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Message
1
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.
Posted on 2/23/15 at 3:04 pm to WG_Dawg
quote:
you just quoted the same thing...
That's because I thought you may have mis-read it the first time since I didn't include anything in that post that made any reference to my personal opinion on the matter.
Posted on 2/23/15 at 3:05 pm to Ace Midnight
By always switching you increase your odds from 33% to 50%
Posted on 2/23/15 at 3:07 pm to Monk
quote:
since I didn't include anything in that post that made any reference to my personal opinion on the matter.
Which is why I asked what your opinion was.
quote:
I thought you were going to reference the time that she stated that there is a 50/50 chance of flipping heads even after 9 heads in a row.
I recall that caused a stir.
quote:
Are you saying you disagree with that?
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 3:11 pm to WG_Dawg
To me, it depends on perspective and what exactly is being asked.
If the question is what is the chance of flipping heads even after flipping 9 heads in a row, the answer is 50/50.
If the question is what are the chances of flipping heads 10 consecutive times in a row?
Different answer.
It's a serial solution versus a single solitary flip of the coin.
If the question is what is the chance of flipping heads even after flipping 9 heads in a row, the answer is 50/50.
If the question is what are the chances of flipping heads 10 consecutive times in a row?
Different answer.
It's a serial solution versus a single solitary flip of the coin.
Posted on 2/23/15 at 3:13 pm to Monk
quote:
If the question is what is the chance of flipping heads even after flipping 9 heads in a row, the answer is 50/50.
If the question is what are the chances of flipping heads 10 consecutive times in a row?
Different answer.
Agree. I always get a kick out of people who think the odds of H/T varies on one single coin flip based on what's happened in previous flips. I'm horrendous at math and statistics, but one thing I DO know is that it's still 50/50 per flip ha.
Posted on 2/23/15 at 3:18 pm to Monk
quote:
If the question is what are the chances of flipping heads 10 consecutive times in a row?
Different answer
The probability of flipping any sequence of 10 flips is: 1/1024
Posted on 2/23/15 at 3:31 pm to CockHolliday
quote:Your error is assuming that all 4 of these scenarios have an equal chance of occurring. I think the problem is better to frame from the original choice. Since the scenarios would play themselves out the same way for all three doors chosen, we will just assume door 1 is chosen.
If Door 1 contains the car:
1) Guest picks Door 1, host opens Door 2. If guest switches, lose. If guest stays, win.
2) Guest picks Door 1, host opens Door 3. Switch: lose. Stay: win.
3) Guest picks Door 2, host opens Door 3. Switch: win. Stay: lose.
4) Guest picks Door 3, host opens Door 2. Switch: win. Stay: lose.
Choose Door 1 and Switch: 2 wins; 1 loss
1.) 1/3 chance car is behind door one (either door can be open; it doesn't matter). Lose
2.) Car is behind door two (door three is open). Win
3.) Car is behind door 3 (door two is open). Win
Choose door 1 and Stay: 1 win; 2 loses
1.) 1/3 chance car is behind door one (either door can be open; it doesn't matter). Win
2.) Car is behind door two (door three is open). Lose
3.) Car is behind door 3 (door two is open). Lose
Basically, choosing the correct door on one's first choice is still 1/3 and it being behind one of the other two doors is 2/3. When you switch, you are basically choosing the other 2 doors (even though we know one doesn't have the car, as shown).
I think a common error is that there are technically 4 possibilities in the above scenarios, 2 of those possibilities occur when you choose the correct door immediately; however, those 2 possibilities can only add up to 1/3. For example, if you chose the correct door (1/3 chance), and each door then has a is a 1/2 chance of being opened, then choosing the correct door and door 2 being open is actually 1/6 (1/3*1/2); same for door 3. In other words, the odds either door 2 or door 3 is chosen must add up to 1/3.
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 3:36 pm to buckeye_vol
(no message)
This post was edited on 2/23/15 at 3:37 pm
Posted on 2/23/15 at 3:37 pm to Monk
Thanks Monk for proving my point.
In your explanation, you are absolutely correct--but only because you consider it all one game.
However, the problem is always described by asking what the odds are of choosing the car once one door is eliminated (and is always a goat). That IS a new game, and the probabilities are 50/50. If mathematicians don't like the fact that people perceive it correctly to be a new game, that is fine, they can continue to run it all together as one multi-stage probability exercise. But that is mere slight of hand, not math.
In your explanation, you are absolutely correct--but only because you consider it all one game.
However, the problem is always described by asking what the odds are of choosing the car once one door is eliminated (and is always a goat). That IS a new game, and the probabilities are 50/50. If mathematicians don't like the fact that people perceive it correctly to be a new game, that is fine, they can continue to run it all together as one multi-stage probability exercise. But that is mere slight of hand, not math.
Posted on 2/23/15 at 3:43 pm to link
quote:But it is not a new game, and the odds aren't 50/50. The odds that it is behind the door not originally chosen is 2/3. Just because there are 2 choices, doesn't mean they are equal.
However, the problem is always described by asking what the odds are of choosing the car once one door is eliminated (and is always a goat). That IS a new game, and the probabilities are 50/50.
quote:Bayes' Theorem and conditional probability are most surely math. Just because it's counter-intuitive doesn't make it a sleight of hand. It has very practical uses. In fact,the same formulation was used to find the wreckage of the Air France flight that crashed in 2009. A sleight of hand doesn't do that.
But that is mere sleight of hand, not math.
This post was edited on 2/23/15 at 3:46 pm
Posted on 2/23/15 at 3:45 pm to Monk
quote:But you are basically choosing 2 doors when you switch (2/3 chance). So it is technically two choices, but the odds are still 2/3 to 1/3.
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.
Posted on 2/23/15 at 3:47 pm to buckeye_vol
I have a better idea.
Take any two doors, where you know that there is a car behind one and a goat behind the other. Pick one. Switch or don't switch, I don't care. Record your results over 100 tries. (Hint: if you flip a coin this will be the same, and easier).
The point is: If you accept carrying the initial 3-door probabilities forward, the mathematical proofs are correct. But that is a lie; the problem has been changed to eliminate one non-car door. That is what people intuitively understand, and why they say that it is a 50/50 shot at that point.
Take any two doors, where you know that there is a car behind one and a goat behind the other. Pick one. Switch or don't switch, I don't care. Record your results over 100 tries. (Hint: if you flip a coin this will be the same, and easier).
The point is: If you accept carrying the initial 3-door probabilities forward, the mathematical proofs are correct. But that is a lie; the problem has been changed to eliminate one non-car door. That is what people intuitively understand, and why they say that it is a 50/50 shot at that point.
Posted on 2/23/15 at 3:52 pm to buckeye_vol
quote:
But you are basically choosing 2 doors when you switch (2/3 chance). So it is technically two choices, but the odds are still 2/3 to 1/3.
That's how I like to look at it. There is no chance involved in Montel's decision. Deliberately showing you a goat and allowing you to switch your door for his remaining door is equivalent to not showing you a door and letting you trade your door for his two, at least one of which, of course, will conceal a goat.
Posted on 2/23/15 at 3:56 pm to buckeye_vol
No matter which door you initially pick, there is always at least one door the host can open to show a goat. So why are your odds always better if you switch your original pick? In the case where the host opens door 3 and shows a goat, why does switching from door 1 to door 2 double your odds? Wouldn't saying that the same would be true had you picked door 2 originally and switched your pick to door 1 be contradictory?
Posted on 2/23/15 at 3:56 pm to link
quote:
Thanks Monk for proving my point.
Thanks, but I think you made your point about as clearly as can be done in this situation. I just tried to conceptually elaborate on what I thought was your point.
I can't really say that the original solution involves treating it as one game or running "it all together as one multi-stage probability exercise."
In the linked article, they explain that
"The short answer is that your initial odds of winning with door #1 (?) don’t change simply because the host reveals a goat behind door #3; instead, Hall’s action increases the odds to ? that you’ll win by switching."
To me, they are admitting that the "initial" 1/3 odds do not change simply by dropping the goat. I read it has saying that the Host allows you an extra chance of winning by giving you another bite at the apple when you consider that the Host will never drop a car so you effectively get a 2nd chance to pick the car, with 1 less goat to choose from.
Posted on 2/23/15 at 4:00 pm to link
quote:Well this is irrelevant to the scenario. If this was the game, then why didn't Monty Hall just present two doors?
Take any two doors, where you know that there is a car behind one and a goat behind the other. Pick one. Switch or don't switch, I don't care. Record your results over 100 tries. (Hint: if you flip a coin this will be the same, and easier).
quote:But the probability doesn't change to 50/50 just because there are two choices. It's 66/33.
The point is: If you accept carrying the initial 3-door probabilities forward, the mathematical proofs are correct. But that is a lie; the problem has been changed to eliminate one non-car door.
Here is my question to you. If you were to play the game 100 times, and decided to switch, how many times would you win?
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