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Author | Topic: Counter-Intuitive Science | ||||||||||||||||||||||||
slevesque Member (Idle past 4668 days) Posts: 1456 Joined: |
Ok, I'll give more reasoning:
(1/3) You pick the door with the car : Presentator will open a door with a goat (sit.A) (2/3) You pick a door with a goat : Presentator will open a door with a goat (sit.B) or with the car (sit.C) Now, this is the full scenario when the presentator does not know where the care is. But since I said the presentator did reveal a goat, it means you are in either Sit.A or Sit.B, and it is pretty straightforward to see that, once a goat is revealed, the odds of your door having the car is 1/2.
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cavediver Member (Idle past 3671 days) Posts: 4129 From: UK Joined: |
Ok, I'll give more reasoning: No, your reasoning is fine.
But since I said the presentator did reveal a goat, Ah, did you? Here's your original statement:
slev writes: While in the second (seemingly identical) scenario he does not know where the prize is, then it doesn't matter if you change or not, you'll have the same odds of winning. What I'm stressing here is the mistake in giving a definitive answer to a loosely defined question. I know it sounds anal (does that phrase translate??) but I see it as an endemic problem and an obstical to critical thinking. If you clearly state the question as: "the presenter doesn't know where the car is, and he opens one of the two doors at random, and he reveals a goat, then should you swap?" then fair enough, the answer is obviously - it doesn't matter. BUT if you ask: "the presenter doesn't know where the car is, and he opens one of the two doors at random, then should you swap?" then it is no wonder that people are confused!
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slevesque Member (Idle past 4668 days) Posts: 1456 Joined: |
It is because I said this:
Because imagine two almost identical scenario where you have a choice between door A,B,C. You choose A, the presentator opens door C and chose a goat in both scenarios. I thought it was pretty clear I was talking about when you find yourself in one of the situations where he reveals a goat and not in the situation where he reveals a car. AbE When you say, ''I know it sounds anal'' and ask if it translates, you are talking translate from what to what ? From engliush to french, from french to english, from UK english to US english ? I didn't quite understand Edited by slevesque, : No reason given.
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RAZD Member (Idle past 1432 days) Posts: 20714 From: the other end of the sidewalk Joined: |
I think I agree with Panda: how does the host not knowing the answer change this picture:
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slevesque Member (Idle past 4668 days) Posts: 1456 Joined: |
See post no136
A third of the time, he opens a goat and you chose the car. (Sit.A) A third of the time, he opens a goat and you chose a goat (Sit.B) A third of the time, he opens a car and you chose a goat (Sit.C) Since the presentator opens a goat, you know you are in situation A or B, and you know each situation is just as likely as the other (33% each), then you also know that since you are as likely to be in sit.A then B, then you are just as likely to have the car behind your door then behind the other door. Hence a 50/50 chance
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Panda Member (Idle past 3740 days) Posts: 2688 From: UK Joined: |
RAZD writes:
I think the reasoning is: I think I agree with Panda: how does the host not knowing the answer change this picture:Since there is now a chance of Monty winning the car, your chance of winning the car (on the right hand side of your diagram) must be lower. I have given this more thought, and I am leaning towards the 50:50 answer.i.e. You stand a 1:3 chance of winning if you stick, 1:3 chance of winning if you switch (and Monty has a 1:3 of winning). To use the normal teaching method for elucidating the standard Monty Hall question: Imagine that there were 100 doors...You randomly choose a door (e.g. Door 15). Monty randomly chooses 98 doors (e.g. Doors 1-14 and Doors 16-99). If you applied the reasoning of your image, then we would have a 99:100 chance of winning if we switch.But that is incorrect: Monty has a 98:100 chance of winning. Whether you switch or not, you would have a 1:100 chance of winning the car. If Monty was to actually open each of his doors (instead of just 'choosing' them), the the odds would only change if he choose a door with a car behind - i.e. you chance of winning is zero.
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NoNukes Inactive Member |
Panda writes: I think the reasoning is:Since there is now a chance of Monty winning the car, your chance of winning the car (on the right hand side of your diagram) must be lower. This reasoning cannot be correct. The question asks what one should do when Monty shows a goat. In other words, Monty's possibility of showing the car should be removed from the considered outcomes, and probabilities would be calculated based on the remaining possible outcomes. It doesn't matter whether Monty picked the goat randomly, by ESP, or by being told which doors hide goats. Slevesque's overlooks that when you initially pick a car, there are two ways for Monty to randomly pick the goat. If you switch your pick in either of those two situations, you lose. Edited by NoNukes, : Address slevesque's post
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Panda Member (Idle past 3740 days) Posts: 2688 From: UK Joined: |
NoNukes writes:
I think I was getting myself mixed up. The question asks what one should do when Monty shows a goat. In other words, Monty's possibility of showing the car should be removed from the considered outcomes, and probabilities would be calculated based on the remaining possible outcomes. If Monty randomly chooses a door (and doesn't open it) then switching is a 50:50 choice.If Monty randomly opens a door with a goat, then you should switch. The chance of you winning the car has dropped (because Monty can now win it).The chance that switching is favourable has stayed the same (if Monty doesn't open the 'car' door). Considering that I am posting this as I run out the door from work again: expect adjustments later!
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ringo Member (Idle past 439 days) Posts: 20940 From: frozen wasteland Joined: |
slevesque writes:
In any case, Open the Goat! would be an excellent name for a TV show. A third of the time, he opens a goat and you chose the car. "I'm Rory Bellows, I tell you! And I got a lot of corroborating evidence... over here... by the throttle!"
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Panda Member (Idle past 3740 days) Posts: 2688 From: UK Joined: |
ringo writes:
I can already hear the crowd clapping and shouting "OPEN THE GOAT! OPEN THE GOAT! OPEN THE GOAT!" and the THWUCK! as Oprah cleaves open the goat with one swing of her mighty machete. In any case, Open the Goat! would be an excellent name for a TV show.The crowd roars!! ...covering up the small, quiet sound of children crying. No doubt the parents will reprimand them later for being 'soft'. But I digress...
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slevesque Member (Idle past 4668 days) Posts: 1456 Joined: |
It doesn't matter whether Monty picked the goat randomly, by ESP, or by being told which doors hide goats. We're in a counter-intuitive thread ? remember The reality is that it does matter, as I have shown twice previously. But let's take a real life example and hopefully it will seem clearer: Everyone knows the game show deal or no deal, right ? 26 cases, each with a money prize in it ranging from a 1 cent to 1M$. You pick a case at the beginning of the show, etc. etc. Now let's suppose you're at the game show, and you go all the way to the grand finale between the last two cases. In one of either your chosen case, or the last case left standing, is 1M$. In the other, 1 penny. Now the host at this point asks you: ''Do you want to change your case ?'' Now, while considering this important question, you remember the Mounty Hall problem. ''My box definitely has a 1/26 chance of having the 1M$, I should change it for the other who has 25/26 chances''. But as you are about to say yes, another thought pops up: ''Yeah but by the same reasoning, my box only has a 1/26 chance of having the penny. This means I should keep it since the other has a 25/26 chance of having that penny''. In the face of these two contradicting conclusions, you face becomes becomes red as the mental battle rages in your head as to why this is the case. And while the host asks for medical attention, since you obviously aren't feeling red, you realize that it's all a matter of what you focus your reasoning on: if you focus on the million, then you should change, but if you focus on the penny, you should keep. Realizing you obviously cannot will the million to be wherever you want it to be, you come to the conclusion that the reasoning is flawed, and that most probably, to change or not to change, that is not the question. You see, this is because the million is not intrinsically any more special then the penny, if the boxes are open at random. You can't focus a ''mounty hall type'' reasoning on the million any more then you can on the penny. But, if the boxes are opened by someone who knows where the million is, and on purpose never opens it, leaving it to be the last case standing, then it becomes special. Then, it already possesses a 25/26 chance of having the million in the eyes of the contestant , even before any cases are yet open, because he knows that whatever cases are open, at the end of the day out of the 25 boxes, one will remain, and the million will not have been opened.
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NoNukes Inactive Member |
slevesque writes: The reality is that it does matter, as I have shown twice previously. But let's take a real life example and hopefully it will seem clearer: I believe you are correct. The difference is explained in excruciating detail on the Wiki page.
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RAZD Member (Idle past 1432 days) Posts: 20714 From: the other end of the sidewalk Joined: |
Hi slevesque,
A third of the time, he opens a goat and you chose the car. (Sit.A) You win if you don't change and lose if you do.
A third of the time, he opens a goat and you chose a goat (Sit.B) You lose if you don't change and win if you do.
A third of the time, he opens a car and you chose a goat (Sit.C) You lose if you don't change and lose if you do (which may seem pointless unless you a filled with boundless optimism ...). ∑ 4 losing situations and 2 winning situations.
Since the presentator opens a goat, you know you are in situation A or B, and you know each situation is just as likely as the other (33% each), then you also know that since you are as likely to be in sit.A then B, then you are just as likely to have the car behind your door then behind the other door. Hence a 50/50 chance But you are forgetting to include the results of sit C in your calcs. Just because you see the car, it doesn't mean you don't lose. Enjoy. by our ability to understand Rebel American Zen Deist ... to learn ... to think ... to live ... to laugh ... to share. Join the effort to solve medical problems, AIDS/HIV, Cancer and more with Team EvC! (click)
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slevesque Member (Idle past 4668 days) Posts: 1456 Joined: |
But you are forgetting to include the results of sit C in your calcs. Just because you see the car, it doesn't mean you don't lose. I'm not forgetting to include sit.C, it's just that nothing worth of interest comes out of doing so. Including sit.C is just like evaluating all the outcomes when no door is yet opened. All it tells us is that when no one chose a door, no one opened a door, you have a 1 in 3 chance of winning the car. Which is self evident. What is interesting is when comparing Sit.A+Sit.B with the Mounty Hall problem, and how it seems counter-intuitive that even though physically the same things happened (you picked, he opened a goat, you change ?) the fact that the presentator opened the goat on purpose or on accident hs a direct impact on how the probabilities are distributed between the two remaining doors.
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Panda Member (Idle past 3740 days) Posts: 2688 From: UK Joined: |
RADZ writes:
The thing to remember is that we are only interested in answering the question: "Should I swap or stay?". But you are forgetting to include the results of sit C in your calcs. Just because you see the car, it doesn't mean you don't lose. If Monty chooses the car, then the question is moot; the answer is neither "swap" nor "stay".
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