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Message
re: Marilyn vos Savant and the history of the Montel Hall question
Posted on 2/23/15 at 3:02 pm to Big Scrub TX
Posted on 2/23/15 at 3:02 pm to Big Scrub TX
I have to disagree with the "always switch" conclusion. The correct answer about the probability is that it is 50/50.
The problem is not in the math--it is that the setup and explanation of the problem always given is a cheat. The problem is changed in the middle, but the original probabilities are not. That is why it is so counterintuitive.
Those who discuss the probability in terms of 33% and 66% after the first door is eliminated are carrying forward into the new problem (where only two doors exist) information from the original problem, and thus not establishing anything more earth shattering than the fact that we originally had 66% goats and 33% cars to choose from.
In fact, the reason people commonly say the answer is 50/50 is that most people intuitively accept the initial premise that a door is simply removed from the problem--leaving two doors, known to hide a goat and a car. This is clearly a 50/50 choice.
You can appreciate this truth if you consider the person who comes into the game at the point where the contestant is told that there is goat behind one of the doors he didn't pick. That door is effectively removed from the game, as if it never existed. The new person faces simply two doors, with no knowledge about what has gone on before, and the certainty that there is one goat and one car. It does not matter which of the two doors he picks (or whether he "picks" one then "switches", or doesn't switch.) In the end, he chooses one door--and has a 50/50 shot at getting it right. Thus, for him, it doesn't matter if he switches or not--his probability is always 50/50.
Again, the problem with this riddle is that the mathematical explanations always start with the premise that you must carry forward the old 33/66 probabilities from the first problem into the second. That is the cheat--you don't. The Monty Hall problem really is a 33/66 probability problem changed into a 50/50 problem, but discussed mathematically (incorrectly)after the basic premise has been changed as if it is still a 33/66 problem.
The problem is not in the math--it is that the setup and explanation of the problem always given is a cheat. The problem is changed in the middle, but the original probabilities are not. That is why it is so counterintuitive.
Those who discuss the probability in terms of 33% and 66% after the first door is eliminated are carrying forward into the new problem (where only two doors exist) information from the original problem, and thus not establishing anything more earth shattering than the fact that we originally had 66% goats and 33% cars to choose from.
In fact, the reason people commonly say the answer is 50/50 is that most people intuitively accept the initial premise that a door is simply removed from the problem--leaving two doors, known to hide a goat and a car. This is clearly a 50/50 choice.
You can appreciate this truth if you consider the person who comes into the game at the point where the contestant is told that there is goat behind one of the doors he didn't pick. That door is effectively removed from the game, as if it never existed. The new person faces simply two doors, with no knowledge about what has gone on before, and the certainty that there is one goat and one car. It does not matter which of the two doors he picks (or whether he "picks" one then "switches", or doesn't switch.) In the end, he chooses one door--and has a 50/50 shot at getting it right. Thus, for him, it doesn't matter if he switches or not--his probability is always 50/50.
Again, the problem with this riddle is that the mathematical explanations always start with the premise that you must carry forward the old 33/66 probabilities from the first problem into the second. That is the cheat--you don't. The Monty Hall problem really is a 33/66 probability problem changed into a 50/50 problem, but discussed mathematically (incorrectly)after the basic premise has been changed as if it is still a 33/66 problem.
5
Posted on 2/23/15 at 3:11 pm to link
quote:
After the 20 trials at the dining room table, the problem also captured Mr. Hall's imagination. He picked up a copy of Ms. vos Savant's original column, read it carefully, saw a loophole and then suggested more trials.
On the first, the contestant picked Door 1.
"That's too bad," Mr. Hall said, opening Door 1. "You've won a goat."
"But you didn't open another door yet or give me a chance to switch."
"Where does it say I have to let you switch every time? I'm the master of the show. Here, try it again."
On the second trial, the contestant again picked Door 1. Mr. Hall opened Door 3, revealing a goat. The contestant was about to switch to Door 2 when Mr. Hall pulled out a roll of bills.
"You're sure you want Door No. 2?" he asked. "Before I show you what's behind that door, I will give you $3,000 in cash not to switch to it."
"I'll switch to it."
"Three thousand dollars," Mr. Hall repeated, shifting into his famous cadence. "Cash. Cash money. It could be a car, but it could be a goat. Four thousand."
"I'll try the door."
"Forty-five hundred. Forty-seven. Forty-eight. My last offer: Five thousand dollars."
"Let's open the door."
"You just ended up with a goat," he said, opening the door. The Problem With the Problem
Mr. Hall continued: "Now do you see what happened there? The higher I got, the more you thought the car was behind Door 2. I wanted to con you into switching there, because I knew the car was behind 1. That's the kind of thing I can do when I'm in control of the game. You may think you have probability going for you when you follow the answer in her column, but there's the pyschological factor to consider."
He proceeded to prove his case by winning the next eight rounds. Whenever the contestant began with the wrong door, Mr. Hall promptly opened it and awarded the goat; whenever the contestant started out with the right door, Mr. Hall allowed him to switch doors and get another goat. The only way to win a car would have been to disregard Ms. vos Savant's advice and stick with the original door.
Was Mr. Hall cheating? Not according to the rules of the show, because he did have the option of not offering the switch, and he usually did not offer it.
And although Mr. Hall might have been violating the spirit of Ms. vos Savant's problem, he was not violating its letter. Dr. Diaconis and Mr. Gardner both noticed the same loophole when they compared Ms. vos Savant's wording of the problem with the versions they had analyzed in their articles.
"The problem is not well-formed," Mr. Gardner said, "unless it makes clear that the host must always open an empty door and offer the switch. Otherwise, if the host is malevolent, he may open another door only when it's to his advantage to let the player switch, and the probability of being right by switching could be as low as zero." Mr. Gardner said the ambiguity could be eliminated if the host promised ahead of time to open another door and then offer a switch.
This post was edited on 2/23/15 at 3:13 pm
Posted on 2/23/15 at 3:34 pm to link
quote:
Again, the problem with this riddle is that the mathematical explanations always start with the premise that you must carry forward the old 33/66 probabilities from the first problem into the second. That is the cheat--you don't. The Monty Hall problem really is a 33/66 probability problem changed into a 50/50 problem, but discussed mathematically (incorrectly)after the basic premise has been changed as if it is still a 33/66 problem.
Conceptually, I agree. It is never a 1/3 proposition since the host is always going to drop a goat for you. In reality, it is always a 50/50 proposition once you accept that the Host will, no matter what you pick initially, always drop a goat.
In the end, you ultimately only get to choose between one goat or one car.
The Host never eliminates the car and therefore it is possible for the contestant to change a goat for a car or a car for a goat but it is impossible for the contestant to change a goat for a goat.
Posted on 2/23/15 at 4:54 pm to link
quote:
I have to disagree with the "always switch" conclusion. The correct answer about the probability is that it is 50/50.
Not treading all this but the real answer is:
Door A (your initial selection) = 1/3 chance of winn
Doors B & C (initially not selected) = 2/3 chance of winning.
Show host eliminates Door C (knowing it does not have the car), therefore door B now has the full 2/3 chance of winning.
Door A has 33.3% chance.
Door B has 66.7% chance.
If the car was randomly assigned after the elimination of Door C, the chances would revert to 50/50, but it doesn't happen that way.
Posted on 2/24/15 at 8:40 am to link
quote:
I have to disagree with the "always switch" conclusion. The correct answer about the probability is that it is 50/50.
The problem is not in the math--it is that the setup and explanation of the problem always given is a cheat. The problem is changed in the middle, but the original probabilities are not. That is why it is so counterintuitive.
Those who discuss the probability in terms of 33% and 66% after the first door is eliminated are carrying forward into the new problem (where only two doors exist) information from the original problem, and thus not establishing anything more earth shattering than the fact that we originally had 66% goats and 33% cars to choose from.
In fact, the reason people commonly say the answer is 50/50 is that most people intuitively accept the initial premise that a door is simply removed from the problem--leaving two doors, known to hide a goat and a car. This is clearly a 50/50 choice.
You can appreciate this truth if you consider the person who comes into the game at the point where the contestant is told that there is goat behind one of the doors he didn't pick. That door is effectively removed from the game, as if it never existed. The new person faces simply two doors, with no knowledge about what has gone on before, and the certainty that there is one goat and one car. It does not matter which of the two doors he picks (or whether he "picks" one then "switches", or doesn't switch.) In the end, he chooses one door--and has a 50/50 shot at getting it right. Thus, for him, it doesn't matter if he switches or not--his probability is always 50/50.
Again, the problem with this riddle is that the mathematical explanations always start with the premise that you must carry forward the old 33/66 probabilities from the first problem into the second. That is the cheat--you don't. The Monty Hall problem really is a 33/66 probability problem changed into a 50/50 problem, but discussed mathematically (incorrectly)after the basic premise has been changed as if it is still a 33/66 problem.
Yeah i'm no mathematician but this seemed pretty obvious to me.
The odds change when the game show host changed them. Why would you stick with the original data when forming your answer?
Posted on 2/24/15 at 8:48 am to link
I haven;t read all the responses but in this game it boils down to your only having two strategy choices in the beginning.
1 - you can pick any one door and stick with it. ( = 1/3 chance of success)
2 - you can pick two doors and stick with it. (to do this one, you 'pick' one and switch to the other two = 2/3 chances of success)
No. 2 is really saying "I'll eliminate that door and take whatever is behind the other two."
1 - you can pick any one door and stick with it. ( = 1/3 chance of success)
2 - you can pick two doors and stick with it. (to do this one, you 'pick' one and switch to the other two = 2/3 chances of success)
No. 2 is really saying "I'll eliminate that door and take whatever is behind the other two."
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