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

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The forum discussion revolves around a humorous limit problem involving the expression \(\lim_{n\rightarrow\infty}\frac{\sin x}{n}=6\), which is presented as a joke rather than a serious mathematical inquiry. Participants share various math-related jokes and puns, including playful interpretations of integrals and humorous anecdotes from calculus classes. Notable jokes include the "log cabin" pun and a limerick involving integrals, showcasing the blend of humor and mathematics that resonates with both students and enthusiasts.

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  • #91
My first post here :)

Q : Why didn't didn't Cauchy like taking his dog for a walk?

A : It kept leaving residues at all the poles.

http://movies.collegehumor.com/items/2005/05/collegehumor.149448.wmv song cracks me up. I expected it to be cheesy as hell, but the amount of puns they fit in is very impressive.
 
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  • #92
What's purple and commutes?

An abelian grape.

What's purple, and worshipped by only a limited number of people?

A finitely venerated abelian grape.
 
  • #93
For this statistical problem, why are we going all the way into Bayes theorem? We can just distinguish between the three available cases, being:

A goes free
B goes free
C goes free

Each is just as likely until the guard points out that B isn't going free, so then you have:

A goes free
C goes free

being equally likely.

But let's assume the probability is still a third. Given that we know that B is not going free and that the sum of all the probabilities should equal to 1...

P(A) + P(B) + P(C) = 1
1/3 + 1/3 + 0 = 1
2/3 = 1
contradiction

This sounds a lot like that case with the chance of the 'other child being a boy' is really non-intuitive.
 
  • #94
Galileo said:
Not so much a joke as a brainteaser.
Three prisoners, strangers to each other, were suspects of a murder case. One day they came to hear that a sentence has been drawn. Two of them have been found guilty and will be executed, but they don't know which of the two . One guy, a statistician, figures his chances for survival are 1/3, 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 2 of us will executed, that means at least one of the other guys will be. I don't know them or anything, surely you can point to one which is guilty?". The guard sees no harm in that and points one of the prisoners, "He is guilty".
"Thanks!", proclaims the statistician, "my chances have just increased to 1/2".

No no no. Think of it like a bag of balls, hahaha :P. What I mean is there are three balls in his bag...odd..., the stats guy is cupping one ball in his hand... Since balls are indistinguishable he removes one of the others, ouch, and then he has two balls so he's happy! see?... oh let's face it I only put this post up for the crummy balls gag...

But seriously I'm still confused by this, emm its a 0.5 probability all along isn't it? He knows one of the others is guilty without asking the guard. The probability that either he or one of the other two prisoners is going to be executed is equal to one. Since this system has only two balls, about and both are equally likely to be innocent then the chances of innocence are 0.5 for each. He wrongly calculated his chances of survival in the first place, and asking the guard further compunded his stupidity.
 
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  • #95
AlphaNumeric said:
My first post here :)
Q : Why didn't didn't Cauchy like taking his dog for a walk?
A : It kept leaving residues at all the poles.
http://movies.collegehumor.com/items/2005/05/collegehumor.149448.wmv song cracks me up. I expected it to be cheesy as hell, but the amount of puns they fit in is very impressive.
Nice video lol.
 
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  • #96
alfredblase said:
No no no. Think of it like a bag of balls, hahaha :P. What I mean is there are three balls in his bag...odd..., the stats guy is cupping one ball in his hand... Since balls are indistinguishable he removes one of the others, ouch, and then he has two balls so he's happy! see?... oh let's face it I only put this post up for the crummy balls gag...
But seriously I'm still confused by this, emm its a 0.5 probability all along isn't it? He knows one of the others is guilty without asking the guard. The probability that either he or one of the other two prisoners is going to be executed is equal to one. Since this system has only two balls, about and both are equally likely to be innocent then the chances of innocence are 0.5 for each. He wrongly calculated his chances of survival in the first place, and asking the guard further compunded his stupidity.

Actually posting that just compounds your stupidity (joking). For anybody who cannot see, the main man still has a 1/3 chance of survival, the known guilty guy has a 0 chance and the other a 2/3 chance. It has already been explained so I won't bother.
 
  • #97
Joffe said:
Actually posting that just compounds your stupidity (joking). For anybody who cannot see, the main man still has a 1/3 chance of survival, the known guilty guy has a 0 chance and the other a 2/3 chance. It has already been explained so I won't bother.

Hold on, how is the other guy any different than the current one? In the context of the problem the only difference is who asked the question!
 
  • #98
exactly (all I wanted to say was exactly)
 
  • #99
Alkatran said:
Hold on, how is the other guy any different than the current one? In the context of the problem the only difference is who asked the question!

The guy who asked the question differs from the other because he was not involved in the selection. Imagine it this way, there are a thousand people and all but one are guilty, I could ask the guard which 998 of the others are guilty and it would narrow it down to me and one other guy, obviously the other guy has a much greater chance of running free.
 
  • #100
no, your argument is lame
 
  • #101
alfredblase said:
no, your argument is lame
Only if you 'lame' you mean 'correct'

This is a version of the Monty Hall problem, explained here:

http://www.comedia.com/hot/monty.html
 
  • #102
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.

Before he asks the guard he knows one of the other prisoners is guilty. He can discard that prisoner from the system. He is left with himself and one of the other prisoners; 1 prisoner + 1 prisoner = 2 prisoners. Since the chances of survival are equal for both, and since one of them is going to be executed, then the probability of survival = 1/2.

He could have calculated all of this BEFORE asking the guard. Now when he asks the guard all he is doing is discarding one of the prisoners, something he could have done before asking him. So the calculation of his chances of survival remains the same, i.e he has a fifty percent chance of survival. His chances of survival were 0.5 all throughout the story. Nuff said.
 
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  • #103
Joffe said:
The guy who asked the question differs from the other because he was not involved in the selection.

I understand now. A has more information, but it doesn't help his odds, it helps the other inmate's odds. To make sure of this, I wrote a simple little program:

Code: 
int main() {
  int a, b, n;
  int data[3];
  int asum = 0, osum = 0;
#define size 1000
  for (b = 0; b < size; b++)  {
    for (a = 0; a < 3; a++)
      data[a] = 0;
    for (a = 0; a < 2; a++)
      do {
        n = (int)(rand() * 2);
      } while (data[n] = 1)
    
    osum += (data[2] == 1) ? data[1] : data[2];
    aSum += data[0]
  }
  printf("Asker chance: %f, other chance: %f\n", asum, osum)
  return 0;
}

The output makes it pretty clear: the asker dies 2/3 of the time and the one who isn't pointed to dies 1/3 of the time.
 
  • #104
When I first read the explanation I thought it meant C+log cabin i.e. a place where Dutchmen live.
 
  • #105
OMG this is a great "brain teaser" and I think it contains something very deep about statistics. Like a good little physicist after my ally alkatran left me all alone I decided to experiment. The results are conclusive. I describe the experiment in the following paragraph. Please repeat the experiment yourself.

I made three squares and labelled two with a G for guilty, and one with an I for innocent, then folded them, then put them in a hat. There were three very little squares in the hat. So if I were to pick one of the squares and assign that to the main man, and I repeated the experiment 100 times, roughly 1/3 times the main man would be innocent.

Now I take one of the squares out and assign it to the main man. I don't look at the paper. Now I play the part of the guard. So I look at both papers remaining and remove one with G on it from the hat. Now I put the main mans paper back in the hat along with the other paper the guard didnt remove. What am I left with? One paper with I and one paper with G. This would be the outcome every time no matter how many times I repeat the excersise. If I were to then remove one paper and assign it to the main man, and repeated the experiment 100 times, roughly half the time the main would be guilty and roughly half the time innocent. There's no argument against this experiment is there??

So wow, statistics is odd.
 
  • #107
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...
 
  • #108
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.
 
  • #109
alfredblase said:
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. :biggrin:

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".
 
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  • #110
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!
 
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  • #111
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.
 
  • #112
alfredblase said:
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.

alfredblase said:
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.
 
  • #113
alfredblase said:
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?
 
  • #114
:blushing: 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 probability 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 :-p

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
 
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  • #115
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).
 
  • #116
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.
 
  • #117
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. :)
 
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  • #118
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.
 
  • #119
How many people got executed while you were collecting your overwhelming experimental evidence?
 
  • #120
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".
 

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