Do you know the correct answer?

[Tom Montgomery]

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Suppose you're a contestant on the game show Let's Make A Deal. You are given the choice of three doors: behind one door is a car; behind the other two, goats.

You pick a door, say #1, and the host (who knows what's behind all the doors) opens another door, say #3, revealing a goat. Then he says, "Before I open the door with the car... you may stick with your first choice of door #1 or you may now switch your choice to door #2. "

Is it to your advantage to switch to door #2?

No fair searching the internet for the answer. Give your reasoning. Enjoy.
.

[Peerie Maa]

It makes no odds. It is a 50:50 chance of winning either way.
Before the host revealed the goat it was a 1 in 3 chance, now the odds are improved, but you still have no way of knowing which is the best bet.

I am reminded of the hoary old conundrum involving two doors, each with a guard. One of the guards is known to be a liar, but you do not know which. You are allowed to ask only one guard one question. What question do you put to determine which door to go through?

[Tom Montgomery]

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Anyone else care to provide an answer?

The problem was originally posed (and solved) in a letter by Steve Selvin to the magazine American Statistician in 1975. It later became famous as a question from reader Craig F. Whitaker's letter quoted in Marilyn vos Savant's "Ask Marilyn" column in Parade magazine in 1990. At the age of ten Marilyn vos Savant registered the highest ever IQ score of 228. After vos Savant's published solution, approximately 10,000 readers, including nearly 1,000 with PhDs, wrote in reply to the magazine, most of them calling vos Savant wrong. Many of the letter writers chastised vos Savant for irresponsibly promoting mathematical illiteracy.

There are only two possible answers 1) No, switching is not advantageous; or 2) Yes, switching is advantageous. You have a 50-50 chance of hitting on the correct answer. But You also must supply your reasoning, as did Nick.

I'll supply the answer with the reasoning after a few more of you give your own.
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[ron ll]

As counterintuitive as it sounds, there is an advantage to switching. Unfortunately I don’t remember the reasoning.

[CWSmith]

The only advantage to switching would be if the host sometimes offered the chance to switch and sometimes did not. There could be a bias in that. Otherwise, I agree with Nick and it's 50:50.

[Tom Montgomery]

^
ron ll is correct. The answer is indeed counter-intuitive which makes it an interesting puzzle.

Intuition says that when it was revealed that an unchosen door did not contain the prize the odds changed to 50-50 and so switching the choice of doors is not advantageous.

But the fact is that the odds never change. Given a choice of three doors, there is a 2/3 chance the prize is behind one of the other two. That does not change when it is revealed that one of the unchosen two doors does not have the prize. The odds remain only 1 in three that the prize is behind your chosen door. And the odds also remain 2 in 3 that the prize is behind one of the other two doors.

[Ron Williamson]

It also depends if you prefer goats to cars.
Just sayin
R

[CWSmith]

Tom, your counting statistics are not correct.

Take, for instance, the first line. "Monty shows the prize is not behind door 2 or 3." This should actually be two separate lines "Monty shows the prize is not behind door 2." and then as another outcome "Monty shows the prize is not behind door 3."

Try it this way and I think you will find it remains a 50:50 outcome.

[Tom Montgomery]

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From Wikipedia (I changed one sentence to bold):

Vos Savant wrote in her first column on the Monty Hall problem that the player should switch. She received thousands of letters from her readers – the vast majority of which, including many from readers with PhDs, disagreed with her answer. During 1990–1991, three more of her columns in Parade were devoted to the paradox. Numerous examples of letters from readers of vos Savant's columns are presented and discussed in The Monty Hall Dilemma: A Cognitive Illusion Par Excellence. The discussion was replayed in other venues (e.g., in Cecil Adams' "The Straight Dope" newspaper column) and reported in major newspapers such as The New York Times.

In an attempt to clarify her answer, she proposed a shell game to illustrate: "You look away, and I put a pea under one of three shells. Then I ask you to put your finger on a shell. The odds that your choice contains a pea are 1/3, agreed? Then I simply lift up an empty shell from the remaining other two. As I can (and will) do this regardless of what you've chosen, we've learned nothing to allow us to revise the odds on the shell under your finger." She also proposed a similar simulation with three playing cards.

Vos Savant commented that, though some confusion was caused by some readers' not realizing they were supposed to assume that the host must always reveal a goat, almost all her numerous correspondents had correctly understood the problem assumptions, and were still initially convinced that vos Savant's answer ("switch") was wrong.

When first presented with the Monty Hall problem, an overwhelming majority of people assume that each door has an equal probability and conclude that switching does not matter. Out of 228 subjects in one study, only 13% chose to switch. In his book The Power of Logical Thinking, cognitive psychologist Massimo Piattelli Palmarini writes: "No other statistical puzzle comes so close to fooling all the people all the time [and] even Nobel physicists systematically give the wrong answer, and that they insist on it, and they are ready to berate in print those who propose the right answer." Pigeons repeatedly exposed to the problem show that they rapidly learn to always switch, unlike humans.

Most statements of the problem, notably the one in Parade, do not match the rules of the actual game show and do not fully specify the host's behavior or that the car's location is randomly selected. Krauss and Wang conjecture that people make the standard assumptions even if they are not explicitly stated.

Although these issues are mathematically significant, even when controlling for these factors, nearly all people still think each of the two unopened doors has an equal probability and conclude that switching does not matter. This "equal probability" assumption is a deeply rooted intuition. People strongly tend to think probability is evenly distributed across as many unknowns as are present, whether it is or not.

The problem continues to attract the attention of cognitive psychologists. The typical behavior of the majority, i.e., not switching, may be explained by phenomena known in the psychological literature as:

1. The endowment effect, in which people tend to overvalue the winning probability of the already chosen – already "owned" – door.
2. The status quo bias, in which people prefer to stick with the choice of door they have already made.
3. The errors of omission vs. errors of commission effect, in which, all other things being equal, people prefer to make errors through inaction (Stay) as opposed to action (Switch).

Experimental evidence confirms that these are plausible explanations that do not depend on probability intuition. Another possibility is that people's intuition simply does not deal with the textbook version of the problem, but with a real game show setting. There, the possibility exists that the show master plays deceitfully by opening other doors only if a door with the car was initially chosen. A show master playing deceitfully half of the times modifies the winning chances in case one is offered to switch to "equal probability".

[Tom Montgomery]

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I came across "The Monty Hall Problem" yesterday and found it fascinating. I too first thought that there was no advantage in switching doors.

[CWSmith]

Probability is the counting of possible outcomes. You have lumped multiple outcomes together and called them one.

Vos Savant wrote in her first column on the Monty Hall problem that the player should switch. She received thousands of letters from her readers – the vast majority of which, including many from readers with PhDs, disagreed with her answer.

Trust the PhDs, Tom. Savant is wrong.

[CWSmith]

I think the chart I posted shows all the possible outcomes.

Yes, but it does not count them correctly.

[Tom Montgomery]

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It is called "The Monty Hall Problem" and there is no longer any doubt but that switching is advantageous. I'm not a professional mathematician. But I trust their conclusions.

[CWSmith]

Be well, Tom. I have no skin in the game. I stand by my answer.

[ron ll]

Schrodinger’s goat.

[Peerie Maa]

By opening a door, the host changed the game. It went from one in three to one in two. Different game, different probabilities.
This is a non sequitur, the car is where it is:

Most statements of the problem, notably the one in Parade, do not match the rules of the actual game show and do not fully specify the host's behavior or that the car's location is randomly selected.

[Paul Pless]

42 ​...

[Tom Montgomery]

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George Sachs, PhD in Mathematics of George Mason University wrote to vos Savant: "You blew it! Let me explain: If one door is shown to be a loser, that information changes the probability of either remaining choice — neither of which has any reason to be more likely — to 1/2. As a professional mathematician, I’m very concerned with the general public’s lack of mathematical skills. Please help by confessing your error and, in the future, being more careful."

Paul Erdős, one of the most prolific mathematicians in history, remained unconvinced until he was shown a computer simulation demonstrating vos Savant's predicted result.

Magicians and con men rely on the psychology of the average human being to guess the odds incorrectly. The actual Monty Hall figured out the correct answer.

Mr. Hall said he was not surprised at the experts' insistence that the probability was 1 out of 2. "That's the same assumption contestants would make on the show after I showed them there was nothing behind one door," he said. "They'd think the odds on their door had now gone up to 1 in 2, so they hated to give up the door no matter how much money I offered. By opening that door we were applying pressure. We called it the Henry James treatment. It was 'The Turn of the Screw.' "

Mr. Hall said he realized the contestants were wrong, because the odds on Door 1 were still only 1 in 3 even after he opened another door. Since the only other place the car could be was behind Door 2, the odds on that door must now be 2 in 3.

Sitting at the dining room table, Mr. Hall quickly conducted 10 rounds of the game as this contestant tried the non-switching strategy. The result was four cars and six goats. Then for the next 10 rounds the contestant tried switching doors, and there was a dramatic improvement: eight cars and two goats. A pattern was emerging.

"So her answer's right: you should switch," Mr. Hall said, reaching the same conclusion as the tens of thousands of students who conducted similar experiments at Ms. vos Savant's suggestion. That conclusion was also reached eventually by many of her critics in academia, although most did not bother to write letters of retraction. Dr. Sachs, whose letter was published in her column, was one of the few with the grace to concede his mistake.

"I wrote her another letter," Dr. Sachs said last week, "telling her that after removing my foot from my mouth I'm now eating humble pie. I vowed as penance to answer all the people who wrote to castigate me. It's been an intense professional embarrassment."

Actually, many of Dr. Sachs's professional colleagues are sympathetic. Persi Diaconis, a former professional magician who is now a Harvard University professor specializing in probability and statistics, said there was no disgrace in getting this one wrong.

"I can't remember what my first reaction to it was," he said, "because I've known about it for so many years. I'm one of the many people who have written papers about it. But I do know that my first reaction has been wrong time after time on similar problems. Our brains are just not wired to do probability problems very well, so I'm not surprised there were mistakes."

https://www.nytimes.com/1991/07/21/u...nd-answer.html

[L.W. Baxter]

what if there are two contestants, each chooses a different door.

do they both benefit by switching after the third door is revealed.

[Peerie Maa]

Mr. Hall said he realized the contestants were wrong, because the odds on Door 1 were still only 1 in 3 even after he opened another door. Since the only other place the car could be was behind Door 2, the odds on that door must now be 2 in 3.

This is untrue. Proving that the car is not behind door 3 does not move the car from door 1 to door 2. Opening door 3 does not affect what may or may not be behind doors 1 & 2.

[CWSmith]

I have tried to show how the table in #6 is wrong, but I can't produce an image that is small enough to post.

All you need to do is to break every line that lists 2 outcomes (like "2 or 3") into two separate lines and you get 50:50 outcomes.

[David G]

I've encountered the problem before. What it shows is that statistics are tricksy. The average person is not going to answer this problem on the fly. It requires running some simulations.

These days, it's less critical to understand how statistics can be used to mislead. Because the liars are no longer trying to be 'serious' so much as just trotting out the BigLie and other Goebbels techniques. But it never hurts to read (or revisit) --

https://www.goodreads.com/book/show/...ith_Statistics

How to Lie with Statistics -- Darrell Huff


Darrell Huff runs the gamut of every popularly used type of statistic, probes such things as the sample study, the tabulation method, the interview technique, or the way the results are derived from the figures, and points up the countless number of dodges which are used to fool rather than to inform.

[Peerie Maa]

@ CW
If you go Advanced there is a table function that allows you to construct your table within the post

Contestant pick	Car behind	Host opens	Possible door	odds
1	1	2	1 or 3	50-50
1	2	3	1 or 2	50-50
1	3	2	1 or 3	50-50

[CWSmith]

Okay, Nick, I'll see if this old dog can learn a new trick:

Contestant Picks this Door	Prize is Behind This Door	Monty Shows This Door	Stays	Switches
1	1	2	Win	Lose
1	1	3	Win	Lose
1	2	3	Lose	Win
1	3	2	Lose	Win
2	1	3	Lose	Win
2	2	1	Win	Lose
2	2	3	Win	Lose
2	3	1	Lose	Win
3	1	2	Lose	Win
3	2	1	Lose	Win
3	3	1	Win	Lose
3	3	2	Win	Lose
		Win %	0.5	0.5

Okay, Tom, I told you what I thought was wrong with your chart. Now I'd like you to tell me what is wrong with mine.

Edit - Sorry. I had a typo that I corrected.

[peb]

I am answering without scrolling down the page and viewing (I know you guys have to trust me).
So, either I picked the car (a 33% chance) or I picked the goat (a 66% chance)
If I picked the car, changing my pick gives me a 0% chance of winning.
If I picked the goat, changing my pick gives me a 100% chance of winning (the host had to pick the other goat, so its out of the picture).

So I would switch.

[peb]

@ CW
If you go Advanced there is a table function that allows you to construct your table within the post

Contestant pick	Car behind	Host opens	Possible door	odds
1	1	2	1 or 3	50-50
1	2	3	1 or 2	50-50
1	3	2	1 or 3	50-50

The reason why this is not the same is because its a two step process where the host takes one goat out of the picture after the contestant eliminated one door from consideration. Its a two step process, the only way to lose if you switch is if you picked the car in the first place, but you only had a 33% chance of doing that. The host has to open the door with a goat, one of them is gone. But that is AFTER I have already had a 66% chance of removing the other goat from the picture. This is not even conter-intuitive in my mind. Its pretty simple. As long as I did NOT pick the car, I am guaranteed to win. But I had a 33% chance of picking the car.

You are simply betting you were wrong on your first guess. Thats a pretty good bet.

[Peerie Maa]

The reason why this is not the same is because its a two step process where the host takes one goat out of the picture after the contestant eliminated one door from consideration. Its a two step process, the only way to lose if you switch is if you picked the car in the first place, but you only had a 33% chance of doing that. The host has to open the door with a goat, one of them is gone. But that is AFTER I have already had a 66% chance of removing the other goat from the picture. This is not even conter-intuitive in my mind. Its pretty simple. As long as I did NOT pick the car, I am guaranteed to win. But I had a 33% chance of picking the car.

You are simply betting you were wrong on your first guess. Thats a pretty good bet.

You are correct, it is a two-step process.
Step 1, pick a door.
Step 2 host shows you that the car is behind the door you picked or the third door.
The odd are now between your door and the third door. 50-50.
By opening a door, the host in effect starts a new game with only two doors.

[Tom Montgomery]

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There are three doors. The gift is behind one of the three. You are allowed to choose one door. You have a 1 in 3 chance of winning. The odds are 2 in 3 that the gift is behind one of the other two doors. Showing that one of the other two doors is not the winner does not change the fact that you have a 1 out of 3 chance of winning if you stick with your initial choice.

Monty Hall on Let's Make A Deal took advantage of human psychology by offering higher and higher amounts of cash to get the contestent to switch his choice of door. Usually the higher the amount of cash offered to switch the more resistant the contestant became to do so. Hall, of course, knew the odds and also where the gift was. Sometimes a recalcitrant contestant won the gift. More often he did not. Hall was usually offering cash to give the gift away. He knew the contestant was unlikely to accept the offer.

[Peerie Maa]

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There are three doors. The gift is behind one of the three. You are allowed to choose one door. You have a 1 in 3 chance of winning. The odds are 2 in 3 that the gift is behind one of the other two doors. Showing that one of the other two doors is not the winner does not change the fact that you have a 1 out of 3 chance of winning if you stick with your initial choice.

Monty Hall on Let's Make A Deal took advantage of human psychology by offering higher and higher amounts of cash to get the contestent to switch his choice of door. Usually the higher the amount of cash offered to switch the more resistant the contestant became to do so. Hall, of course, knew the odds and also where the gift was. Sometimes a recalcitrant contestant won the gift. More often he did not. Hall was usually offering to give the gift away. He knew the contestant was unlikely to accept the offer.

This must be a cross post with my last. The odd change when one door is removed from the game. You have new information, and it is in effect a new game.
Monty did con his punters because he knew human nature and how to play on it, but the probabilities are as CW and I state.

[peb]

You are correct, it is a two-step process.
...
By opening a door, the host in effect starts a new game with only two doors.

Yes, this is the reason, but not only is he starting a new game, he is starting a game that has a 66% chance of him showing you the correct answer!!! The fact that so many mathematicians got this wrong is really surprising.

[Too Little Time]

I came across "The Monty Hall Problem" yesterday and found it fascinating. I too first thought that there was no advantage in switching doors.

I think the real question is if the show producers knew the answer or not at the time.

There is certainly a correct answer to the problem. But it depends on if you believe the choice is between 2 doors or 3 doors. That is if we disregard the first door and just consider the choice between 2 doors or if we include the goat behind the open door.

[Tom Montgomery]

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The new info gained when showing that one of the unchosen doors does not contain the prize increases your odds of winning... if you choose to switch from your first choice. Staying with your first choice keeps your odds of winning the gift at only 1 in 3. You could win. More often you lose. Remember... the game show host is only going to open a door that is a loser.

Here is another explanation of the problem and its solution. Follow this link and you can actually play the game over and over for yourself: https://betterexplained.com/articles...-hall-problem/

[CWSmith]

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There are three doors. The gift is behind one of the three. You are allowed to choose one door. You have a 1 in 3 chance of winning.

That sort of logic only applies of Hall does not open any doors.

Like I said, look at my table on post #24 and tell me where I went wrong. I've already told you what is wrong with your table in post #6.

[Peerie Maa]

Yes, this is the reason, but not only is he starting a new game, he is starting a game that has a 66% chance of him showing you the correct answer!!! The fact that so many mathematicians got this wrong is really surprising.

How can you get a 60% probability out of only two alternatives?
Inquiring minds and all that.

[CWSmith]

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Staying with your first choice keeps your odds of winning the gift at only 1 in 3.

No, that only applies if Hall does not open a door.

It is interesting that Wolfram also gets it wrong online. They are normally very sharp.