Register | Sign In


Understanding through Discussion


EvC Forum active members: 65 (9162 total)
4 online now:
Newest Member: popoi
Post Volume: Total: 915,815 Year: 3,072/9,624 Month: 917/1,588 Week: 100/223 Day: 11/17 Hour: 0/0


Thread  Details

Email This Thread
Newer Topic | Older Topic
  
Author Topic:   Counter-Intuitive Science
RAZD
Member (Idle past 1405 days)
Posts: 20714
From: the other end of the sidewalk
Joined: 03-14-2004


Message 151 of 182 (601159)
01-18-2011 11:35 PM
Reply to: Message 149 by slevesque
01-17-2011 6:48 PM


Re: Counter-Intuitive Math
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):
Casedoor 1door 2door 3hoststayswitch
Sit A1cargoatgoatdoor 2winlose
Sit A2cargoatgoatdoor 3winlose
Sit C1goatcargoatdoor 2loselose
Sit B1goatcargoatdoor 3losewin
Sit B2goatgoatcardoor 2losewin
Sit C2goatgoatcardoor 3loselose
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

we are limited in our ability to understand
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)

This message is a reply to:
 Message 149 by slevesque, posted 01-17-2011 6:48 PM slevesque has replied

Replies to this message:
 Message 152 by slevesque, posted 01-19-2011 2:02 AM RAZD has seen this message but not replied

  
slevesque
Member (Idle past 4640 days)
Posts: 1456
Joined: 05-14-2009


Message 152 of 182 (601181)
01-19-2011 2:02 AM
Reply to: Message 151 by RAZD
01-18-2011 11:35 PM


Re: Counter-Intuitive Math
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)

This message is a reply to:
 Message 151 by RAZD, posted 01-18-2011 11:35 PM RAZD has seen this message but not replied

  
cavediver
Member (Idle past 3643 days)
Posts: 4129
From: UK
Joined: 06-16-2005


Message 153 of 182 (601191)
01-19-2011 5:01 AM
Reply to: Message 138 by slevesque
01-16-2011 4:54 PM


Re: Counter-Intuitive Math
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"

This message is a reply to:
 Message 138 by slevesque, posted 01-16-2011 4:54 PM slevesque has replied

Replies to this message:
 Message 154 by slevesque, posted 01-19-2011 3:08 PM cavediver has not replied

  
slevesque
Member (Idle past 4640 days)
Posts: 1456
Joined: 05-14-2009


Message 154 of 182 (601266)
01-19-2011 3:08 PM
Reply to: Message 153 by cavediver
01-19-2011 5:01 AM


Re: Counter-Intuitive Math
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

This message is a reply to:
 Message 153 by cavediver, posted 01-19-2011 5:01 AM cavediver has not replied

Replies to this message:
 Message 155 by Noetherian Atheist, posted 01-19-2011 8:27 PM slevesque has not replied

  
Noetherian Atheist
Junior Member (Idle past 4552 days)
Posts: 7
From: London
Joined: 08-19-2010


Message 155 of 182 (601320)
01-19-2011 8:27 PM
Reply to: Message 154 by slevesque
01-19-2011 3:08 PM


Re: Counter-Intuitive Math
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.

This message is a reply to:
 Message 154 by slevesque, posted 01-19-2011 3:08 PM slevesque has not replied

  
Noetherian Atheist
Junior Member (Idle past 4552 days)
Posts: 7
From: London
Joined: 08-19-2010


Message 156 of 182 (601325)
01-19-2011 8:51 PM
Reply to: Message 1 by Dr Adequate
01-08-2011 11:29 AM


More counterintuitive Maths
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.

This message is a reply to:
 Message 1 by Dr Adequate, posted 01-08-2011 11:29 AM Dr Adequate has not replied

Replies to this message:
 Message 157 by slevesque, posted 01-19-2011 10:24 PM Noetherian Atheist has not replied
 Message 158 by slevesque, posted 01-20-2011 5:27 PM Noetherian Atheist has not replied
 Message 176 by Kaichos Man, posted 03-05-2011 5:45 AM Noetherian Atheist has not replied

  
slevesque
Member (Idle past 4640 days)
Posts: 1456
Joined: 05-14-2009


Message 157 of 182 (601341)
01-19-2011 10:24 PM
Reply to: Message 156 by Noetherian Atheist
01-19-2011 8:51 PM


Re: More counterintuitive Maths
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.

This message is a reply to:
 Message 156 by Noetherian Atheist, posted 01-19-2011 8:51 PM Noetherian Atheist has not replied

Replies to this message:
 Message 161 by RAZD, posted 01-20-2011 7:48 PM slevesque has replied

  
slevesque
Member (Idle past 4640 days)
Posts: 1456
Joined: 05-14-2009


Message 158 of 182 (601461)
01-20-2011 5:27 PM
Reply to: Message 156 by Noetherian Atheist
01-19-2011 8:51 PM


Re: More counterintuitive Maths
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

This message is a reply to:
 Message 156 by Noetherian Atheist, posted 01-19-2011 8:51 PM Noetherian Atheist has not replied

Replies to this message:
 Message 159 by Panda, posted 01-20-2011 6:27 PM slevesque has not replied
 Message 160 by cavediver, posted 01-20-2011 6:57 PM slevesque has replied

  
Panda
Member (Idle past 3712 days)
Posts: 2688
From: UK
Joined: 10-04-2010


Message 159 of 182 (601468)
01-20-2011 6:27 PM
Reply to: Message 158 by slevesque
01-20-2011 5:27 PM


Re: More counterintuitive Maths
slevesque writes:
Anyone else can solve these problems, see if I'm at least half-right ?
I'll have a go at writing some code to do 1,000,000 iterations and I'll post the results.
I should get the time and inclination to do it tomorrow.

This message is a reply to:
 Message 158 by slevesque, posted 01-20-2011 5:27 PM slevesque has not replied

  
cavediver
Member (Idle past 3643 days)
Posts: 4129
From: UK
Joined: 06-16-2005


Message 160 of 182 (601473)
01-20-2011 6:57 PM
Reply to: Message 158 by slevesque
01-20-2011 5:27 PM


Re: More counterintuitive Maths
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.

This message is a reply to:
 Message 158 by slevesque, posted 01-20-2011 5:27 PM slevesque has replied

Replies to this message:
 Message 164 by slevesque, posted 01-20-2011 9:19 PM cavediver has not replied
 Message 170 by Noetherian Atheist, posted 01-21-2011 7:40 PM cavediver has not replied

  
RAZD
Member (Idle past 1405 days)
Posts: 20714
From: the other end of the sidewalk
Joined: 03-14-2004


Message 161 of 182 (601478)
01-20-2011 7:48 PM
Reply to: Message 157 by slevesque
01-19-2011 10:24 PM


Re: More counterintuitive Maths
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.

we are limited in our ability to understand
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)

This message is a reply to:
 Message 157 by slevesque, posted 01-19-2011 10:24 PM slevesque has replied

Replies to this message:
 Message 163 by arachnophilia, posted 01-20-2011 8:34 PM RAZD has seen this message but not replied
 Message 165 by slevesque, posted 01-20-2011 9:31 PM RAZD has seen this message but not replied

  
RAZD
Member (Idle past 1405 days)
Posts: 20714
From: the other end of the sidewalk
Joined: 03-14-2004


Message 162 of 182 (601480)
01-20-2011 7:55 PM
Reply to: Message 150 by Panda
01-17-2011 6:59 PM


Re: Counter-Intuitive Math
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.

we are limited in our ability to understand
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)

This message is a reply to:
 Message 150 by Panda, posted 01-17-2011 6:59 PM Panda has replied

Replies to this message:
 Message 167 by Panda, posted 01-21-2011 6:59 AM RAZD has seen this message but not replied

  
arachnophilia
Member (Idle past 1343 days)
Posts: 9069
From: god's waiting room
Joined: 05-21-2004


Message 163 of 182 (601481)
01-20-2011 8:34 PM
Reply to: Message 161 by RAZD
01-20-2011 7:48 PM


Re: More counterintuitive Maths
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.

אָרַח

This message is a reply to:
 Message 161 by RAZD, posted 01-20-2011 7:48 PM RAZD has seen this message but not replied

  
slevesque
Member (Idle past 4640 days)
Posts: 1456
Joined: 05-14-2009


Message 164 of 182 (601492)
01-20-2011 9:19 PM
Reply to: Message 160 by cavediver
01-20-2011 6:57 PM


Re: More counterintuitive Maths
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.

This message is a reply to:
 Message 160 by cavediver, posted 01-20-2011 6:57 PM cavediver has not replied

  
slevesque
Member (Idle past 4640 days)
Posts: 1456
Joined: 05-14-2009


Message 165 of 182 (601495)
01-20-2011 9:31 PM
Reply to: Message 161 by RAZD
01-20-2011 7:48 PM


Re: More counterintuitive Maths
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.

This message is a reply to:
 Message 161 by RAZD, posted 01-20-2011 7:48 PM RAZD has seen this message but not replied

  
Newer Topic | Older Topic
Jump to:


Copyright 2001-2023 by EvC Forum, All Rights Reserved

™ Version 4.2
Innovative software from Qwixotic © 2024