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Marilyn vos Savant and the history of the Montel Hall question
Posted on 2/22/15 at 8:57 pm
Posted on 2/22/15 at 8:57 pm
quote:
When vos Savant politely responded to a reader’s inquiry on the Monty Hall Problem, a then-relatively-unknown probability puzzle, she never could’ve imagined what would unfold: though her answer was correct, she received over 10,000 letters, many from noted scholars and Ph.Ds, informing her that she was a hare-brained idiot.
What ensued for vos Savant was a nightmarish journey, rife with name-calling, gender-based assumptions, and academic persecution.
LINK
Posted on 2/22/15 at 9:29 pm to Big Scrub TX
I vaguely remember the controversy. The Monty Hall Problem is pretty counter-intuitive and, in fairness (it was touched upon in the article) in "real life" it works a little different than it was presented in the original hypo.
Still, a resounding victory for the genius and proof that the Internet didn't invent the shrill response crowd, it just perfected it.
Still, a resounding victory for the genius and proof that the Internet didn't invent the shrill response crowd, it just perfected it.
Posted on 2/22/15 at 10:09 pm to Big Scrub TX
quote:I remember Montel Hall. He hosted this weird game show where if you wanted to be picked as a contestant you had to dress up as an interracial transsexual unwed mother
Montel Hall
Posted on 2/23/15 at 11:47 am to Big Scrub TX
quote:
Marilyn vos Savant
She was pretty good-looking back in the day.
I've always had a thing for really smart chicks.
Dumb ones too, come to think of it.
Posted on 2/23/15 at 12:09 pm to Big Scrub TX
Doesn't it matter more whether the opened door was chosen at random or picked specifically because it did NOT have the car? This issue confuses me even after reading the details.
Posted on 2/23/15 at 12:56 pm to Big Scrub TX
Assuming the game host would never open a door containing the car, there are 12 possible scenarios here (4 separate ones for each door containing the car):
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.
Theoretically there would be 2 more scenarios for Door 1 containing the car, but those would involve opening Door 1 which we've already ruled out for purposes of the contest. So far, switching is 50/50.
If Door 2 contains the car:
1) Guest picks Door 1, host opens Door 3. Switch: win. Stay: lose.
2) Guest picks Door 2, host opens Door 1. Switch: lose. Stay: win.
3) Guest picks Door 2, host opens Door 3. Switch: lose. Stay: win.
4) Guest picks Door 3, host opens Door 1. Switch: win. Stay: lose.
Still 50/50 by switching vs. staying.
If Door 3 contains the car:
1) Guest picks Door 1, host opens Door 2. Switch: win. Stay: lose.
2) Guest picks Door 2, host opens Door 1. Switch: win. Stay: lose.
3) Guest picks Door 3, host opens Door 1. Switch: lose. Stay: win.
4) Guest picks Door 3, host opens Door 2. Switch: lose. Stay: win.
Once again, 50/50.
What am I missing here?
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.
Theoretically there would be 2 more scenarios for Door 1 containing the car, but those would involve opening Door 1 which we've already ruled out for purposes of the contest. So far, switching is 50/50.
If Door 2 contains the car:
1) Guest picks Door 1, host opens Door 3. Switch: win. Stay: lose.
2) Guest picks Door 2, host opens Door 1. Switch: lose. Stay: win.
3) Guest picks Door 2, host opens Door 3. Switch: lose. Stay: win.
4) Guest picks Door 3, host opens Door 1. Switch: win. Stay: lose.
Still 50/50 by switching vs. staying.
If Door 3 contains the car:
1) Guest picks Door 1, host opens Door 2. Switch: win. Stay: lose.
2) Guest picks Door 2, host opens Door 1. Switch: win. Stay: lose.
3) Guest picks Door 3, host opens Door 1. Switch: lose. Stay: win.
4) Guest picks Door 3, host opens Door 2. Switch: lose. Stay: win.
Once again, 50/50.
What am I missing here?
Posted on 2/23/15 at 1:16 pm to Big Scrub TX
I don’t follow her logic. We’re ultimately presented with information that a prize is behind one of two doors. Why should it matter which one is called door number 1, 2, or 3? Why should it matter that the original question was posed as if the odds were 1 in 3?
Let’s work it backwards. There are two doors. A prize is behind one. The odds of correctly guessing which door the prize is behind is 50/50, right? You choose one. I ask you if you want to change your mind. If you do and choose the other door, did the odds of correctly guessing suddenly change? Or is it still 50/50 odds? If so, explain. What about if I suddenly say, “Would you change your mind if I told you there was another door that did NOT have a prize behind it, would you change your mind?” Would that additional question and fact suddenly change the odds to 1 out of 3? What if I said there were 100 additional doors that a prize was NOT behind? Does that make the original question anything different than a 50/50 chance?
I still don’t understand how her answer is regarded as correct.
Let’s work it backwards. There are two doors. A prize is behind one. The odds of correctly guessing which door the prize is behind is 50/50, right? You choose one. I ask you if you want to change your mind. If you do and choose the other door, did the odds of correctly guessing suddenly change? Or is it still 50/50 odds? If so, explain. What about if I suddenly say, “Would you change your mind if I told you there was another door that did NOT have a prize behind it, would you change your mind?” Would that additional question and fact suddenly change the odds to 1 out of 3? What if I said there were 100 additional doors that a prize was NOT behind? Does that make the original question anything different than a 50/50 chance?
I still don’t understand how her answer is regarded as correct.
Posted on 2/23/15 at 1:57 pm to Big Scrub TX
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.
I recall that caused a stir.
Posted on 2/23/15 at 2:25 pm to Big Scrub TX
I learned something from TD.com today.
Posted on 2/23/15 at 2:30 pm to Big Scrub TX
quote:
You are the goat!
Glenn Calkins
Western State College
Posted on 2/23/15 at 2:57 pm to Big Scrub TX
This actually a fairly smart question for fairly smart people.
The answer is counterintuitive - but it is correct from a probability standpoint.
However, it is the nature of the 1/3 and of the laws of probability itself.
The original choice is a 1 in 3. The fact that it eliminates one of the 2 bad choices inherently increases your odds - you cannot switch from a bad answer to a bad answer but only from bad to good or good to bad.
All of the failures occur on the initial choice (picking the right one) - all of the successes occur on the switches (switching from the bad one, with the other bad one eliminated).
Weird, but it works.
(Because they never show you the good choice if you pick one of the bad - that's the other factor that makes this work.)
The answer is counterintuitive - but it is correct from a probability standpoint.
However, it is the nature of the 1/3 and of the laws of probability itself.
The original choice is a 1 in 3. The fact that it eliminates one of the 2 bad choices inherently increases your odds - you cannot switch from a bad answer to a bad answer but only from bad to good or good to bad.
All of the failures occur on the initial choice (picking the right one) - all of the successes occur on the switches (switching from the bad one, with the other bad one eliminated).
Weird, but it works.
(Because they never show you the good choice if you pick one of the bad - that's the other factor that makes this work.)
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 2/24/15 at 4:06 am to Big Scrub TX
This again proves my point that most scholars are huge aholes. I worked in academia for a while and if it wasn't for the benefits I wanted to beat the ever living crap out of PhDs and professors on a daily basis.
Posted on 2/24/15 at 10:30 pm to Big Scrub TX
Enjoyable read. I remembered the hullabaloo about the problem.
Posted on 2/24/15 at 10:47 pm to Big Scrub TX
What I love about this is that ANYONE can prove out for themselves that switching wins 2/3 of the time by using nothing more than three pieces of paper to represent the three doors and following the rules of the problem and playing the game themselves. No math involved, just play the game.
Yet, people STILL insist it's 50/50. Bless their hearts.
Yet, people STILL insist it's 50/50. Bless their hearts.
This post was edited on 2/24/15 at 10:49 pm
Posted on 2/25/15 at 6:07 am to Big Scrub TX
The issue that is problematic is that most view this as a show of probability when it is more a show of profitability. Analytic thought process requires one to consider all alternatives including those outside the norm.
In some ways you need to understand this game show is like being at a casino. You are playing against the house and the game is rigged to ensure the house stays profitable.
There is a final component in this show - it is called profitability- Though the show is designed to be entertaining and on some level challenges the math component of our intelligence - ultimately this show is about marketing. IF the show is not successful in viewer ratings then securing sponsors would falter and the show would not be a business success.
The fact that this show has endured time indicates how strong the "house" component is integrated into the show.
A solid contestant could increase his odds of winning by never forgetting the show is designed for the house to not win but be profitable.
You gotta know when to holdem and when to foldem!
In some ways you need to understand this game show is like being at a casino. You are playing against the house and the game is rigged to ensure the house stays profitable.
There is a final component in this show - it is called profitability- Though the show is designed to be entertaining and on some level challenges the math component of our intelligence - ultimately this show is about marketing. IF the show is not successful in viewer ratings then securing sponsors would falter and the show would not be a business success.
The fact that this show has endured time indicates how strong the "house" component is integrated into the show.
A solid contestant could increase his odds of winning by never forgetting the show is designed for the house to not win but be profitable.
You gotta know when to holdem and when to foldem!
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