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Article I came across and thought it was interesting. The article discusses the "Monty Hall problem", but also discusses the feedback Marilyn vos Savant got after her coverage of it. Amusingly, the Times put this in the "Goats" collection.
Warning: The article is filled with advertisements.
Link (http://"http://www.nytimes.com/1991/07/21/us/behind-monty-hall-s-doors-puzzle-debate-and-answer.html?src=pm").
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The article (from more than twenty years ago) does eventually get around to making the simple point that the whole argument swings on: Monty knows.
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Suppose you're the producer of LMAD and you plan to play this deal once per week for the next 39 weeks. How many cars should you budget for and expect to give away?
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Suppose you're the producer of LMAD and you plan to play this deal once per week for the next 39 weeks. How many cars should you budget for and expect to give away?
I budget for 20 and hope like hell that I only give away 13.
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Suppose you're the producer of LMAD and you plan to play this deal once per week for the next 39 weeks. How many cars should you budget for and expect to give away?
I budget for 20 and hope like hell that I only give away 13.
That might be a problem. If the game is presented as we typically understand "The Monty Hall Problem", and if people do what they're supposed to do, the odds say you're going to lose 26 cars. Now, some people aren't going to do what they're supposed to, and Monty might employ some of his patented smooth-talking to play with the odds a bit, but the basic, dry game is a 2/3 win probability when played correctly.
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That might be a problem. If the game is presented as we typically understand "The Monty Hall Problem", and if people do what they're supposed to do, the odds say you're going to lose 26 cars. Now, some people aren't going to do what they're supposed to, and Monty might employ some of his patented smooth-talking to play with the odds a bit, but the basic, dry game is a 2/3 win probability when played correctly.
And here's where I become disappointed in myself. When the professor of the statistics/probability class did this, I demonstrated why people were falling for the fallacy, and just now I did it myself, reducing the problem to what happens after a door is opened, as opposed to taking into account the previous event that sets up the door being opened.
Budget for 26, hope that some of the contestantry don't get it, fall for the fast-talk or just are plain unlucky.
/Does knowing the way to get to the right answer but putting down something else count for partial credit?
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You don't budget and hope, you budget. The correct answer is 26.
This is where knowing the odds of the problem has utility. The worst-case scenario (or best-case, depending on your point of view) is that the contestant will switch every time, theoretically resulting in 66.7% wins. In actual practice you might budget for 27 cars to allow for a little variance in the outcomes.
The classic Monty Hall problem does not provide for Monty offering inducements to switch or not switch.
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In actual practice you might budget for 27 cars to allow for a little variance in the outcomes
I'm not mathematically sound to figure it out, but to those that are, what would the standard deviation be?
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what would the standard deviation be?
0.48
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Every time I read about the Monty Hall Problem, I come out of it with a headache*. Except for this time...it always fascinates me, and I always forget the "right" answer. Speaking of which, I forgot that back then it was still possible to get snarky with a columnist, just with snail mail instead of the online comments section. I'm bit surprised that college professors and Ph.D.'s would be such twats about it though**.
Does anyone know if Wayne's version uses the scenario, or is there any rhyme or reason to the deals there?
*/There's a reason I majored in journalism :-P
**//At least one apologized
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I'm bit surprised that college professors and Ph.D.'s would be such twats about it though**.
Why would you be surprised? They're paid to hold that position of authority and they probably don't get much blowback from students or other faculty, so when they're told "erm, excuse me professor, but you in fact have that precisely backwards", it doesn't surprise me one bit.
/Next time this topic comes up I'll say that you have a 2/3rds chance of winning if you stick.
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I'm going to blow the lid off the Monty Hall problem.
There is what I call "House Odds" outlined above. Given n trials and assuming the player switches every time, the odds are that the car will be won 66.7% of the time.
Then there are what I call "Player's Odds". To illustrate, suppose I have a fair die and give you six rolls of that die. What will be the distribution of the six rolls? Theoretically it will be
1 - 2 - 3 - 4 - 5 - 6
Now wipe the slate clean. I give you one roll of the die. What will the distribution of outcomes be? The answer is:
? ? ?
because the number of trials is less than the number of possible outcomes. On LMAD each player receives just one trial.
How's that headache, Brandon?
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I'm going to blow the lid off the Monty Hall problem.
...
because the number of trials is less than the number of possible outcomes. On LMAD each player receives just one trial.
This was the same problem people had with Deal or No Deal. People tried to use pot odds to determine whether to deal or not. Doesn't always work.
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Now wipe the slate clean. I give you one roll of the die. What will the distribution of outcomes be? The answer is:
? ? ?
because the number of trials is less than the number of possible outcomes. On LMAD each player receives just one trial.
But you can still apply knowledge of probability to know that your range is one through six, and that for a fair die you should get each number 1/6th of the time. Even though the die is rolled one time you still can know what to expect, if not the result.
LMAD has a game where you have a certain number of rolls to score 10 points. For the first prize you have four rolls, then three rolls, then two. Are you asserting that since a player has but one trial they should ignore the math required to compute the likelihood of acquiring the number of points needed in the rolls given and just fly by the seat of their pants?
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I'm going to blow the lid off the Monty Hall problem....
because the number of trials is less than the number of possible outcomes. On LMAD each player receives just one trial.
This was the same problem people had with Deal or No Deal. People tried to use pot odds to determine whether to deal or not. Doesn't always work.
Well, Deal or No Deal is a lot more complicated an exercise than the MHP. The fact that the LMAD contestant gets one trial doesn't change anything. The odds are what the odds are. The only question is whether you want a better chance to win the car, or a worse chance to win the car. You still might not win the car, but just because it's a single trial doesn't suddenly make it a 50/50 proposition.
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Even though the die is rolled one time you still can know what to expect, if not the result.
Here's your one roll. What is the outcome? All you know is that you have a 5/6 chance of being wrong.
just because it's a single trial doesn't suddenly make it a 50/50 proposition.
The house odds are 66.7% assuming the contestant switches every time, regardless of the number of trials. If the player doesn't switch every time the odds of winning are 33.3%. The distinction has to be made between the house odds and the player's odds.
It is never an incorrect strategy to switch.
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Even though the die is rolled one time you still can know what to expect, if not the result.
Here's your one roll. What is the outcome? All you know is that you have a 5/6 chance of being wrong.
Right, but so what? There are n different results for an n-sided die. The outcome is unknown until the event is carried out.
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Now to blow the lid off my blowing the lid off.
In all of the circumlocutious explanations of the MHP I've seen, I've never seen it explained in the following simple manner. Suppose the doors are laid out such that the car is behind door 2. Doors 1 and 3 conceal zonks. Now consider the following truth table:
PICK REVEAL REMAINING SWITCH NO SWITCH
1 3 1 OR 2 2 (Win) 1 (Lose)
2 1 2 OR 3 3 (Lose) 2 (Win)
3 1 2 OR 3 2 (Win) 3 (Lose)
By switching, the only way to lose is to pick the door with the car and switch, giving a 1/3 chance of losing and a 2/3 chance of winning. By not switching, the only way to win is to pick the door with the car, giving a 1/3 chance of winning.
Simple.
/If the contestant picks door 2, either door 1 or 3 can be revealed. It works out the same in either case.
//Headache any better, Brandon?
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PICK REVEAL SWITCH NO SWITCH
1 3 2 (Win) 1 (Lose)
2 1 3 (Lose) 2 (Win)
2 3 1 (Lose) 2 (Win)
3 1 2 (Win) 3 (Lose)
There's nothing more fun in a math class than those extension problems. The table above looks like the probability of winning after a switch is 1/2, but it's still 2/3. Who wants to take a crack at explaining why? (This will be on the final.)
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Because two of the outcomes are mutually exclusive, right?
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/If the contestant picks door 2, either door 1 or 3 can be revealed. It works out the same in either case.
//Headache any better, Brandon?
No. Where's my Advil?
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There's nothing more fun in a math class than those extension problems. The table above looks like the probability of winning after a switch is 1/2, but it's still 2/3. Who wants to take a crack at explaining why? (This will be on the final.)
Let me see if I'm way off base here.
At the very beginning, you have a 1/3 chance of picking the right one, which means you have a 2/3 chance of picking the wrong one. Revealing one of the wrong doors does not change that at all. You're still 1/3 of the time right, 2/3 of the time wrong. The only difference is now you know which one to switch to if you want to switch, and since you're wrong 2/3 of the time at the start, you should always switch.
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PICK REVEAL SWITCH NO SWITCH
1 3 2 (Win) 1 (Lose)
2 1 3 (Lose) 2 (Win)
2 3 1 (Lose) 2 (Win)
3 1 2 (Win) 3 (Lose)
There's nothing more fun in a math class than those extension problems. The table above looks like the probability of winning after a switch is 1/2, but it's still 2/3. Who wants to take a crack at explaining why? (This will be on the final.)
You're double counting the door 2 pick. There are only three options: door 1, 2 or 3. If you want to examine the outcome of picking 2 and revealing 3, here's how to do it:
PICK REVEAL REMAINING SWITCH NO SWITCH
1 3 1 OR 2 2 (Win) 1 (Lose)
2 3 1 OR 2 1 (Lose) 2 (Win)
3 1 2 OR 3 2 (Win) 3 (Lose)
This is better analyzed as a logic problem than as a math problem.
/Brandon: We could analyze this some more until your headache is better.
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Let me see if I'm way off base here.
At the very beginning, you have a 1/3 chance of picking the right one, which means you have a 2/3 chance of picking the wrong one. Revealing one of the wrong doors does not change that at all. You're still 1/3 of the time right, 2/3 of the time wrong. The only difference is now you know which one to switch to if you want to switch, and since you're wrong 2/3 of the time at the start, you should always switch.
Without switching your chances of winning the car are 1/3. Switching inverts the logic of the outcomes and gives you the 2/3 odds. Switching gives you the logical NOT (inverse) of the outcomes.
/Need more Advil?
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At the very beginning, you have a 1/3 chance of picking the right one, which means you have a 2/3 chance of picking the wrong one. Revealing one of the wrong doors does not change that at all. You're still 1/3 of the time right, 2/3 of the time wrong. The only difference is now you know which one to switch to if you want to switch, and since you're wrong 2/3 of the time at the start, you should always switch.
You've got it.
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You're double counting the door 2 pick.
Not when you count the pick and reveal as the experiment. Then there are two outcomes associated with picking the correct door: either one of the two remaining doors can be revealed by the host. Thus, there are 4 elements in the sample space. The catch is that the two middle rows in my table each have probability of 1/6: 1/3 you'll pick Door 2 and 1/2 Monty will reveal 1 or 3.
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You're double counting the door 2 pick.
Not when you count the pick and reveal as the experiment. Then there are two outcomes associated with picking the correct door: either one of the two remaining doors can be revealed by the host. Thus, there are 4 elements in the sample space. The catch is that the two middle rows in my table each have probability of 1/6: 1/3 you'll pick Door 2 and 1/2 Monty will reveal 1 or 3.
I don't see a bright future for you in the fields of statistics or logic.
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You're double counting the door 2 pick.
Not when you count the pick and reveal as the experiment. Then there are two outcomes associated with picking the correct door: either one of the two remaining doors can be revealed by the host. Thus, there are 4 elements in the sample space. The catch is that the two middle rows in my table each have probability of 1/6: 1/3 you'll pick Door 2 and 1/2 Monty will reveal 1 or 3.
I don't see a bright future for you in the fields of statistics or logic.
I'll pass that along to my graduate advisor in the math department.
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You're double counting the door 2 pick.
Not when you count the pick and reveal as the experiment. Then there are two outcomes associated with picking the correct door: either one of the two remaining doors can be revealed by the host. Thus, there are 4 elements in the sample space. The catch is that the two middle rows in my table each have probability of 1/6: 1/3 you'll pick Door 2 and 1/2 Monty will reveal 1 or 3.
I don't see a bright future for you in the fields of statistics or logic.
I'll pass that along to my graduate advisor in the math department.
If I were you, I'd keep it a secret.
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You're double counting the door 2 pick.
Not when you count the pick and reveal as the experiment. Then there are two outcomes associated with picking the correct door: either one of the two remaining doors can be revealed by the host. Thus, there are 4 elements in the sample space. The catch is that the two middle rows in my table each have probability of 1/6: 1/3 you'll pick Door 2 and 1/2 Monty will reveal 1 or 3.
I don't see a bright future for you in the fields of statistics or logic.
I'll pass that along to my graduate advisor in the math department.
If I were you, I'd keep it a secret.
Now hold on a minute here...I've taught the Monty Hall problem in various math classes for years, and, in all three textbooks that I have used that explained how it works, what Dale wrote is the exact explanation they use. You can even go more general and show that, even if the probability that Monty opens either of the two doors in the "bad" case is not equal, you will still get the same 2/3 probability winning when switching.
It's basically an exercise in conditional probability and/or in using tree diagrams and the "multiplication rule."
Anthony
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Now hold on a minute here...I've taught the Monty Hall problem in various math classes for years, and, in all three textbooks that I have used that explained how it works, what Dale wrote is the exact explanation they use. You can even go more general and show that, even if the probability that Monty opens either of the two doors in the "bad" case is not equal, you will still get the same 2/3 probability winning when switching.
Take it a step further. Let's say there are 1000 doors. The contestant picks a door. Monty opens up 998 other doors, all showing zonks. The probability the car resides behind the unrevealed door not chosen by the contestant is 999/1000. The probability the car is behind the the unpicked door is (x-1)/x, where x is the number of doors in play.
Mythbusters tackled The Monty Hall Problem beautifully about 6 months ago:
http://www.youtube.com/watch?v=Ytv-_RB4xWI
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If I were you, I'd keep it a secret.
You know full well you don't get full credit unless you show your work. Care to explain why my method is incorrect?
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I completely understand why it's better to switch, but I've always thought that if I were ever actually on a game show where it was used, I'd be stubbornly apprehensive about switching. That little thought of "But what if I DID pick the right one the first time?" would make the decision a little tougher. But that's just how my mind works...lol.
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Take it a step further. Let's say there are 1000 doors. The contestant picks a door. Monty opens up 998 other doors
Monty opens only 27 of those 1000 doors, it being a 30-minute show. Monty has plum run out of time. The stagehands are about to go into overtime. We have proven nothing.
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If I were you, I'd keep it a secret.
You know full well you don't get full credit unless you show your work. Care to explain why my method is incorrect?
I already did.
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Take it a step further. Let's say there are 1000 doors. The contestant picks a door. Monty opens up 998 other doors
Monty opens only 27 of those 1000 doors, it being a 30-minute show. Monty has plum run out of time. The stagehands are about to go into overtime. We have proven nothing.
You and your realism. We're doing a 24-hour marathon. After every 100 doors are opened, the audience is entertained by naked dancing girls. There won't be a dry seat in the house.
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I already did.
You explained why yours was correct. You didn't explain why mine was incorrect. As a logician, you should be the first to realize those two situations aren't equivalent.
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You're double counting the door 2 pick.
Not when you count the pick and reveal as the experiment. Then there are two outcomes associated with picking the correct door: either one of the two remaining doors can be revealed by the host. Thus, there are 4 elements in the sample space. The catch is that the two middle rows in my table each have probability of 1/6: 1/3 you'll pick Door 2 and 1/2 Monty will reveal 1 or 3.
To clarify, you're both right. Chris correctly mentions that by counting the "pick door 2" event twice the probability appears to be 1/2. Dale is correct, however, in acknowledging that there are two different events that can happen after picking door 2, each with probability 1/6, as opposed to the probability of 1/3 when picking door 1 or door 3.
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Alternate explanation?
Let p = Probability of winning
Let q = Probability of not winning
As I recall from my statistics and probability classes, p + q =1.
If p = 1/3, then 1/3 + q = 1 -> q = 1 - 1/3 -> q = 2/3? In selecting your original door, your change of winning is p. By switching, your probability of winning is now q (or, rather, not p).
Just checking...
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Alternate explanation?
Let p = Probability of winning
Let q = Probability of not winning
As I recall from my statistics and probability classes, p + q =1.
If p = 1/3, then 1/3 + q = 1 -> q = 1 - 1/3 -> q = 2/3? In selecting your original door, your change of winning is p. By switching, your probability of winning is now q (or, rather, not p).
Just checking...
Perfectly (and simply) accurate.
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The contestant has only three options for the original pick. The truth table must reflect that. Each door can be one of two states: CAR or ZONK. Here is the truth table reworked to reflect that, assuming the car is behind door 2:
PICK REVEAL REMAINING SWITCH NO SWITCH
1 ZONK CAR OR ZONK CAR ZONK
2 ZONK CAR OR ZONK ZONK CAR
3 ZONK CAR OR ZONK CAR ZONK
Note that the "REVEAL" and "REMAINING" columns are the same across the contestant's three possible picks.
Mike will now post the truth table for 1,000 doors which will include naked dancing girls.
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Mike will now post the truth table for 1,000 doors which will include naked dancing girls.
No. The dancing girls are staying with me. Giggity.
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Mike will now post the truth table for 1,000 doors which will include naked dancing girls.
No. The dancing girls are staying with me.
You may have your dancing girls provided you can satisfy 10 per night. I know you can do it.
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From Comics I Don't Understand, "You Can't Fight Monty Hall (http://"http://comicsidontunderstand.com/wordpress/2012/06/12/doors/")." Convincingly demonstrating why not switching is a bad idea.
-3M
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Does anyone know if Wayne's version uses the scenario, or is there any rhyme or reason to the deals there?
In the Big Deal Wayne always reveals one the of the lessor deals first. One of the players who most likely knew the puzzle asked Wayne "Can I switch now?" Without hesitation he answered, "No"
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You may have your dancing girls provided you can satisfy 10 per night. I know you can do it.
Mike's got a great lasagna recipe.
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Does anyone know if Wayne's version uses the scenario, or is there any rhyme or reason to the deals there?
In the Big Deal Wayne always reveals one the of the lessor deals first. One of the players who most likely knew the puzzle asked Wayne "Can I switch now?" Without hesitation he answered, "No"
I don't know if he still does this (I haven't watched in a while) but in the early going, after revealing one of the lesser deals, he'd say something like "Now you have a fifty-fifty chance". It's a small thing, but that fallacy lies at the heart of why some people don't understand the problem.