Ask Marilyn
Nevertheless, I welcome your comments on it.
Article: THE MONTY HALL PROBLEM.
Quote: WizardOkay, I just finished a rather lengthy article on this topic. Hopefully this will be my last word on the topic and I can just refer people to my article from now on.
Nevertheless, I welcome your comments on it.
Article: THE MONTY HALL PROBLEM.
Mice article. Clearly laid out and easy to follow. Thanks for doing this.
Quote: unJonMice article. Clearly laid out and easy to follow. Thanks for doing this.
Thank you!
I say this because, yes, some people WOULD think those odds improved, but would agree that their original guess is still 1%.Quote:Finally, to this question, one analogy that has helped some is to use 100 doors instead of one. If you know the host will reveal 98 goats, does it sound reasonable to think your chances of winning having guessed correctly went from 1% to 50%? No, they stayed at 1% and you would have a 99% chance of winning the car by swapping.
I might also add that there’s a 99% chance that the car is in a door not chosen and that doesn’t change as doors are opened.
Also, in the first paragraph, you called out Richard Brodie. Um, who?
I think the puzzle may be considered an IQ test. Some, like Marilyn, get it right away. Others get it after a while, and some never get it at all. I have found it frustrating explaining it to people who don't get it. They still don't get it.
Some object to the condition that Monty opens a door and offers a choice whether the contestant has chosen correctly or not on the ground that it doesn't relate to what Monty did on his show. Indeed it doesn't. Monty never offered a switch. It is merely a condition of the mathematical problem. To address this concern I propose the following:
Let s be the probability that Monty offers the switch if the contestant has chosen correctly and t the probability that Monty offers the switch if the contestant has chosen incorrectly. 1-s-t would be the probability that Monty does not offer the switch at all.
What is the contestant's best strategy in terms of these parameters?
Hint: It would be a mixed strategy, but with what probability?
We'll say the contestant always chooses Door 1. We could go through the same reasoning if he chose door 2 or 3, but permuting the doors really doesn't change the logic or the result.
Let's say that it's a game between the contestant and Monty. Monty wins if he opens the right door or the contestant opens the wrong door. C means contestant stays with Door 1 and Cs means contestant switches.
Car is behind Door 1 Monty opens Door 2 C wins Cs loses
Car is behind Door 1 Monty opens Door 3 C wins Cs loses
Car is behind Door 2 Monty opens Door 2 C loses Cs loses
Car is behind Door 2 Monty opens Door 3 C loses Cs wins
Car is behind Door 3 Monty opens Door 2 C loses Cs wins
Car is behind Door 3 Monty opens Door 3 C loses Cs loses
Out of 6 possible outcomes there are 2 C wins and 2 Cs wins, so the probability of winning is 1/3 for either strategy.
The Wizard gets 1/2.
Another take is that if the contestant never switches it doesn't matter what Monty does, so the Pr of winning is 1/3.
If he always switches he wins if his initial choice was wrong and Monty also chooses incorrectly. The probabilities are 2/3 and 1/2, so the net probability is 1/3.
If a game were played according to these rules there would have to be some sort of outcome that would maintain the pace of the show, not merely play the game over as the Wizard's analysis implies. I would suggest an alternate prize: "I'm sorry, you didn't win the car, but (curtain) here is an all expense paid vacation in Detroit, Michigan courtesy of Greyhound Bus Lines."
1/3 of the time contestant picks car first. Monty will always open a goat. Not switching gets car 100% and goat 0%.
2/3 of the time contestant picks goat first. 50% of those times Monty opens a goat. Not switching gets car 0% of the time and switching gets car 100% of the time. The other 50% of this 2/3 case, Monty opens the car door and whether the contestant switches or not, he gets the car 0% of the time.
Adding up probability of getting car by not switching:
1/3*100% + 2/3*50%*0% + 2/3*50%*0% = 1/3
Adding up probability of getting car by switching:
1/3*0% + 2/3*50%*100% + 2/3*50%*0% = 1/3
So the contestant does the same whether he switches or not.
And the Wizard get 1/2 because he takes the question as presented by Marilyn at face value: Monty opened a door (randomly) that had a goat. So you toss out the 2 cases where Monty opened the door. In other words, given that Monty randomly opened a door with a goat, the contestants starting chances of 1/3 just jumped to 1/2. Because Monty’s actions in this case provide new Bayesian information that his actions don’t If Monty doesn’t act randomly.
Quote: unJon
And the Wizard get 1/2 because he takes the question as presented by Marilyn at face value: Monty opened a door (randomly) that had a goat. So you toss out the 2 cases where Monty opened the door. In other words, given that Monty randomly opened a door with a goat, the contestants starting chances of 1/3 just jumped to 1/2. Because Monty’s actions in this case provide new Bayesian information that his actions don’t If Monty doesn’t act randomly.
I follow you up to this point. What does "face value" mean? Usually it means "superficially without considering the real meaning."
You are saying that if Monty randomly opens a door and shows a goat it provides more information than if he intentionally does the same thing. Why? The information provided to the contestant is what he sees and hears and it is the same in either case.
Quote: netzerI follow you up to this point. What does "face value" mean? Usually it means "superficially without considering the real meaning."
You are saying that if Monty randomly opens a door and shows a goat it provides more information than if he intentionally does the same thing. Why? The information provided to the contestant is what he sees and hears and it is the same in either case.
I’m saying that the problem, as given, starts after Monty has opened a door and shown a goat. So if Monty opens a random door, you can throw out the two cases where Monty opens a door and shows a car, because that’s not what happened. The contestant now has a 1/2 chance to win a car, because he survived the trial of Monty not showing a car. In other words, Monty opening a door randomly provides information that lets us update the probability of winning.
That’s in contrast to the case where Monty always chooses to open a door showing a goat. In this assumption, Monty opening a door provides no information useable to update probabilities since in all cases he opens a door with a goat.
Quote: unJonI’m saying that the problem, as given, starts after Monty has opened a door and shown a goat. So if Monty opens a random door, you can throw out the two cases where Monty opens a door and shows a car, because that’s not what happened. The contestant now has a 1/2 chance to win a car, because he survived the trial of Monty not showing a car. In other words, Monty opening a door randomly provides information that lets us update the probability of winning.
This is the popular but incorrect solution. The player's probability of being correct starts at 1/3 and the only way it can change is of Monty opens a door and shows the car , in which case it goes to zero.
I would like to assign some homework:
http://priceonomics.com/the-time-everyone-corrected-the-worlds-smartest/
Replace the "z" with "." and paste it as a URL.
[edit by mod(OnceDear) Link fixed as a courtesy. Site not tested or endorsed]
Quote: netzerQuote: unJonI’m saying that the problem, as given, starts after Monty has opened a door and shown a goat. So if Monty opens a random door, you can throw out the two cases where Monty opens a door and shows a car, because that’s not what happened. The contestant now has a 1/2 chance to win a car, because he survived the trial of Monty not showing a car. In other words, Monty opening a door randomly provides information that lets us update the probability of winning.
This is the popular but incorrect solution. The player's probability of being correct starts at 1/3 and the only way it can change is of Monty opens a door and shows the car , in which case it goes to zero.
I would like to assign some homework:
priceonomicszcom/the-time-everyone-corrected-the-worlds-smartest/
Replace the "z" with "." and paste it as a URL.
We are talking past each other. My homework to you is to read this thread and think on the difference between these problems:
Contestant A picks a door:
1) Monty, knowing what is behind the doors, then always opens a door not picked and it’s always a goat.
2) Monty then randomly opens an unpicked door. It happens to be a goat.
This thread was mostly about 1, which is the classic 1/3 and 2/3 problem.
You came to this thread and started discussing 2, incorrectly thinking it’s the same problem as 1. It’s not.
Quote: unJonYou came to this thread and started discussing 2, incorrectly thinking it’s the same problem as 1. It’s not.
All right, please explain the difference so that I can understand it. My IQ is only 165, somewhat less than Marilyn's.
A few years ago I put this problem on a web site with an explanation that I thought anyone could understand. How wrong I was! One reader wrote in to say that his teacher said that 1/2 was the correct probability. How do you tell a student his teacher is wrong?
I came to this thread out of worry that the Wizard thought Marilyn was wrong. My mind is now at ease. You will note that you and I arrived at the same answer for the random choice problem, 1/3 and 1/3. Are you changing your answer?
Quote: netzerAll right, please explain the difference so that I can understand it. My IQ is only 165, somewhat less than Marilyn's.
A few years ago I put this problem on a web site with an explanation that I thought anyone could understand. How wrong I was! One reader wrote in to say that his teacher said that 1/2 was the correct probability. How do you tell a student his teacher is wrong?
I came to this thread out of worry that the Wizard thought Marilyn was wrong. My mind is now at ease. You will note that you and I arrived at the same answer for the random choice problem, 1/3 and 1/3. Are you changing your answer?
1/3 for for A to 1/3 for door C is the chance before Monty opens a random unpicked door. After he randomly opens B and it happens to have a goat, you have to be a good Bayesian and update the probabilities.
If Monty not randomly always opens a goat, a good Bayesian knows the probabilities remain 1/3 for Door A.
The key are the italicized words.
Quote: unJonIf Monty not randomly always opens a goat, a good Bayesian knows the probabilities remain 1/3 for Door A
A good Bayesian knows how to write the Bayes formula, plug in the data, and calculate the result. Over to you!
Quote: netzerA good Bayesian knows how to write the Bayes formula, plug in the data, and calculate the result. Over to you!
Wow, you really want me to spoon feed it to you?
Hypo 1: Contestant picks door A. Monty always opens whichever of door B or C contains a goat (and if both do, he randomly picks one). This is the classic problem on Ask Marilyn.
Let P(X) be the probability Door X contains a car.
Let P(m) be the probability that Monty opens a door with a goat.
Here P(m) is 1. Because Monty always opens an unpicked door with a goat.
P(A|m) = P(m|A)*P(A)/P(m)
P(A|m) = 1 * 1/3 / 1 = 1/3
Hypo 2: Contestant picks door A. Monty randomly opens either B or C, irrespective of the prize behind the door. He picks B and it has a goat. This is the hypo where Wiz gets 1/2 and you aren’t (yet) following.
Here P(m) = 2/3. Because 1/3 of the time Monty actually opens a door with a car behind it. Recall you got to this same place because in 2 of your 6 scenarios, Monty opened the door with a car. Agreed?
P(A|m) = P(m|A)*P(A)/P(m)
P(A|m) = 1 * 1/3 / 2/3 = 1/2
Note that P(m|A) is 1 because given A has the car then Monty has a 100% of getting a goat (because he only picks between B and C, which both have goats).
Three cups. One has a prize under it and other two empty.
A third party (I’ll nominate the Wizard) randomly puts the prize under one of the cups. Neither of us know the location.
I’ll choose one of the cups.
You then pick and reveal the contents of one of the remaining two cups I didn’t pick. If you reveal the prize, we start again and no money changes hands.
If your chosen cup is empty, then we bet whether I picked the cup with a prize. I’ll put up $1 for every $1.5 you put up. Since you believe I only have a 1/3 chance of being right, you should think the odds heavily favor you.
I've been working on the Wizard's coin toss problem. I sense a screw-up there as the answer is not independent of p. Apparently he thinks well of you, as he expected you to show up there and put a timer on your response. I'll study your bet proposition and get back to you. In the meantime perhaps Crystalmath would like to jump in. He has been following the thread. Don't forget that your first solution agreed with mine: 1/3 and 1/3 for switch and no switch if Monty chose at random.
I still agree that’s true, before Monty randomly opens a door and shows a goat.Quote: netzerunJon:
Don't forget that your first solution agreed with mine: 1/3 and 1/3 for switch and no switch if Monty chose at random.
Quote: netzerunJon:
I've been working on the Wizard's coin toss problem. I sense a screw-up there as the answer is not independent of p. Apparently he thinks well of you, as he expected you to show up there and put a timer on your response. I'll study your bet proposition and get back to you. In the meantime perhaps Crystalmath would like to jump in. He has been following the thread. Don't forget that your first solution agreed with mine: 1/3 and 1/3 for switch and no switch if Monty chose at random.
I agree with unJon. It depends on whether or not the host’s choice is random. If the choice is random, and he reveals a goat, the initial door has a 50% chance of winning. Bayes agrees too.
As for the coin problem, the answer is independent of p, except when p=0 or p=1; in those cases the answer is 2. For all other cases, the answer is
Quote: billryanJames wins the showcase and picks Door #1. He has one third chance to win. That days filming ends so they pick it up tomorrow. James is busy so his identical twin Jeff takes his place, but James forgets to tell him which box he picked. Monty opens Door B and asks Jeff if he wants to switch. As Jeff doesn't know which box James chose, it's 50-50 ,correct? Just as he is going to take Door 1, a power outage occurs forcing the taping to be canceled. When it resumes, neither Brother can attend. They pick you to attend and make the choice. When James chose, it was one in three. When Jeff chose it was one in two. Both picked Door 1. What is your best choice?
In the original game, before any power outage or absence, we had a 3 door probability issue. The two doors that were rejected are identified and identifiable. If there was a powere outage and the two unopened doors got shuffled, then the probability of selecting the good door would be 50/50 because knowledge has been removed. Anyone keeping an eye on the outage (shuffle tracking) would have the 66.7 / 33.33 probabilities.
Quote: unJon
You then pick and reveal the contents of one of the remaining two cups I didn’t pick. If you reveal the prize, we start again and no money changes hands.
That didn't take long. What you are describing is the second game the Wizard analyzes in his article. I have no problem with his answer but I don't think it is a game worth analyzing because it would not be good show business.
I urge you to get back to the thread about the unbalanced coin. It is turning into a real doozy and I think the Wizard has underestimated the complications.
Quote: netzerI have to go do some yard work now.
Is it common to do yard work at 9:30pm in South Africa?
Hilarious. I’d say this whole netzer detour was a waste of time, but it was fun to pull out Bayes theorem.Quote: CrystalMathIs it common to do yard work at 9:30pm in South Africa?
Quote: unJonHilarious. I’d say this whole netzer detour was a waste of time, but it was fun to pull out Bayes theorem.
It is better to adopt a respectful tone until you know who you are dealing with. Get back to the Wizard's biased coin problem and see what joy awaits you.
/45916/marilynvossavant/22-sunday-column-2/
