Is the Monte Hall Problem Really Just a Coin Toss?

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In summary: I mean 'you'... have a perfect record in this game.In summary, the Monty Hall problem is a game show scenario where contestants must choose between three doors, one of which has a car behind it and the other two have goats. After the contestant makes their initial choice, the host opens one of the remaining doors to reveal a goat and then offers the contestant the chance to switch their choice. Despite seeming like a 50/50 chance, it is actually more beneficial to switch to the remaining unopened door, with a 2/3 probability of winning. This has been demonstrated through mathematical solutions, experimental evidence, and logical reasoning.
  • #1
FallenApple
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Ok so here is the problem.

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?


If I know that the car is not in the third door, then with the knowledge, the probability space collapses to being only one of two possibilities, current door or other door. It might as well be completely random between the two. So it's 1/2, 1/2I am unconvinced by mathematical solutions.

This is a question of philosophy here. If I know its not in the third door, then how is it any different than there being no third option in the first place?So I flip a coin. I know its going to be H or T.

This is the same as me knowing that there isn't a third option begin with. So it's 1/2, 1/2.So what if I flip a coin and cover it with a cup? Well, I know it is going to be H or T once I lift the cup since I have the knowledge that there isn't a third option.
 
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  • #2
FallenApple said:
I am unconvinced by mathematical solutions.
So what would you be convinced by? MythBusters did an experimental confirmation also.
 
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  • #4
jedishrfu said:
Wikipedia has a good writeup and references for this problem that worth reading:

https://en.m.wikipedia.org/wiki/Monty_Hall_problem

In the end, though youll have to convince yourself of its counter intuitive answer.

Of course it is the counter-intuitive aspect that makes it interesting, But if you go to the "Simple Solutions" section of that wiki entry:

https://en.wikipedia.org/wiki/Monty_Hall_problem#Simple_solutions

how can the OP argue the validity? It is a very simple logic chart of each possibility. Very simple to do, it is the basic method used for a problem like this, and the result is inarguable. Though I will admit, even after seeing that, I have a little problem 'getting my head around it'. A good example of how we cannot trust our instincts. A good book on the subject (IMO) is "Thinking, Fast and Slow" by Daniel Kahneman.
 
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  • #5
It's not a question of philosophy, since it is a concrete problem with a definite answer. Your argument seems to assume that whenever there are two possibilities the chances are always 50/50, but that is certainly not the case.

There are pages on the web where you can test the problem experimentally, for example http://www.advanced-ict.info/javascript/gameshow.html and https://www.damninteresting.com/monte-hall-simulator/ .
 
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  • #6
FallenApple said:
Ok so here is the problem.

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?


If I know that the car is not in the third door, then with the knowledge, the probability space collapses to being only one of two possibilities, current door or other door. It might as well be completely random between the two. So it's 1/2, 1/2
You are missing the critical point here. Initially there was 1/3 chance that you had the car and 2/3 that the car was behind the other two doors. When the host opens a door, he will not open a door with the car so he has filtered those two doors down to one. So the door he didn't pick now has all 2/3 probability from those two doors. You should switch.

An extreme example makes it more obvious. Suppose there is a bucket of sand with a small diamond. You take a pinch of sand from the bucket. The odds that your pinch has the diamond is tiny and the diamond is almost certainly still in the bucket. Suppose the host takes a filter that is too small for the diamond to pass and empties all but a pinch from the bucket through the filter. The diamond is almost certainly still in the bucket where a remaining pinch of sand is. So you should switch your original pinch of sand for the one in the bucket. Notice how critical the filtering process is. If the host had just dumped all the sand but one pinch, there would be no reason to think it is still in the bucket. So it would not help to switch. (It wouldn't hurt either.)

In the original problem, if the host just randomly opens a door then he has not filtered anything. If the car is not behind the opened door, that was just luck and it just increases the odds of both your door and the third door to 1/2 each. Then there is no advantage or disadvantage to switching.
 
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  • #7
I like FactChecker's bucket of sand analogy. But to keep it a little closer to the Monty Hall example, consider if Monty showed you 1000 doors. Just like in the game, one randomly has a car, 999 are empty.

So you choose door #1, and then Monty opens all the other doors, except for say, #42, and asks if you want to switch doors. Now you must consider - why did he choose to leave door #42 closed, out of 999 choices?

There are only two possibilities- he is either 'goofing' with you (knows you have the car in #1), or #42 contains the car. But... there was only a 1/1000 chance that #1 has the car, so there is only a 1/1000 chance that he is goofing with you. And a 999/1000 chance that the car is in door #42.

So once again, #42 is the (most) correct answer. Switch doors!

There, now I think I finally got my head around it!
 
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  • #8
I wonder if 999 typical goats are worth more than 1 typical car. I do like the example either way.
 
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  • #9
NTL2009 said:
I like FactChecker's bucket of sand analogy. But to keep it a little closer to the Monty Hall example, consider if Monty showed you 1000 doors. Just like in the game, one randomly has a car, 999 are empty.

So you choose door #1, and then Monty opens all the other doors, except for say, #42, and asks if you want to switch doors. Now you must consider - why did he choose to leave door #42 closed, out of 999 choices?

There are only two possibilities- he is either 'goofing' with you (knows you have the car in #1), or #42 contains the car. But... there was only a 1/1000 chance that #1 has the car, so there is only a 1/1000 chance that he is goofing with you. And a 999/1000 chance that the car is in door #42.

So once again, #42 is the (most) correct answer. Switch doors!

There, now I think I finally got my head around it!

That's my favourite explanation as well.
 
  • #10
FactChecker said:
In the original problem, if the host just randomly opens a door then he has not filtered anything. If the car is not behind the opened door, that was just luck and it just increases the odds of both your door and the third door to 1/2 each. Then there is no advantage or disadvantage to switching.

The critical point being that in this case, on average, 1 in 3 games is spoiled by the host revealing the car. This never happens in the real game.
 
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  • #11
PeroK said:
The critical point being that in this case, on average, 1 in 3 games is spoiled by the host revealing the car. This never happens in the real game.
Yes. This brings up another unstated assumption in the Monte Hall puzzle -- that he must always offer a switch. Otherwise he can change the odds. If he is allowed to, he might only offer a switch when he knows that your door has the prize. That makes it always bad to switch. People would catch on to that and never switch. Or he can offer a switch a certain percentage of the time that your door has the prize. The right percentage would bring the odds that a switch helps back to even. Other percentages can bring the odds to anything between 2/3 and 0. That opens the door to game theory.
 
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  • #12
FallenApple said:
I am unconvinced by mathematical solutions.
This is a great example where Bayes' Theorem helps you to avoid deceptive intuition. It leads to the correct answer in small steps like supplying the correct value for prob( prize behind door 2 | Monte opened door 2) = 0.
 
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  • #13
I loved this problem and after M. vos Savant (the smartest person in the world) agreed with the OP I used to make a fortune in bars offering $6 if I guessed wrong against $5 if I was right. The guessing 3 consecutive coin tosses game was even a better winner.
 
  • #14
1) Marilyn Vos Savant is not the smartest person in the world, even if at some point she scored a record on some IQ test she once took.

2) She described the solution to the Monte Hall problem correctly. Which is contradicted by the original poster. So in what sense is it true that she "agreed with the original poster"?
 
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  • #15
zinq said:
1) Marilyn Vos Savant is not the smartest person in the world, even if at some point she scored a record on some IQ test she once took.

2) She described the solution to the Monte Hall problem correctly. Which is contradicted by the original poster. So in what sense is it true that she "agreed with the original poster"?
1) I was being facetious. 2) You're right I erred, thank you. I was thinking of a different error she made, my bad.
 
  • #16
FallenApple said:
Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?

Take the goat. Cheaper to run.
 
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  • #17
I don't know the mathematical proof I think I can help you understand it here.
Think about a million doors. The car is behind only one door. And Monty Hall gives you a chance to try your luck. You pick door no. 27. The chance of your winning is 0.0000001. But that's only in theory that you have a chance of winning. But do you feel that you have any chance of winning? Well, practically you'll no for sure that you're going to lose and there is almost practically 100% chance that the car is hidden behind one of these doors: 1,2,3...26,28,29...1000000. Let's say Monty Hall gives you the chance to choose one option: (a) choose your own door, (b) choose all the other doors except your chosen one, and if the car is behind anyone of them, then you will win. Obviously, you'll choose (b). But in the actual problem, Monty Hall doesn't allow you to choose the other group, instead he opens all the wrong doors in the other group and makes you choose between you own door and the door which came out to be the 'correct one' among the other group doors. Well, there's almost 100% chance that the car is behind that door.
Now, if we talk about three doors. When you pick one of them, you know, in you heart, that there's a 2/3 chance that the cars is behind one of the two other doors. Now, Monty Hall removes one incorrect door from the other two doors and you have the option to choose between the remaining doors. Now you decide whether you pick your original door (which had a 1/3 chance of winning) or the 'chosen one' from the other group.
 
  • #18
I also use the explanation @FactChecker laid out at the start of post #6, but I express it a little differently.
MH always opens a door, and always can do so without risking revealing the prize. Therefore it provides no information as to whether your original choice was correct, thus it must still have odds of 1/3.
 
  • #19
Just know you're in good company. Even Paul Erdős, after reading through the mathematical reasoning, was not convinced until, if I recall correctly, he watched a computer simulation of numerous trials.
 
  • #20
"This is the same as me knowing that there isn't a third option begin with. So it's 1/2, 1/2."

The fundamental error here is assuming that, just because there are two options, their probabilities must be equal. That does not follow, and is not the case here.
 

Related to Is the Monte Hall Problem Really Just a Coin Toss?

What is the Monte Hall Problem and why is it confusing?

The Monte Hall Problem is a probability puzzle named after the TV game show host, Monte Hall. It is a statistical paradox that involves selecting one of three doors to win a prize. The confusion comes from the counterintuitive answer, which is different from what most people expect.

What is the correct solution to the Monte Hall Problem?

The correct solution to the Monte Hall Problem is to switch doors after the host reveals an empty door. This increases the chances of winning from 1/3 to 2/3.

Why is it hard to accept the correct solution to the Monte Hall Problem?

It is hard to accept the correct solution because it goes against our intuition and common sense. Most people assume that switching doors does not change the probability of winning, but in this case, it does.

Is there a mathematical proof for the solution to the Monte Hall Problem?

Yes, there is a mathematical proof for the solution to the Monte Hall Problem. It involves using conditional probability and Bayes' theorem to show that switching doors increases the chances of winning.

Are there real-life applications of the Monte Hall Problem?

Yes, the Monte Hall Problem has real-life applications in various fields, such as game theory, statistics, and decision-making. It helps to understand the importance of using logical reasoning and mathematical calculations to make informed choices.

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