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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 11:43 am to Walking the Earth
quote:
proof that the Internet didn't invent the shrill response crowd, it just perfected it
Good point.
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 11:52 am to Walking the Earth
quote:
in "real life" it works a little different than it was presented in the original hypo.
The crux of this argument is that a door containing nothing will always be removed first and the person opening the doors know which is which.
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 12:17 pm to USMCTiger03
quote:
Doesn't it matter more whether the opened door was chosen at random or picked specifically because it did NOT have the car?
Yes, there is no randomness factored in.
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 1:23 pm to Willie Stroker
quote:
I still don’t understand how her answer is regarded as correct.
I'm with you since I"m not a math/probability guy AT ALL, but how someone explained it to me was this.
At the start when you pick a door, you have a 33% chance of guessing correctly. After 1 door is revealed to be nothing, now your chances of picking the correct door are 50/50 between car and goat, so you're "supposed" to switch. I don't really get it, but that's how it was told to me.
Posted on 2/23/15 at 1:33 pm to WG_Dawg
quote:
I'm with you since I"m not a math/probability guy AT ALL, but how someone explained it to me was this.
At the start when you pick a door, you have a 33% chance of guessing correctly. After 1 door is revealed to be nothing, now your chances of picking the correct door are 50/50 between car and goat, so you're "supposed" to switch. I don't really get it, but that's how it was told to me.
Look at it this way:
There are 3 doors. You pick door 1. The host says, "You can either keep door 1 or have BOTH of the the other doors."
You know for sure that at least one of these other two doors have a goat behind it, but you will still make the switch because two doors are better than one.
The actual problem follows the same logic.
You pick a door, and then are given the option of keeping your door or switching to the other 2 doors. The only difference is that this time they show you that one of the other two doors has a goat behind it...but you already knew with certainty that one of these two doors had a goat behind it, so him showing it to you doesn't change anything.
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 2:23 pm to Monk
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.
Are you saying you disagree with that?
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 2:52 pm to CockHolliday
quote:
What am I missing here?
I'm guessing it has to do with the fact that you are considering 4 options instead of focusing on only 3 doors/options.
If you pick the right door initially, the host has 2 options on doors to open. As you basically noted, if you pick wrong initially, the host only has 1 door to open.
In your test, you are allowing the contestant to pick the correct door twice. In reality, there are 3 doors and he gets 1 initial pick, regardless which of the 2 goat doors the host decides to open next. So 1 and 2 in your test is really 1 OR 2 so ONE win, not two.
Posted on 2/23/15 at 2:54 pm to WG_Dawg
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.
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.)
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