Can You Solve This Hilarious Limit Problem Involving Sine and Infinity?

[benorin]

Kronecker Delta Function Song

 

[rachmaninoff]

Just to clarify, a 2-teapot is homeomorphic to the surface of a torus. If you remove the lid, it's the surface of a double torus*.

*which has very interesting homotopy groups, I'm trying to calculate them right now...

 

[Your.Master]

Actually, you can argue with your experiment, because you did it wrong.

To explain your problem: IF the main man is guilty (1/3 of the time), then only one of the other guys is guilty, and the guard HAS to point to him. He will point at him 100% of the time. So 1/3 of the time this situation occurs, and the not-pointed-to guy gets off free. Your experiment, however, does not account for the fact that the guilty man must always be chosen: your experiment has the guard choosing among the other two ~50/50 ALL the time.

Let us analyse that. This case, which obviously occurs 1/3 of all times (and as is borne out by your experiment with no removals), leads to the not-pointed-to, not-main guy being innocent. The main guy is, by assumption, guilty.

IF the main man is not guilty (2/3 of the time), then the other guys are BOTH guilty. The guard can point at either of the other men. Now the odds are 50/50 which of these two will be selected.

Let us analyse this. This case, which occurs the other 2/3 of all times, leads to the not-pointed-to, not-main-guy as being guilty. The main guy is, by assumption, not guilty.

THEREFORE, the not-pointed-to guy is guilty 1/3 of the time, the pointed-to guy is guilty ALL the time, and the main man is guilty 2/3 of the time.

Or think of it like this, in a much less-rigorous way: all three have a 2/3 chance of being guilty to start with. When the guy is pointed-to, he gets an additional 1/3 chance of being guilty. Somebody has to donate that 1/3 chance, and that has the other guy, since the guard was NEVER going to point at you so you weren't involved in that selection.

Or simply: If main guy is guilty, then the guy not pointed to is always innocent. Therefore, if you are NOT the main guy AND the guard doesn't point at you, you've increased the chances that YOU are innocent -- if you were guilty, he'd point at you half the time, if you were not guilty, he'd NEVER point at you, and combine half with never and you get something LESS than half.

A more analogous experiment that you could try is to pick the innocent guy out of the hat FIRST, then remove one of the guilty, non-main guys, then see how often the main guy was innocent and how often the not-pointed-to not-main guy was innocent. After all, in the example, the guard knew which one was innocent even before the main guy asked.

Even better, if you know anything about computer programming, you can make a simple program to convince you. The program randomly determines who is guilty, and you play the role of the main guy. You hit enter, the computer tells you (choosing at random if necessary) one of the other guys who is guilty, then you hit enter again and see who is actually guilty. There's one already made here:

http://chrisc.freeshell.org/random/pages/montyhall.html

ALTHOUGH given your bias against random webpages it could just be a conspiracy to give you the wrong answer and not ACTUALLY an instantiation of the code shown on that page.

-----

Before ignoring a website as "random" you might want to read what it says -- it does not matter how credible that site is, if the reasoning is sound.

But maybe you trust wikipedia more? http://en.wikipedia.org/wiki/Monty_Hall_problem

Or a math site?
http://mathforum.org/dr.math/faq/faq.monty.hall.html
http://mathworld.wolfram.com/MontyHallProblem.html

Or the site whose very URL refers to the problem in question?
http://www.montyhallproblem.com/

Honestly, I'm not sure what kind of website you were expecting. The CIA?

In general, don't believe everything you see on the Internet, but also don't disbelieve everything just because it's on the Internet, at least not without a second look when a majority of people are telling you you are wrong.

 

[BobG]

bah not convinced, I trust my argument and the judgment of Alkatran over yours, Joffe's and that random webpage. I sahll repeat my argument for the readers benefit.

But, surely you have to trust Marilyn Mach vos Savant! She holds the record for the highest IQ score, ever.

See the discussion under http://www.wiskit.com/marilyn.html or check the link titled "Behind Monty Hall's Doors" towards the bottom of the page.

Edit: Oops, my bad. I inadvertantly directed you to the "Marilyn is Wrong!" website with link above. I meant to direct you to the http://www25.brinkster.com/ranmath/marlright/ website - check out "The Monty Hall Problem".

 

[alfredblase]

Ok I read a few of those sites, I can see the argument but I disagree with it. And you can't argue that someone wrote a program "proving" anything. The program is based on your understanding of the problem. If you have a preconcieved way of setting up the problem in programing language, then of course the program will give you the answer you imagined it would. I think one of the links from the wikipedia article you gave me, although arguing against my opinion, illustrates how a reasonable argument could be set up arguing for my point of view.

http://www.angelfire.com/trek/nfold/monty.html

This is my version of his over easy proof. The player chooses A at first.

.....A...B...C
case 1...0...(0...1)
case 2...0...(1...0)
case 3...1...(0...0)
case 4...1...(0...0)

case 1: Monty opens B, switch succeeds
case 2: Monty opens C, switch succeeds
case 3: Monty opens B, switch fails
case 4: Monty opens C, switch fails

In two out of the four cases the switch succeeds, and in 2/4 cases the switch fails. So there.

Anyway, the ONLY way to really know for sure is to carry out an EXPERIMENT like physicists do. Not using a computer program but setting up the scenario physically say 1000 times and seeing the results. I still can't see how my experiment wasn't ok; I think you misunderstood what I wrote or something... it seems perfectly simple to me.

I mean FGS! if someone offers you a choice between two doors (that is what is done when you are offered the chance to switch), one of which hides a goat and the other a car what are the chances you will make the right decision??! COME ON people!

 

[Galileo]

You shouldn't take into account the action of the gameshow host. Which door he opens doesn't matter, but which door YOU choose does, since those 3 choices have equal probabilities.

This has been debated to death, the chances of winning are 2/3 when you switch and the argument is very very simple:
Suppose you decide to switch.
Picking the right door means you will lose, since you then switch to a wrong door.
Picking a wrong door means you win, since you will switch to the correct door.
Because you have 2/3 chance your initial guess is wrong, you have a 2/3 chance of winning if you switch.

 

[LeonhardEuler]

Ok I read a few of those sites, I can see the argument but I disagree with it. And you can't argue that someone wrote a program "proving" anything. The program is based on your understanding of the problem. If you have a preconcieved way of setting up the problem in programing language, then of course the program will give you the answer you imagined it would. I think one of the links from the wikipedia article you gave me, although arguing against my opinion, illustrates how a reasonable argument could be set up arguing for my point of view.
http://www.angelfire.com/trek/nfold/monty.html
This is my version of his over easy proof. The player chooses A at first.
.....A...B...C
case 1...0...(0...1)
case 2...0...(1...0)
case 3...1...(0...0)
case 4...1...(0...0)
case 1: Monty opens B, switch succeeds
case 2: Monty opens C, switch succeeds
case 3: Monty opens B, switch fails
case 4: Monty opens C, switch fails
In two out of the four cases the switch succeeds, and in 2/4 cases the switch fails. So there.

I don't really understand what those 1's and 0's are supposed to represent and I don't know what exactly you mean by 'succeed' and 'fail', but in any event I believe you are mistaken. Monty will never open a door that contains the prize, and he can't change where it is. Surely you agree that the odds of the player chosing the prize on the first try are 1/3. Clearly whatever anybody does after that can not affect the fact that there was a 1/3 chance that the prize was behind that door.

Now what happens when the host opens a door? Suppose he chooses one of the other doors at random. Then if there is no prize behind the door he opens, you have gained information about the probability of there being a prize behind your door: If the prize wasn't behing your door it would be less likely for the host to choose a door without the prize behind it than if it was. This changes the probability that a prize is behind your door.

But suppose we know beforehand that the host will pick a door that does not have the prize behind it. Obviously the chances of picking the door with the prize on the first try are 1/3. Obviusly this does not change when another door is revealed not to have the prize, since we knew that was going to happen in the first place. Now the chances that your door has the prize are still 1/3 while the chances that the prize is in the open door have dropped to zero. Since there are no other possibilities, the probability that the other door has the prize must be 2/3.

Anyway, the ONLY way to really know for sure is to carry out an EXPERIMENT like physicists do. Not using a computer program but setting up the scenario physically say 1000 times and seeing the results. I still can't see how my experiment wasn't ok; I think you misunderstood what I wrote or something... it seems perfectly simple to me.

Actually, that is not true. The only way to know for sure is to come up with a mathematical proof. There is always a finite probability that that statistical data are misleading you when you carry out a process a finite number of times. The probability may get very small that you are being mislead, but you will never know for certain.

 

[Your.Master]

Ok I read a few of those sites, I can see the argument but I disagree with it. And you can't argue that someone wrote a program "proving" anything. The program is based on your understanding of the problem. If you have a preconcieved way of setting up the problem in programing language, then of course the program will give you the answer you imagined it would.

...

Anyway, the ONLY way to really know for sure is to carry out an EXPERIMENT like physicists do. Not using a computer program but setting up the scenario physically say 1000 times and seeing the results. I still can't see how my experiment wasn't ok; I think you misunderstood what I wrote or something... it seems perfectly simple to me.

This is frustrating -- these are world-class mathematicians and professionals who have laid this out point by point.

First of all, you can't disagree with mathematical proof, and if you do, you aren't reading it right. And yes, you can argue that somebody wrote a program proving stuff. It's identical to carrying out a physical experiment. What's the difference? Anyway, the experiment should carry no preconceptions if done properly and exactly as outlined in the problem. But this is solvable by pure math, anyway.

Your experiment wasn't okay because you didn't account for the fact that the guard will NEVER choose the guy who is innocent, even if it isn't the main guy. You just chose at random.

Your wikipedia example fails; here's why:

.....A...B...C
case 1...0...(0...1)
case 2...0...(1...0)
case 3...1...(0...0)
case 4...1...(0...0)
case 1: Monty opens B, switch succeeds
case 2: Monty opens C, switch succeeds
case 3: Monty opens B, switch fails
case 4: Monty opens C, switch fails
In two out of the four cases the switch succeeds, and in 2/4 cases the switch fails. So there.

Hello? Look at cases 3 and 4...they are the *same*. it should read:

case 1...0...(0...1)
case 2...0...(1...0)
case 3...1...(0...0)
case 1: Monty opens B, switch succeeds
case 2: Monty opens C, switch succeeds
case 3: Monty opens either B or C (at random), switch fails

What you seem to be missing is that Monty chooses which door to open AFTER the door that the car is behind is determined -- AFTER the cases. The fact that he has a choice after the chooser chose the car door does not magically make it more likely that the chooser chose the car door in the first place.

I mean FGS! if someone offers you a choice between two doors (that is what is done when you are offered the chance to switch), one of which hides a goat and the other a car what are the chances you will make the right decision??! COME ON people!

If somebody offered me a choice of TWO doors, then it's 50/50. I was not offered a choice of two doors. I was offered a choice of THREE doors.

I already knew that at least one of the other doors had a goat behind it. The guy revealed it, but that did not tell me anything new -- I already knew that one of those doors had a goat behind it, and those doors are basically indistinguishable. Essentially, when I'm switching at the end, I'm saying "I think there's a car behind one of the two doors I *didn't* pick in the first place".

Imagine this. Say there are a billion doors. You choose one, Monty Hall opens nine hundred ninety nine million, nine hundred ninety nine thousand, nine hundred ninety eight doors with goats behind it. You now have your door, and another door. Do you *really* believe that this proves that your door is right 50% of the time? Because what that means is that ANY guess you make at the billion doors, any at all, is right 50% of the time (because no matter what you guess, the requisite number of other doors can ALWAYS be opened...it does not even matter if they ARE opened or not). Does that make sense? Do you win the lottery 50% of the time?

Or put it this way. A billion doors. You choose one. Monty Hall ALWAYS says, you are either absolutely right and you chose the correct door, OR, it is this other door that he points out. You can then choose the other door that he points out.

Is it really 50/50? Remember, 100% of the time, when you make the initial choice, Monty Hall reduces it to two. Do you really guess the right door the first time, out of a billion, half the time?

If you do, can I have your lottery numbers?

 

[alfredblase]

ok guys and gals, I am going to bow down. I like Galileo's explanation and it sounds very reasonable and simple to me. Since his reasoning, as far as I can percieve, is at least as valid as mine, and since I have trustworthy accounts from trustworthy sources that there is overwhelming experimental evidence in the form of computer program code and readouts, I will accept the very high probabilty that my experiment was faulty, and that the orthodox arguments are correct. This also means that I have realized that an unbiased program can reproduce the real thing (in terms of statistics).

I apologise if I have caused any of you tear your hair out in frustration :shy:

I appreciate your patience and repeated, good natured efforts to make me see the light. hehe :tongue2:

Here's a crummy maths joke to compensate

I will prove that girls are evil:

Girls mean money*time
time equals money
money is the root of all evil
therefore girls are evil

 

[Your.Master]

No harm done. I was also sort of impatient with you...it's exam time, you see.

In the spirit of your joke:

Stupid people will always make more money.

Proof:

By definition,

Power = Work / Time

Rearranging:

Time = Work / Power

Time IS Money, and Knowledge IS Power, so substituting:

Money = Work / Knowledge

Therefore, as knowledge goes to 0, money goes to infinity, regardless of the work being done (if Work > 0 and it is impossible to have negative knowledge).

 

[Cybersteve]

Monty Hall problem. Non-mathematical(ish) explanation.

The key to the answer to this problem is that Monty knows the answer.

When you make your choice of 3 doors you are in effect dividing the doors into two groups - yours and Monty's.

He has more twice as many doors, so it is twice as likely that he has the correct door.

He will always show the goat, so by switching you get to choose the best of his group - which is twice as likely to have contained the goat as your original selection.

 

[Number Lover]

I remember hearing a story on another forum at one point where a high school math club of some sort made t-shirts which imitated the anti-drug commercials. Here's the design: http://img236.imageshack.us/img236/6195/integrals9tr6uo.jpg

I found it extremely funny the first time I saw it. :)

 

[alfredblase]

Your experiment wasn't okay because you didn't account for the fact that the guard will NEVER choose the guy who is innocent, even if it isn't the main guy. You just chose at random.

Your.Master

Although I still admit the orthodox explanation is the correct one due to overwhelming experimental evidence, the statement concerning where you think my experiment went wrong is incorrect. As the guard I did not just choose one of the papers at random, (after removal of the randomly chosen main man paper). As the guard I looked at both of the other prisoner's papers and removed one that was guilty. So we have still to ascertain where my experiment went wrong.

 

[Martin Rattigan]

How many people got executed while you were collecting your overwhelming experimental evidence?

 

[Martin Rattigan]

Actually the easiest way to convince people the argument is wrong is to increase the number of people. Would so many people take it seriously if it read it read:

"A thousand prisoners, strangers to each other, were suspects of a murder case. One day they came to hear that a sentence has been drawn. Nine hundred and ninety nine of them have been found guilty and will be executed, but they don't know which nine hundred and ninety nine. One guy, a statistician, figures his chances for survival are 1/1000, so he goes to the bars of his cell and hails the guard: "Hey psst, do you know which of us has been sentenced?".
"Eh, yes.", says the guard, "But I'm not allowed to tell you.".
"Tell you what", says the guy, "I already know that 999 of us will executed, that means at least 998 of the other guys will be. I don't know them or anything, surely you can point to 998 which are guilty?". The guard sees no harm in that and points 998 of the prisoners, "They are guilty".
"Thanks!", proclaims the statistician, "my chances have just increased to 1/2".

 

[JasonRox]

I remember hearing a story on another forum at one point where a high school math club of some sort made t-shirts which imitated the anti-drug commercials. Here's the design: http://img236.imageshack.us/img236/6195/integrals9tr6uo.jpg

I found it extremely funny the first time I saw it. :)

That's a nice one.

For everyone else, keep the jokes coming.

 

[alfredblase]

The experimental evidence I keep banging on about is in the form of computer programs apparently written by many in this forum. One of the major advances that Newton made in science, in my opinion, is that he was a pioneer of the system of looking at experimental results, and THEN making arguments, laws, equations, to fit the theory, not the other way round. You guys seem to think your argument would hold true if experiments told you otherwise. However it seems that experiment bears out your argument, and so it is the correct argument. That is the only reason I accept it over my old argument, which to me seems as logical and reasonable as the argument I now believe in.

 

[Joffe]

http://img207.imageshack.us/img207/4050/unfindxlol12fb.png

 

[Reshma]

Angles and Angels

Not exactly a joke...but happened often when I was in grade 3 when we were introduced to a bit "advanced" geometry like points, segments, angles etc.

A common typo made by students is instead of writing "Angle ABC" they write "Angel ABC".

 

[benorin]

Another limerick:

Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed?

 

[Rach3]

That's not a limerick.

 

[BSMSMSTMSPHD]

Descartes walks into a bar.

The waitress says, "can I get you something?"

Descartes says, "I think not," and promptly disappears.

 

[BSMSMSTMSPHD]

Here's one that we actually saw in a pre-calculus class.

Upon learning that $$\lim_{x\rightarrow 8} \frac{1}{(x-8)^2} = \infty$$

the class was asked to consider $$\lim_{x\rightarrow 3} \frac{1}{(x-3)^2}$$.

One student raised his hand and said, "m".

Of course, I guess the answer could also be "$$\omega$$"

 

[rhj23]

Here's one that we actually saw in a pre-calculus class.

Upon learning that $$\lim_{x\rightarrow 8} \frac{1}{(x-8)^2} = \infty$$

the class was asked to consider $$\lim_{x\rightarrow 3} \frac{1}{(x-3)^2}$$.

One student raised his hand and said, "m".

Of course, I guess the answer could also be "$$\omega$$"

if it was intended to be an omega that's pretty clever..

 

[BSMSMSTMSPHD]

Originally Posted by rhj23
if it was intended to be an omega that's pretty clever..

Yeah - I never thought of that. The actual joke requires a sideways 5, but I didn't have enough imagination to create that symbol.

 

[benorin]

http://www.maths.uwa.edu.au/~berwin/humour/invalid.proofs.html#TopPageAnchor

 

[benorin]

Q: How do you make one burn?
A: Differentiate a log fire!

 

[Nimz]

These integrals are sweet!
$$\int dy$$
$$\int_{-C}^y dt$$

Q: What is a proof that is based on two prior results called?
A: A dilemma.


"Once I thought I was capable of making mistakes, but then I found out I was wrong."

 

[Edgardo]

Why are mathematicians convergent?
Because they are monotonic and bounded!

 

[BSMSMSTMSPHD]

Funny, I've always seen myself as just being Cauchy...

 

[ajsingh]

I really like the following one which I got from SimonSingh.net. Not sure if its a "bad math joke" :)

Two mathematicians are in a bar. The first one
says to the second that the average person
knows very little about basic mathematics.
The second one disagrees, and claims that most
people can cope with a reasonable amount
of mathematics.

The first mathematician goes off to the washroom,
and in his absence the second calls over the waitress.
He tells her that in a few minutes, after his friend has
returned, he will call her over and ask her a question.
All she has to do is answer "one third x cubed."

She repeats "one thir -- dex cue"?

He repeats, "one third x cubed".

She asks, "one thir dex cubed?"

"Yes, that's right," he says.

So she agrees, and goes off mumbling to herself,
"one thir dex cubed...".

The first guy returns and the second proposes a bet
to prove his point, that most people do know something
about basic mathematics. He says he will ask the blonde waitress
an integral, and the first laughingly agrees. The second man
calls over the waitress and asks "what is the integral of x squared?".

The waitress says "one third x cubed" and whilst walking away,
turns back and says over her shoulder "plus a constant!"

 

[tony873004]

What's the square root of 69?

8-something.

 

[Curious3141]

What's the square root of 69?

8-something.

That's so wrong! :rofl:

 

[Nimz]

That's terrible.

Conversation Starter #176
One theme that comes up in many of my lectures is "The Fourth Dimension: It's not just 'time' anymore." To illustrate, I tell the story of the herb distributor who wanted to use a vector space to keep track of orders. There would be one coordinate for parsley, one for sage, one for rosemary, and one for oregano--because the fourth dimenstion isn't just thyme anymore. The reaction of the class? Well, let's just say their groans give my story yet another dimension.
-Tom Banchoff, Brown University
(From 777 Mathematical Conversation Starters)

Keep 'em coming

 

[eljose]

-What,s purple and Commutative?----> An Abelian Grape

-What,s the contour integral over all Western Europe?---> 0 because all the "Poles" live in Eastern Europe.