|
Register | Sign In |
|
QuickSearch
EvC Forum active members: 65 (9162 total) |
| |
popoi | |
Total: 915,817 Year: 3,074/9,624 Month: 919/1,588 Week: 102/223 Day: 0/13 Hour: 0/0 |
Thread ▼ Details |
|
Thread Info
|
|
|
Author | Topic: Counter-Intuitive Science | ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
RAZD Member (Idle past 1405 days) Posts: 20714 From: the other end of the sidewalk Joined: |
Sorry slevesque, I still do not agree.
I'm not forgetting to include sit.C, it's just that nothing worth of interest comes out of doing so. But it is of interest to me: it means there is a lose-lose case that cannot be omitted from the calculations of all the possibilities. Once the door is opened on the car, there is zero possibility of you winning. This is what changes the original results. Probability calculations are (properly) based on evaluation of the relative success of each and every one of the possibilities (this is why probability calculations cannot be made without knowing all the possibilities), so let's look at the possibilities (your initial pick is door 1, as per the original game analysis, but here the host randomly picks door 2 or door 3):
It appears that you have a 1/3 possibility of winning whether you switch or not (unless you want to win the goat). This is the way the game appears at first. Now it could be argued that in the original game the host eliminates Sit C1 and Sit C2, and this leaves you with a 50:50 chance ... what am I missing? Enjoy. Edited by RAZD, : added last P 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)
|
||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
slevesque Member (Idle past 4641 days) Posts: 1456 Joined: |
It appears that you have a 1/3 possibility of winning whether you switch or not (unless you want to win the goat). This is the way the game appears at first. Yes, at the very beginning, when no door has been chosen nor opened. You have a 1/3 chance of winning the car (if the presentator will open one at random) However, when the presentator opens a goat, you know you are not in sit. C1 and C2, and so you know your chances just went up to 50/50
Now it could be argued that in the original game the host eliminates Sit C1 and Sit C2, and this leaves you with a 50:50 chance ... what am I missing? It also eliminates one of A1 and A2 (in fact, they become one and the same) Without this, there would be A1,A2,B1,B2 which would mean, even with three doors, that you would be randomnly picking the car half of the time. (which we know is not true, since you will be picking it 1/3 of the time)
|
||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
cavediver Member (Idle past 3644 days) Posts: 4129 From: UK Joined: |
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. Even to this I will object The scenarios are not in anyway identical as in the second situation you are throwing away one possibility. And use of the word "chose" has a different meaning in each situation (in the first, he choses by way of knowledge, in the second, there is no choice - he makes a random guess which provides the choice for him) I repeat, I'm not arguing with your result or your methodology. I am trying to ensure that the question is set up unambiguously to begin with, because it is pointless saying that people are confused by the situation when the very question introduces most of that confusion!
When you say, ''I know it sounds anal'' and ask if it translates... Ha - there's me using one colloquialism to ask about another! That will teach me... I was just asking if "I know it sounds anal" makes sense to you? (given that you have English as a second language) - it means "I know it appears that I am being unreasonably stubborn about this point"
|
||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
slevesque Member (Idle past 4641 days) Posts: 1456 Joined: |
Even to this I will object The scenarios are not in anyway identical as in the second situation you are throwing away one possibility. And use of the word "chose" has a different meaning in each situation (in the first, he choses by way of knowledge, in the second, there is no choice - he makes a random guess which provides the choice for him) I repeat, I'm not arguing with your result or your methodology. I am trying to ensure that the question is set up unambiguously to begin with, because it is pointless saying that people are confused by the situation when the very question introduces most of that confusion! Then How would you formulate it then ? I get what you mean, but I thought maybe the easiest way would be to just see it from the referential of the contestant if you will. From the eyes of the contestant, the two situations are identical. You choose a door, he opens one of the two left and reveals a goat, and he asks you if you want to change ? Now it is just to realize that if someone from behind a curtain behind you would say: ''He deliberately opened the goat, he knows where the car is'' or ''he opened a door randomnly'', then you should then know that the probabilities aren't distributed in the same way in both situations. But if there is a better way to present it, I'm open
Ha - there's me using one colloquialism to ask about another! That will teach me... I was just asking if "I know it sounds anal" makes sense to you? (given that you have English as a second language) - it means "I know it appears that I am being unreasonably stubborn about this point" Ha, didn't know that expression. And know, it would prove difficult to appear unreasonably stuborn about a point with me. Remember, I study Math and Physics, so the mathmatician side of me knows the devil is in the details
|
||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
Noetherian Atheist Junior Member (Idle past 4553 days) Posts: 7 From: London Joined: |
Try this:
Instead of thinking through the various outcomes, consider 2 different contestants who adopt opposing strategies: Messers Stick & Change. Under what circumstances will each win the car? If Mr Stick is playing, he would only win if he choose correctly at the first time of asking - on average 1/3 of the time If Mr Change is playing, then he wins the car if he chooses the wrong door first time around because Monty is forced to open the other goat-door leaving only the car for him to change to (as he always does) - on average 2/3 of the time. Clearly better to be Mr Change. For the avoidance of doubt, this is the original version of the game where Monty does know where the car is & deliberately avoids it (and the contestant understands this). Also anal, I know, but you absolutely have to be clear about who knows what and when to understand puzzles like these. I think it's fairly obvious that where Monty chooses blindly & his door isn't opened before you re-choose, then it doesn't matter whether or not you change doors. This is similar to drawing lots: it doesn't matter which order everyone chooses. If Monty chooses blindly and his door is opened before you get to change, then again it doesn't matter. Mr Stick is still going to win only when he chooses the right door to start with - 1/3 of the time. Mr Change will win when he chooses an incorrect door to start with (prob = 2/3) AND Monty chooses the other goat's door (prob = 1/2). Overall probability = 1/3. We can ignore the cases where Monty opens the car's door because either strategy has an equal probability of success (0). If one strategy were better than the other, it would be because it succeeds more often in the cases when Monty does not choose the car.
|
||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
Noetherian Atheist Junior Member (Idle past 4553 days) Posts: 7 From: London Joined: |
I saw a couple of good examples in my undergraduate days:
One involves 3 plain cards with faces coloured red/red, red/green & green/green. Now if one card is chosen at random and placed on a table so that you can see one side (red say), what it the probability that the other side is also red? Secondly, perhaps more of a paradox, but anyway... draw a circle (doesn't matter how big) and then inscribe an equilateral triangle (inside the circle with its corners touch the circle). What is the probability that, for a straight line drawn through the circle, the part of the line which is inside the circle is longer than the length of the equilateral triangle's sides? This one is not obvious, but the point is that there are perfectly reasonable arguements which appear to demonstrate that the probability is both 1/2 & 1/3. Finally, and it's quite complicated - also >20 years ago so I can't fully remember, but an undergraduate problem I was set was to come up with two (mathematically) continuous lines which would connect diagonally opposite corners of a square without leaving the square and without crossing over one another! To be specific, there is no point which is common to both lines. It involved a function like sinx/x which has a thing called a removable singularity at x=0 - a singularity which you can "remove" by simply redefining it at the singularity. The point being that the function can take a range of values at the singularity and remain continuous. Such redefining allows the 2 lines to cross over at x=0 without touching.
|
||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
slevesque Member (Idle past 4641 days) Posts: 1456 Joined: |
One involves 3 plain cards with faces coloured red/red, red/green & green/green. Now if one card is chosen at random and placed on a table so that you can see one side (red say), what it the probability that the other side is also red? 2/3 ?
Secondly, perhaps more of a paradox, but anyway... draw a circle (doesn't matter how big) and then inscribe an equilateral triangle (inside the circle with its corners touch the circle). What is the probability that, for a straight line drawn through the circle, the part of the line which is inside the circle is longer than the length of the equilateral triangle's sides? This one is not obvious, but the point is that there are perfectly reasonable arguements which appear to demonstrate that the probability is both 1/2 & 1/3. The line can be drawn anywhere in the circle ? I mean, it's length can range from 0 up to D (diameter) ? AbE I get 25% (1/4) Edited by slevesque, : No reason given.
|
||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
slevesque Member (Idle past 4641 days) Posts: 1456 Joined: |
Anyone else can solve these problems, see if I'm at least half-right ?
I have a feeling Noetherian Atheist won't be coming back to give me the answers
|
||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
Panda Member (Idle past 3713 days) Posts: 2688 From: UK Joined: |
slevesque writes:
I'll have a go at writing some code to do 1,000,000 iterations and I'll post the results. Anyone else can solve these problems, see if I'm at least half-right ? I should get the time and inclination to do it tomorrow.
|
||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
cavediver Member (Idle past 3644 days) Posts: 4129 From: UK Joined: |
Anyone else can solve these problems, see if I'm at least half-right ? Well, the 2/3 is obvious What is counterintuuitive is that it could be anything else... and that can be a stumbling block, as the obvious answer is 2/3, so that must be the trap, so what must the real answer be? I'm not sure where you get 0.25 from. I can get 1/3 by considering chords from a fixed point (one of the triangle vertices), and sweeping the chords from the tangent around the circle and back to the tangent. So through the first 60 degrees, the chords grow to the length of the triangle; the second 60 degrees they are longer, and the final 60 degrees they shrink again back to the tangent line (with length 0) For 1/2, same thing but consider the chords sweeping across the circle, staying parallel to the tangent at one of the vertices. The base of the triangle is 3/4 of the way across the circle, so the chords between 1/4 of the way and 3/4 of the way across must be longer than the base. So 1/2 are longer. The thing is, you can't use lines to space fill ! Lines are not infinitessimally thin, they have zero width. Hence the "paradox". The whole question is ill-defined anyway. You cannot just "randomly" draw lines across circles without giving some concept of measure, some distribution function. There is no obvious "uniform distribution" that one could just assume is meant by the questioner.
|
||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
RAZD Member (Idle past 1405 days) Posts: 20714 From: the other end of the sidewalk Joined: |
Hi slevesque,
One involves 3 plain cards with faces coloured red/red, red/green & green/green. Now if one card is chosen at random and placed on a table so that you can see one side (red say), what it the probability that the other side is also red?
2/3 ? Isn't this similar to your last version of the Monty game? Once you have a card on the table (with any one card being chosen 1/3rd of the time), there are only two cards that can have the face that is up (ie it cannot be the green/green card), and one of them the same on the bottom and the other is not: 50:50? 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)
|
||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
RAZD Member (Idle past 1405 days) Posts: 20714 From: the other end of the sidewalk Joined: |
Hi Panda
The thing to remember is that we are only interested in answering the question: "Should I swap or stay?" If Monty chooses the car, then the question is moot; the answer is neither "swap" nor "stay". Curiously, you do not walk away with the car. You still have the option to swap or stay, whether you have a possibility of winning or not. 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)
|
||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
arachnophilia Member (Idle past 1344 days) Posts: 9069 From: god's waiting room Joined: |
RAZD writes: Isn't this similar to your last version of the Monty game? Once you have a card on the table (with any one card being chosen 1/3rd of the time), there are only two cards that can have the face that is up (ie it cannot be the green/green card), and one of them the same on the bottom and the other is not: 50:50? that's what i thought, but i also thought that was too obvious, so i was gonna wait and see what people said. how are people getting 2/3rds? edit: oh, wait. got it. yes, 2/3rds, same as monty hall. Edited by arachnophilia, : No reason given.
|
||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
slevesque Member (Idle past 4641 days) Posts: 1456 Joined: |
Well, the 2/3 is obvious What is counterintuuitive is that it could be anything else... and that can be a stumbling block, as the obvious answer is 2/3, so that must be the trap, so what must the real answer be? It seems so obvious I tell myself exactly the same thing
I'm not sure where you get 0.25 from. I can get 1/3 by considering chords from a fixed point (one of the triangle vertices), and sweeping the chords from the tangent around the circle and back to the tangent. So through the first 60 degrees, the chords grow to the length of the triangle; the second 60 degrees they are longer, and the final 60 degrees they shrink again back to the tangent line (with length 0) For 1/2, same thing but consider the chords sweeping across the circle, staying parallel to the tangent at one of the vertices. The base of the triangle is 3/4 of the way across the circle, so the chords between 1/4 of the way and 3/4 of the way across must be longer than the base. So 1/2 are longer. I had thought about the reasoning for 1/2. But here's how I viewed to to get the 1/4. Each line drawn inside the circle will have a middle point. For example, if the middle point is at the center, then the line is actually the diameter. So I calculated the area inside the circle where the middle point can be which would involve a line longer then the side of the triangle. This area is a smaller circle with a diameter of D/2 (D being the diameter of the bigger circle) at the center of the bigger circle. Which has a fourth of the total area of the circle. AbE I don't feel this is clear. It's a circle in a circle, and whenever the middle point of the line is in the inner circle, the line is longer. Everywhere else in the outer circle it is smaller. Since the area of the inner circle takes 25% of the area of the outer circle, so will the line be longer then the sides of the triangle 25% of the time. Edited by slevesque, : No reason given.
|
||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
slevesque Member (Idle past 4641 days) Posts: 1456 Joined: |
The question that is asked is essentially: Out of the three, what is the chance I get a card that has the same color on both sides ?
Seen this way, it is obvious that the answer is 2/3. Seen another way, there are 6 faces in total (2 per card). half are green, and half are red. Whichever color you choose, you have 2/3 chances to have picked the card single-colored card, since it has 2 of the total three faces of that color. In other words. Suppose you get red. You either picked the red/red card on it's face no1, or on it's face no2 or the red/green card on it's face no1. Idem for green.
|
|
|
Do Nothing Button
Copyright 2001-2023 by EvC Forum, All Rights Reserved
Version 4.2
Innovative software from Qwixotic © 2024