monty hall paradox

[RonaldColeman]

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Heard this today. Interested in hearing what you brilliant doctors think about it. Here it is:

You are on the game show Lets Make a Deal. There are 3 doors: 2 have goats behind them and one has a car behind it. You are given a choice of one door. Upon making your choice, the host opens one of the doors you did not choose, revealing a goat. He now offers you the opportunity to switch your choice to the remaining door. Should you do it?

(By the way, I realize that this isn't a residency issue. I just don't know where else to post it)

 

[penguins]

No...wait.... yes.... no ..... yes
No, my final answer is NO.
Your chances/ odds don't change or anything.

 

[tigershark]

Yes you should switch...increases your chance of winning to 2/3, as opposed to 1/3 if you dont switch.

 

[penguins]

Yes you should switch...increases your chance of winning to 2/3, as opposed to 1/3 if you dont switch.

What? How does it do that? It is still independently a 50/50 chance now that the 3rd is out of the way.
Explain please

 

[BKN]

What? How does it do that? It is still independently a 50/50 chance now that the 3rd is out of the way.
Explain please

Correct, It was always 50/50, since Monte had perfect info and he was going to show you one of the two losers, regardless whether you picked the winner or one of the losers.

 

[tigershark]

What? How does it do that? It is still independently a 50/50 chance now that the 3rd is out of the way.
Explain please

The first time you choose there is a 66% chance you are wrong, that doesnt change just because one goat is revealed.

So then after one goat is revealed there continues to be only a 33% chance you were right, and conversly there is now a 66% chance that by switching you will win the car.

 

[WatchingWaiting]

What? How does it do that? It is still independently a 50/50 chance now that the 3rd is out of the way.
Explain please

No. Actually, at the beginning, the probability of the goods being behind your door was 1/3 and the probability of the goods being behind one of the other two doors was 2/3. However, Monty opened one of the other two doors. The prize does not get redistributed at this point though. So, there is still a 1/3 probability that the prize is behind your door, but a 2/3 probability that the prize is behind the door Monty didn't open. The prize does not get randomly assigned between the two remaining doors (which is why it's not 50-50 at this point).

 

[penguins]

No. Actually, at the beginning, the probability of the goods being behind your door was 1/3 and the probability of the goods being behind one of the other two doors was 2/3. However, Monty opened one of the other two doors. The prize does not get redistributed at this point though. So, there is still a 1/3 probability that the prize is behind your door, but a 2/3 probability that the prize is behind the door Monty didn't open. The prize does not get randomly assigned between the two remaining doors (which is why it's not 50-50 at this point).

I was thinking the same down to the last line. At that point it is 50-50 because it become independent of the 3rd. I think at the end it is 1/2, not 1/3.

 

[tigershark]

I was thinking the same down to the last line. At that point it is 50-50 because it become independent of the 3rd. I think at the end it is 1/2, not 1/3.

That's the fallacy, it's not independent of the third.

 

[penguins]

That's the fallacy, it's not independent of the third.

Okay, agreed that at the begining it is 1/3 and that at the end it is still 1/3.
However, if you re-evaluate after that door has been opened. Then it becomes 1/2, does it not?

I like BKN's answer that it was always 50-50 since the goat door would always be opened. 🙂

 

[penguins]

My husband wants to know that if these same group of doctors go to remove his Left leg one day, what are the odds that the doc will remove the correct leg?
He says we have too much time on our hands. 😛

 

[tigershark]

Okay, agreed that at the begining it is 1/3 and that at the end it is still 1/3.
However, if you re-evaluate after that door has been opened. Then it becomes 1/2, does it not?

I like BKN's answer that it was always 50-50 since the goat door would always be opened. 🙂

No, it does not become 50/50. You have to use all the information you have been given.

This is a classic stats question often presented in introductory stats courses.

 

[penguins]

No, it does not become 50/50. You have to use all the information you have been given.

This is a classic stats question often presented in introductory stats courses.

That is why I said if you re-evaluate and look at the remaining 2 doors as independent of the original 3 door question.
You should never change your answer or door because statistically your odds haven't changed at all.

OP, can we have another one? I loved stats and logic classes. I need another one to keep me entertained tonight.

 

[HamOnWholeWheat]

No, it does not become 50/50. You have to use all the information you have been given.

This is a classic stats question often presented in introductory stats courses.

It does become 50/50 because its now a new question. The elimination of one door changes the assumptions of the original problem.

Using your logic, if you chose door "A" and Monty opened doors "B" and "C" and showed you that they both had goats, you'd still only have a 33% chance of choosing the right door? That's too brilliant for me to understand. If your statistics book says I'm wrong, I don't want to be right.

HamOn

 

[BKN]

No, it does not become 50/50. You have to use all the information you have been given.

This is a classic stats question often presented in introductory stats courses.

I'd like to humbly point out that I have a master's in statistics. This discussion is about conditional probability. Humans, including doctors are terrible at it. There's also a bit of game theory. You are correct, you use all the info you will get. However that will lead to a 50/50 chance of success.

I'll do it again. The first and second decisions are not independent.

At the first call you have an apparent 1 in 3 chance of being correct. But Monte who is playing the crowd for drama has no intention of letting it end after one decision. He always intended to give you additional info. He has already decided to reveal one of the two curtains which did not have the prize. Which one he reveals depends on your first call. If you pick the correct curtain he can reveal either loser, If you pick a loser, he will reveal the other loser. Therefore the real decision for you to make is at the second call. Two curtains left, 1 in 2 chance.

Questions?

p.s. He could also use the variation where he doesn't reveal a curtain, but instead offers you a cash settlement in lieu of you curtain. In that case you can not make any reasonable judgement about the decision, since he has perfect info about the value of the curtain and the settlement and you only know the settlement amount.

 

[penguins]

Thanks, BKN
You nerd, masters in stats! 😀

 

[BKN]

can we have another one? I loved stats and logic classes. I need another one to keep me entertained tonight.

I need to pay you a fee as straight man.

Okay, you are seeing a patient in your office who is a 25 y/o female complaining of nervousness, palpitations and weight loss. Her pulse is 110 and indeed she appears thin.

You examine her thyroid and she has no goiter. The literature says that happens only once in 20 times. But given that last two patients with hyperthyroidism in your office did not have a goiter, what is the probability that this patient has hyperthyroidism?

 

[penguins]

I need to pay you a fee as straight man.

Okay, you are seeing a patient in your office who is a 25 y/o female complaining of nervousness, palpitations and weight loss. Her pulse is 110 and indeed she appears thin.

You examine her thyroid and she has no goiter. The literature says that happens only once in 20 times. But given that last two patients with hyperthyroidism in your office did not have a goiter, what is the probability that this patient has hyperthyroidism?

1 in 20?

 

[BKN]

Thanks, BKN
You nerd, masters in stats! 😀

*humbly adhesive tapes the bridge of his glasses*

Answer to my last post in 1 hour.

 

[penguins]

Wait, is this a trick? Is this a stats question or a reality question. Wait... same thing. Now I am overthinking...

 

[tigershark]

I'd like to humbly point out that I have a master's in statistics. This discussion is about conditional probability. Humans, including doctors are terrible at it. There's also a bit of game theory. You are correct, you use all the info you will get. However that will lead to a 50/50 chance of success.

I'll do it again. The first and second decisions are not independent.

At the first call you have an apparent 1 in 3 chance of being correct. But Monte who is playing the crowd for drama has no intention of letting it end after one decision. He always intended to give you additional info. He has already decided to reveal one of the two curtains which did not have the prize. Which one he reveals depends on your first call. If you pick the correct curtain he can reveal either loser, If you pick a loser, he will reveal the other loser. Therefore the real decision for you to make is at the second call. Two curtains left, 1 in 2 chance.

Questions?

p.s. He could also use the variation where he doesn't reveal a curtain, but instead offers you a cash settlement in lieu of you curtain. In that case you can not make any reasonable judgement about the decision, since he has perfect info about the value of the curtain and the settlement and you only know the settlement amount.

You may have a masters in stats but you're wrong about this. It's only a 50/50 chance if he does not know where the prize is.

He always knew where the prize was and never opened that door.

This is old and has been beaten to death...

http://www.stat.sc.edu/~west/javahtml/LetsMakeaDeal.html

http://math.ucsd.edu/~crypto/Monty/montybg.html

 

[BKN]

You may have a masters in stats but you're wrong about this. It's only a 50/50 chance if he does not know where the prize is.

He always knew where the prize was and never opened that door.

This is old and has been beaten to death...

http://www.stat.sc.edu/~west/javahtml/LetsMakeaDeal.html

http://math.ucsd.edu/~crypto/Monty/montybg.html

Wrong, We were talking about imperfect information (yours) not reality with perfect info (his).

 

[Hitch]

He always intended to give you additional info. He has already decided to reveal one of the two curtains which did not have the prize.

That's just specultation. You cannot possibly know what Monty was thinking. He may be a rather spontaneous fellow.


In any case, changing your choice doesn't effect the odds. The chances of your guess being right went from 1 in 3 to 1 in 2 as soon as monty eliminated the third option. Whether you pick again or stick to your guns, your chances of winning the car are still 50/50. It doesn't make a difference.


Of course you may want to check that there isn't a car behind each door on some kind of trap door/elevator mechanism to whisk it down to a sublevel if that door is chosen, with a goat at the ready in each position to pass it off as a losing choice. Unfortunately, if Monty is that much of a crook, and he may well be, then he'll probably have security personel at the ready to prevent any such inspection of the set. If this is the case then your probability of winning, regardless of which door you choose, is exactly zero.

 

[Hitch]

BTW Tigershark, it's called a paradox because both answers are (sort of) right. At the time of the second choice the odds are better than they were at the time of the first choice. But since the odds changed as time went by, his second choice won't change his chance of winning.

 

[BKN]

I need to pay you a fee as straight man.

Okay, you are seeing a patient in your office who is a 25 y/o female complaining of nervousness, palpitations and weight loss. Her pulse is 110 and indeed she appears thin.

You examine her thyroid and she has no goiter. The literature says that happens only once in 20 times. But given that last two patients with hyperthyroidism in your office did not have a goiter, what is the probability that this patient has hyperthyroidism?

Answer: There are three common replies

1) 1 in 8000 derived from 1/20*1/20*1/20. since the events are independent, the first two events are irrelevant. If you answered this, do not go to Las Vegas to get rich. Past independent events never predict future ones. You're never "due" to win.

2) 1 in 20. This is the probability that if the patient does have hyperthyroidism, that she does not have a goiter. That wasn't the question.

3) The question was what is the probability that the patient has hyperthyroidism. You have to consider all possible conditions that cause nervousness, weight loss and palpitations in 25 you females. Certainly at least as common as thyroid problems in that class would be anxiety neurosis. If you had a probability distribution for hyperthyroid vs nonhyperthyroid in this class of patients and the False Negative Rate for goiter in hyperthyroids (1/20) and the True Negative Rate for goiters in nonhyperthyroids you could revise the probability upon the condition of no goiter. However, without the info all you can say is that whatever the probability before you paplated her neck, it's probably much lower after you palpated and found no goiter.

Part of the example and much of the logic in #3 can be found in a very good old book that I think gives an easier approach to EBM than most out now. It is Medical Decision Making by Harold Sox (published late 80s I think).

 

[tigershark]

BTW Tigershark, it's called a paradox because both answers are right. At the time of the second choice the odds are better than they were at the time of the first choice. But since the odds changed as time went by, his second choice won't change his chance of winning.

This was first presented to me at the harvard business school...a way of opening a lecture demonstrating how hard-headed apparently smart people can be in accepting something counter-intuitive. It is indeed better to switch in this situation, and has been proven countless times, even in scientific journals, just google it and you will find countless explanations.

Bapeswara Rao, V. V. and Rao, M. Bhaskara (1992). "A three-door game show and some of its variants". The Mathematical Scientist 17, no. 2, pp. 89–94

Bohl, Alan H.; Liberatore, Matthew J.; and Nydick, Robert L. (1995). "A Tale of Two Goats ... and a Car, or The Importance of Assumptions in Problem Solutions". Journal of Recreational Mathematics 1995, pp. 1–9.
Joseph Bertrand (1889) Calcul des probabilites

Gardner, Martin (1959). "Mathematical Games" column, Scientific American, October 1959, pp. 180–182. Reprinted in The Second Scientific American Book of Mathematical Puzzles and Diversions.

Mueser, Peter R. and Granberg, Donald (1999), "The Monty Hall Dilemma Revisited: Understanding the Interaction of Problem Definition and Decision Making" (University of Missouri Working Paper 99-06). http://econwpa.wustl.edu:80/eps/exp/papers/9906/9906001.html (retrieved July 5, 2005).

Nahin, Paul J. Duelling idiots and other probability puzzlers. Princeton University Press, Princeton, NJ: 2000, pp. 192-193. (ISBN 0-691-00979-1).

Selvin, Steve (1975a). "A problem in probability" (letter to the editor). American Statistician 29(1):67 (February 1975).

Selvin, Steve (1975b). "On the Monty Hall problem" (letter to the editor). American Statistician 29(3):134 (August 1975).

Tierney, John (1991). "Behind Monty Hall's Doors: Puzzle, Debate and Answer?", The New York Times 21 July 1991, Sunday, Section 1; Part 1; Page 1; Column 5
vos Savant, Marilyn (1990). "Ask Marilyn" column, Parade Magazine p. 12 (17 February 1990). [cited in Bohl et al., 1995]

Adams, Cecil (1990). "On 'Let's Make a Deal,' you pick Door #1. Monty opens Door #2--no prize. Do you stay with Door #1 or switch to #3?", The Straight Dope November 2, 1990. http://www.straightdope.com/classics/a3_189.html (retrieved July 25, 2005).

Tijms, Henk (2004). Understanding Probability, Chance Rules in Everyday Life. Cambridge University Press, New York, pp. 213-215.

 

[penguins]

Answer: There are three common replies

1) 1 in 8000 derived from 1/20*1/20*1/20. since the events are independent, the first two events are irrelevant. If you answered this, do not go to Las Vegas to get rich. Past independent events never predict future ones. You're never "due" to win.

2) 1 in 20. This is the probability that if the patient does have hyperthyroidism, that she does not have a goiter. That wasn't the question.

3) The question was what is the probability that the patient has hyperthyroidism. You have to consider all possible conditions that cause nervousness, weight loss and palpitations in 25 you females. Certainly at least as common as thyroid problems in that class would be anxiety neurosis. If you had a probability distribution for hyperthyroid vs nonhyperthyroid in this class of patients and the False Negative Rate for goiter in hyperthyroids (1/20) and the True Negative Rate for goiters in nonhyperthyroids you could revise the probability upon the condition of no goiter. However, without the info all you can say is that whatever the probability before you paplated her neck, it's probably much lower after you palpated and found no goiter.

Part of the example and much of the logic in #3 can be found in a very good old book that I think gives an easier approach to EBM than most out now. It is Medical Decision Making by Harold Sox (published late 80s I think).

Ha! Knew there was going to be a catch. Hey, at least I didn't say #1 🙂
These guys on this thread are taking it all way to seriously I think.

 

[Hitch]

This was first presented to me at the harvard business school...a way of opening a lecture demonstrating how hard-headed apparently smart people can be in accepting something counter-intuitive. It is indeed better to switch in this situation, and has been proven countless times, even in scientific journals, just google it and you will find countless explanations.

Bapeswara Rao, V. V. and Rao, M. Bhaskara (1992). "A three-door game show and some of its variants". The Mathematical Scientist 17, no. 2, pp. 89–94

Bohl, Alan H.; Liberatore, Matthew J.; and Nydick, Robert L. (1995). "A Tale of Two Goats ... and a Car, or The Importance of Assumptions in Problem Solutions". Journal of Recreational Mathematics 1995, pp. 1–9.
Joseph Bertrand (1889) Calcul des probabilites

Gardner, Martin (1959). "Mathematical Games" column, Scientific American, October 1959, pp. 180–182. Reprinted in The Second Scientific American Book of Mathematical Puzzles and Diversions.

Mueser, Peter R. and Granberg, Donald (1999), "The Monty Hall Dilemma Revisited: Understanding the Interaction of Problem Definition and Decision Making" (University of Missouri Working Paper 99-06). http://econwpa.wustl.edu:80/eps/exp/papers/9906/9906001.html (retrieved July 5, 2005).

Nahin, Paul J. Duelling idiots and other probability puzzlers. Princeton University Press, Princeton, NJ: 2000, pp. 192-193. (ISBN 0-691-00979-1).

Selvin, Steve (1975a). "A problem in probability" (letter to the editor). American Statistician 29(1):67 (February 1975).

Selvin, Steve (1975b). "On the Monty Hall problem" (letter to the editor). American Statistician 29(3):134 (August 1975).

Tierney, John (1991). "Behind Monty Hall's Doors: Puzzle, Debate and Answer?", The New York Times 21 July 1991, Sunday, Section 1; Part 1; Page 1; Column 5
vos Savant, Marilyn (1990). "Ask Marilyn" column, Parade Magazine p. 12 (17 February 1990). [cited in Bohl et al., 1995]

Adams, Cecil (1990). "On 'Let's Make a Deal,' you pick Door #1. Monty opens Door #2--no prize. Do you stay with Door #1 or switch to #3?", The Straight Dope November 2, 1990. http://www.straightdope.com/classics/a3_189.html (retrieved July 25, 2005).

Tijms, Henk (2004). Understanding Probability, Chance Rules in Everyday Life. Cambridge University Press, New York, pp. 213-215.

Are you trying to prove that there are lots of dumb people at harvard? I hope those guys don't ever have to make it as bookies. They'd be broke in a day.

 

[RonaldColeman]

Are you trying to prove that there are lots of dumb people at harvard? I hope those guys don't ever have to make it as bookies. They'd be broke in a day.

tigershark is right. Imagine that instead of 3, you have 10000 different doors. Only one has the car. You pick one door. Now Monty Hall, who knows where the car is, opens 9998 doors, all of which have goats. Would you now switch? What are the odds that you actually made the correct guess on your first shot?

 

[Hitch]

Hmmmmm

 

[RonaldColeman]

Nothing has changed. The situation is exactly the same. Why would you switch? Monty is just ****ing with you. You still have a 50/50 chance, either way.

Sorry, changed my example just before you posted.

 

[Hitch]

I change my answer. Tigershark is right. I was wrong. In any case Monty is a stupid little ****er.

 

[RonaldColeman]

Wait a second. No I changed my answer again. If there's a more than three doors then your chances increase. But not if there's only three. With only three doors and monty obviously going to choos a goat door, his action tell you nothing useful. Both options are equally correct. Alter the sample size and that changes.

The odds that the door you picked is correct are determined before monty reveals a door with a goat. I am assuming that when you make your choice you don't know that monty is going to tell you what is behind one of the doors. With 3 doors, the odds that your choice is correct are 1/3. The odds that the goat is behind one of the other 2 doors is 2/3. So even when he shows you one of the doors, the odds that your initial choice is correct is actually 1/3! You should switch. With 10,000 doors, you would certainly want to switch since your odds would now be only 1/10000.

 

[Hitch]

The odds that the door you picked is correct are determined before monty reveals a door with a goat. I am assuming that when you make your choice you don't know that monty is going to tell you what is behind one of the doors. With 3 doors, the odds that your choice is correct are 1/3. The odds that the goat is behind one of the other 2 doors is 2/3. So even when he shows you one of the doors, the odds that your initial choice is correct is actually 1/3! You should switch. With 10,000 doors, you would certainly want to switch since your odds would now be only 1/10000.

I changed the above answer while you were writing this. I feel fine about getting this wrong. It's not like I have a degree in statistics or anthing like that.

 

[Von Hohenheim]

I'd like to humbly point out that I have a master's in statistics.

You may have a masters in stats but you're wrong about this. It's only a 50/50 chance if he does not know where the prize is.

He always knew where the prize was and never opened that door.

This is old and has been beaten to death...

http://www.stat.sc.edu/~west/javahtml/LetsMakeaDeal.html

http://math.ucsd.edu/~crypto/Monty/montybg.html

Wah ha. BKN gets publicly depanted. Nice to see Mister Charlie ("the man") fall on his ass once in a while. 👍 😀

 

[RonaldColeman]

I changed the above answer while you were writing this. I feel fine about getting this wrong. It's not like I have a degree in statistics or anthing like that.

I didn't get it correct either, so no need to apologize. Apparently there was quite a bit of controversy when this question was initially proposed.

 

[BKN]

tigershark is right. Imagine that instead of 3, you have 10000 different doors. Only one has the car. You pick one door. Now Monty Hall, who knows where the car is, opens 9998 doors, all of which have goats. Would you now switch? What are the odds that you actually made the correct guess on your first shot?

Like I said, humans are terrible at conditional probability, apparently most of you fall in this category.

There are two decision points. At the first there is a prize and two goats. 1/3 chance of success. You make a choice.

Monty tells you nothing about your choice, but reveals one of the wrong choices. Let's say he shows you a goat behind curtain 3. Now you have to make a second choice. Monty has given you additional info for you to condition your answer. Are you going to pick curtain 3? Not unless you have an unusual passion for goats. You now have two curtains and one prize. Probability of success is 1 in 2. It doesn't matter whether you change your choice or not. How hard is that? 🙂

BTW I'm not going to read those articles, since I have other things to do. I suspect they deal with some other things about the game as it was played at the time. There wasn't always two goats, sometime there were two prizes. He didn't always show a loser curtain, sometimes he offered a box, sometimes he offered a cash settlement. The game was rather more complicated than the example presented here.

BTW2 Ronald. your chances were 1 in 10000 in your example after revealing one loser curtain your chances are 1 in 9999 whether you change your curtain or not.

 

[penguins]

BTW2 Ronald. your chances were 1 in 10000 in your example after revealing one loser curtain your chances are 1 in 9999 whether you change your curtain or not.[/QUOTE]

Thank you, I thought so but I have been afraid to speak on this thread today in fear of getting another reading list. (matched my #1 neuro spot today 😀 )

 

[dodo2]

The answer is simple. Open the remaining door if, you do not see a car at the door you opened.

 

[CameronFrye]

Like I said, humans are terrible at conditional probability, apparently most of you fall in this category.

There are two decision points. At the first there is a prize and two goats. 1/3 chance of success. You make a choice.

Monty tells you nothing about your choice, but reveals one of the wrong choices. Let's say he shows you a goat behind curtain 3. Now you have to make a second choice. Monty has given you additional info for you to condition your answer. Are you going to pick curtain 3? Not unless you have an unusual passion for goats. You now have two curtains and one prize. Probability of success is 1 in 2. It doesn't matter whether you change your choice or not. How hard is that? 🙂

BTW I'm not going to read those articles, since I have other things to do. I suspect they deal with some other things about the game as it was played at the time. There wasn't always two goats, sometime there were two prizes. He didn't always show a loser curtain, sometimes he offered a box, sometimes he offered a cash settlement. The game was rather more complicated than the example presented here.

BTW2 Ronald. your chances were 1 in 10000 in your example after revealing one loser curtain your chances are 1 in 9999 whether you change your curtain or not.

Ah yes, the old "Monty Hall paradox". I've seen people come close to fisticuffs over this. When "Ask Marilyn" gave the correct answer to this question, she received over 10,000 responses from angry readers.

I've found the best way to convince people, is just to do it. You can easily set this up with cards, or you can go to one of the websites online that have a program to play this game repeatedly.

I just played the game 202 times:
In 102 games, I did NOT switch: I won 19 times.
In 100 games, I did switch: I won 71 times.

If you play the game enough, the numbers start getting closer to 67% and 33%.

 

[CameronFrye]

Oh, tigershark has already posted a link to play the game:

http://www.stat.sc.edu/~west/javahtml/LetsMakeaDeal.html

 

[robotsonic]

It's amazing that BKN and others are clinging so tightly to the wrong answers. This problem fools a lot of people because the correct answer isn't intuitive, but that doesn't mean that the correct answer is wrong!

Tigershark's first link is to a program that will allow you to play the game yourself. Try sticking with your first choice and then try changing your choice - you will find that you really are better off switching doors.

BKN, I'm sure you are great at other areas in statistics, but you are wrong on this one.

 

[CameronFrye]

Ok, so I just played the game another 1000 times (by repeatedly clicking my mouse). Everytime I kept the same door. The final results: 341 wins out of 1000. Not too far off from 33%.

 

[BKN]

It's amazing that BKN and others are clinging so tightly to the wrong answers. This problem fools a lot of people because the correct answer isn't intuitive, but that doesn't mean that the correct answer is wrong!

Tigershark's first link is to a program that will allow you to play the game yourself. Try sticking with your first choice and then try changing your choice - you will find that you really are better off switching doors.

BKN, I'm sure you are great at other areas in statistics, but you are wrong on this one.

My bad. Jeez, I said it myself, "Monte tells me nothing about my choice". Why didn't I listen. My error was assuming that he was giving me additonal info about the other curtains. Actually, he was telling me that at least one of the two other curtains hid a goat. I already knew that, so why did I think it was something to use to revise? 😳 😳 😡

Crow is being cooked to be eaten.

 

[CameronFrye]

My bad. Jeez, I said it myself, "Monte tells me nothing about my choice". Why didn't I listen. My error was assuming that he was giving me additonal info about the other curtains. Actually, he was telling me that at least one of the two other curtains hid a goat. I already knew that, so why did I think it was something to use to revise? 😳 😳 😡

Crow is being cooked to be eaten.

Don't beat yourself up too much. This problem dates back to at least the 19th century (although it obviously didn't reference Monty Hall) and many famous mathematicians got it wrong at first. (and I definitely didn't get it right at first).

 

[BKN]

Don't beat yourself up too much. This problem dates back to at least the 19th century (although it obviously didn't reference Monty Hall) and many famous mathematicians got it wrong at first. (and I definitely didn't get it right at first).

Easiest way to look at this is Monte is loading the probability of success for the two unpicked curtains (2/3) on a single curtain by making a nonrandom choice of which he reveals.

 

[fuegofrio17]

Or you could look at it as starting off with the House holding a 2:1 advantage over you from the start. You then have the opportunity to switch bets and take the 2:1 house advantage (with a 66% chance of winning) or stick to your guns with your original bet, which has a 33% chance of winning.

Basically, Monty's open door (or flipped card) belongs to the House. So by switching bets, you get the advantages of having 2 cards instead of 1 in a 3-card game.

Interestingly enough, the showing of the donkey really doesn't give you any new or usable information. You already know the house either has A) one donkey or B) two donkeys. The flipping of the single donkey card is irrelevant. Monty is going to show a donkey 100% of the time--it is not an issue of probability. The relevant part of this point in the show, though, is the opportunity to switch bets and claim the House advantage.

 

[BKN]

Or you could look at it as starting off with the House holding a 2:1 advantage over you from the start. You then have the opportunity to switch bets and take the 2:1 house advantage (with a 66% chance of winning) or stick to your guns with your original bet, which has a 33% chance of winning.

Basically, Monty's open door (or flipped card) belongs to the House. So by switching bets, you get the advantages of having 2 cards instead of 1 in a 3-card game.

And that's why he's called Monte.

 

[tigershark]

As you can see, this little question is a great way to zing people who are a little too confident of themselves. I've seen it at several meetings and it always causes a commotion.

BKN was so arrogant he would rather take the time to write out a page of wrongness and thinly veiled insults than click on a link that succintly demonstrates the right answer.

It's a great way of demonstrating how arrogance and overconfidence can cloud your judgment when a problem has a counter-inutuitive solution, even in the face of the explanation.

 

[HamOnWholeWheat]

BKN was so arrogant he would rather take the time to write out a page of wrongness and thinly veiled insults than click on a link that succintly demonstrates the right answer.

And he was the only arrogant one in this thread? 😉

"This was first presented to me at the harvard business school...a way of opening a lecture demonstrating how hard-headed apparently smart people can be in accepting something counter-intuitive."

I love the use of the "..." by the way. It forces the reader to pause and fully reflect on the implication that you were at Harvard Business School.

HamOn