal (B and C).

1/2 of the time he can open B to show a goat
1/2 of the time he can open A to show a goat

STILL, 2/3 of the time, the goat will be in B OR C, which he can't touch for obvious reasons.

The 2/3 probability remains, as seen from the host's perspective, which the player knows.
 
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aSagacious said:
So, once the host opens the door and reveals that it is a dud, he gives you another chance to pick a door. This now becomes an entirely different scenario that needs to be treated as such (therefore disregard the prior scenario completely)

The 1/2 probability that I am describing only pertains to the new choice (not the initial one). In the moment that I am selecting between doors after the goat has been revealed I am now only concerning myself with the remaining two doors. Why do you suggest that the host is 'letting you choose both of the remaining doors' (bolded above)?
Click to expand...

A) It is true that you are only concerning yourself with two doors when deciding whether or not to switch. But it doesn't follow that the odds are 50/50. You have two choices, but information is available that tips the odds in one of the door's favor.

B) The host is letting you choose both doors, in a sense, because one of the two doors not chosen was bound to be a no-prize door anyway. If he had let you open both doors you would have a 1/3 chance of winning with the first door, and a 1/3 chance of winning with the second door, but one of them had to be a no-prize door, and it doesn't matter whether he shows you which one before or after you choose to switch (as long as he means switching to both doors).
 
sEgacious,

Do you understand that your blind insistence on conventional wisdom, if applied to real life situations would lead to failure for you?

When given proof that the technique of switching your answer will lead to success more often you still stubbornly proclaim incorrectly that each door has a 50% probability of having the car behind it? The goal is to have as much control over your outcome as possible, not narrow it down to a "50/50" guess.

As I have said before: switching doors is an acknowledgment that you were most likely wrong in your initial guess. When guessing between three options, you will always be most likely wrong on your first guess, likely correct on your second guess, and definitely correct by your third guess..

When trying to guess which of three doors something is hidden behind, it will take anywhere between 1 to 3 guesses. Sometimes you will get it on your first guess. Sometimes you will get it on your second. Sometimes you will get it on your third. By switching your choice, you are banking on the car being found in what would have been either your 2nd or 3rd guess.

You must realize that there is no way to guarantee success. Following the formula will fail 33% of the time. You aren't ever going to come up with an answer in your head that satisfies you as making sense. In the end, you are still at the mercy of chance/probabilities.

But just because you are frustrated and stubborn, you shouldn't sell yourself short by thinking that 66/33 is the same as 50/50. There is a noticeable difference. I see football coaches and baseball managers make the same mistake as you all the time. It is shooting yourself in the foot.
 
boo5000 said:
After your decision (choice A), there is a 2/3 chance it is behind the other two doors. The host only has two choices to reveal (B and C).
Click to expand...
Correct

boo5000 said:
1/2 of the time he can open B to show a goat
1/2 of the time he can open A to show a goat
Click to expand...
But, regardless of which door you choose initially, the host can and will ALWAYS reveal a goat. After doing so he gives you a new choice.

boo5000 said:
STILL, 2/3 of the time, the goat will be in B OR C, which he can't touch for obvious reasons.
Click to expand...
I'm not sure I know what you're trying to say here.

boo5000 said:
The 2/3 probability remains, as seen from the host's perspective, which the player knows.
Click to expand...

Perhaps I'm misunderstanding the possible actions of the contestant. Here's my perception, please correct me if I am mistaken.

1) Contestant chooses a door
2) Host reveals one of the other doors, always revealing a goat
3) Contestant chooses again between two non-revealed doors

... am I mistaken?
 
If you think like sEgacious, EVERYTHING is 50/50! There is a 50 percent chance of rain today! There is a 50 percent chance of getting into Harvard for crying out loud as long as you apply!! You either get in or you don't! 50%, right?!?
 
I had a professor that would give this problem to every class he had. After 20ish years of teaching, only one student ever figured it out on his own.

To those that are insisting that after opening the door with a goat, there is still a 50:50 chance you will get a car regardless of if you switch, make three cards, one with a car and two with goats. Grab a friend. Perform experiment. Have friend switch for 100 trials, stay for 100 trials. You'll see.
 
How about this-since what is behind the doors is set and doesn't change you can do this:

Door 1: Goat
Door 2: Car
Door 3: Goat

If you pick Door 1, then the host will open door three and switching will cause you to win. Same if you pick Door 3. If you pick Door 2, switching will cause you to lose. So switching gets you 2/3 of the possible winning combinations.

Keep in mind too that the question isn't "What is the probability that the car is behind this door?" but rather "What is the probability that switching is the winning strategy?"
Click to expand...

👍 I think this is the simplest way of explaining it...it was the one that made sense to me. I think the real problem is that it's hard to think of the problem in the overall sense rather than the immediate decision of switching doors, which will lead you to conclude that the odds are 50/50. You have to look all the possible combinations you could have picked....thus, if the host opens a door with a goat there is a 2/3 chance you have also picked a door with a goat and only a 1/3 chance you have picked the door with the car. The card explanation (with the 99 card deck and 1 card deck) helps to explain it a lot too.
 
yankeesmed said:
If you think like sEgacious, EVERYTHING is 50/50! There is a 50 percent chance of rain today! There is a 50 percent chance of getting into Harvard for crying out loud as long as you apply!! You either get in or you don't! 50%, right?!?
Click to expand...

Precisely, and that is why I play the lotto 😀
 
aSagacious said:
Correct


But, regardless of which door you choose initially, the host can and will ALWAYS reveal a goat. After doing so he gives you a new choice.


I'm not sure I know what you're trying to say here.



Perhaps I'm misunderstanding the possible actions of the contestant. Here's my perception, please correct me if I am mistaken.

1) Contestant chooses a door
2) Host reveals one of the other doors, always revealing a goat
3) Contestant chooses again between two non-revealed doors

... am I mistaken?
Click to expand...

Not exactly; contestant chooses whether to stay with original door or switch to new door. Staying with the original door will yield a goat in 33% of situations, switching will yield a car 66% of the time.

If you brought an entirely new contestant in who was not aware of which door was originally chosen, that contestant has a 50/50 shot between the two remaining doors.

The entire problem is dependent on the fact that Monty Hall will not open the door that the contestant has chosen, and he will not open the door with the car behind it.
 
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