Dude... Karma! ?

In the 'palooza ep where Homer is the "fat guy who gets shot with a cannon", there is a scene where he buys a rasta hat from a vendor. As Homer is about to leave, the vendor says something like, "Dude... karma, karma!" Whereupon homer stares at him blankly, then the vendor points to a can marked "tips" and repeats "karma", and Homer, with a look of realization, says something to the effect of "Oh! Karma!" and then turns and walks away without tipping him.

There seemed to be several possible ways to take that. Which one do you see it as?

1. Homer didn't understand until the kid pointed to the can, then Homer understood and simply declined to tip the guy.

2. Homer still didn't understand after the kid pointed to the tip can, but didn't want to appear stupid, so he pretended to understand.

3. Homer is so stupid that he thought he understood what the kid meant when he pointed to the tip can, but obviously didn't understand at all.

• Something else?

I'll pick... door #2

"That Guy" wrote in message news:Nw8%6.63$[email]...

cannon",

about

away

see

when

Sounds like YOU... are working for your KARMA.... SIMPLIFY, MAN!

Seriously, it seems like #1 would be the funniest way to take it.

I'm inclined towards choice 2. Something along the lines of "Do I know what rhetorical is! Eight."

speaking of picking doors, are there any math freaks here?

the monty hall problem is extremely insteresting, and it is simple yet impossible.

here it is:

there are three doors. behind one is a car, and behind 2 are donkeys. you pick a door, then the host of the game show, tells shows you one door with a donkey behind. and then asks you if you want to switch.

do you switch, or not?

you can get seriously bugged out on this one, its better then ganga.(to keep with the posts original theme of rasta)

This problem is fairly simple if you look at it the right way.

The first door you picked has a 1/3 chance of being correct. Monty will always pick a losing door because he can't eliminate the prize door, so If you switch, you have a 2/3 chance of being correct. Here it is broken down:

The numbers are irrelevant because any door could be a winning or losing door, so X will stand for the winning door, Y and Z are losing doors.

Scenario A: You don't switch

• You pick door X. You win
• You pick Y. You lose.
• You pick Z. You lose.

Scenario B: You do switch.

• You pick door X. You switch and lose.
• You pick door Y. Monty eliminates Z, you switch and win.
• You pick Z, Monty eliminates Y, you switch and win.

If you switch, the odds are reversed, because you will win if you initially picked any losing door!

"ben-plot" wrote in message news:[email]...

Well, now you're doomed. This is not a soluble problem. It was posted as an example of an insoluble problem. I've been on more than one group/board that COMPLETELY exhausted this at length. You end up with fifty different solutions, giving several different answers, all of which MAKE SENSE.

Good luck.

--John

you must remember that true probability is distorted by the fact that monty knows, and will always open up a goat. try it with a friend and put a marble 3 different cups. it gets tricky. i did it 100 times each way(switch/no switch) with my friend joe glyn. ill tell you my results later.

"Groucho027" wrote in message news:[email]...

If

down:

initially

an

that

solutions,

They make sense if you think you understand them but don't. This is FAR from unsolvable. I posted the solution. It's fairly simple. Run it by someone who has studied math if you don't believe me.

Yes, of course, you are God and you are obviously right. My actual experience of watching many people who are experienced mathematicians lock horns and after much discussion agree to disagree means nothing in the face of your dazzling brilliance. And pig-headed arrogance like yours does not in any way contribute to the interminable arguments this problem spawns.

--John

Yeah, but the Monty Hall problem is easy to solve (although quite non-intuitive), and the solution has been posted. I haven't met one mathematician who knows anything about stats that hasn't agreed with the posted solution. If Monty has to open up a door with a bad prize after you've chosen a door, the chances of you winning after you switch are 2/3. Hell, run this experiment yourself and you'll see. I did when I first heard this problem.

-- Rick

"Groucho027" wrote in message news:[email]...

will

so

losing

yet

donkeys.

ganga.(to

as

experience

after

dazzling

contribute

Good grief, you moron. Go back to smoking your "ganga" and stop harassing people who aren't on dope. Christ on a crutch, there IS NO CONTROVERSY over this simple problem!

"Groucho027" wrote in message news:[email]...

experience

after

dazzling

contribute

I have posted several links that prove me right. Can you show me even one that contradicts me? This is, as I have said, a VERY SIMPLE problem. I feel like I'm arguing with a 2-year old about what 10 + 10 is. You can tell him it's twenty, you can show him it's twenty, but he's still not going to understand.

http://www.cs.unb.ca/~alopez-o/math-faq/mathtext/node32.html

http://cartalk.cars.com/About/Monty/

http://www.nadn.navy.mil/MathDept/courses/pre97/sm230/MONTYHAL.HTM

http://www.cut-the-knot.com/hall.html

hey man, this problem was even a bitch for erdos. and erdos was the best fuckin mathematician of the 20th century. so dont call me a baby for not trusting you.

A contrary solution? ok...

You pick a door. Say it's door number 1, it doesn't really matter. Monty removes one of the loser doors. Say it's door 2. So there's two doors. You can either stay, which is picking door 1, or you can switch to door 3. You have a choice between one of those 2 choices. One of them wins, one of them loses. Therefore, the odds are fifty-fifty, and it doesn't matter if you switch. Why doesn't this solution work?

--John

"Groucho027" wrote in message news:[email]...

can

have a

loses.

Why

Why doesn't that solution work? The same reason 10 plus 10 equals15 doesn't work: Because it's wrong. Think of it this way. If you pick the correct door and switch, you will be wrong 100% of the time. If you pick the wrong door, Monty eliminates the other wrong door, so if you switch, you win, 100% of the time.

This means that if you pick a wrong door and switch, you will ALWAYS win. Your chances of picking a wrong door are 2/3. Therefore your chances of winning if you switch are 2/3. Your chances of willing if you don't switch are 1/3.

You can try this with a friend and some cards, and prove it to yourself. Have, say one red (winning door) and two black (losing door) cards held so only your friend can see them. You pick one without looking at it. At lease one of the cards he has left is a black one, have him pull out a black one.

Do this twenty times without switching and see how many you win, then do it twenty times switching each time and see how many you win.

You will find that you win roughly twice as often when you switch.

That's not MY solution. It's just one of the other ones that tends to get offered when this question turns up, as it frequently does on message boards and newsgroups. I have no interest in either solution. My point is that this question always spawns interminable arguments of the type, "My answer is right because it is right, Yours is wrong because it is wrong, you are stupid." Which is a good summary of what you just posted.

Oh well. I just hope you keep it one thread. I'll go now. Enjoy your stupid pointless argument that will never be resolved.

--John

"Groucho027" wrote in message news:[email]...

Monty

You

switch.

time.

you

switch

so

black

it

boards

this

right

stupid

Actually, as I have said, the problem is very simple and has been resolved. I posted a bunch of web links that prove me right-- you offer nothing except insults. There is only one resolution to this problem, which you simply cannot, or will not understand. I suppose since you think the problem is "better than ganga" that you cling to your belief like any die-hard drug addict. I feel sorry for people like you.

It's very simple why this solution doesn't work. The whole process of picking a door discovering one that is a not a winner and then choosing to switch or not is considered ONE event in statistical terms. The odds of winning in any single event DO NOT change during the event. Therefore the odds that the door you originally picked is a winner is 1/3. After discovering one of the incorrect doors the odds that the one you originally picked STILL is 1/3. The odds to do not change because one of the doors has been revealed.

This is the statistical fact that That Guy uses in his answer. An answer which I have seen in many places explained in many ways by many people. Any person who has taken even a basic college course on statistics can tell you the answer without any controversy whatsoever.

Kevin

in article [email], Groucho027 at [email] wrote on 7/2/01 4:43 PM:

"Kevin Madden" wrote in message news:B766DD09.12E%[email]...

originally

has

Any

you

Sorry, Kevin, but you're wrong and I can prove it. I believe that you will give me that your chances of picking a winning door is one in three, and your chances of picking a losing door are two in three.

Now here is the scenario if you ALWAYS SWITCH:

You pick any losing door, you will always win. Why? Because Monty eliminates the other losing door, and the only one left to switch to is the winning door. Your chances of picking a losing door are two in three, therefore your chances of winning are two in three if you always switch. Apparently, you slept through your "basic college course on statistics!"

Your faulty assumption is that Monty will pick any door at random. IF this was the case, you'd be 100% correct. However, in this puzzle (and in the game show) Monty always picks a losing door, he will never pick a winning door. That changes the outcome. If you don't believe me and you can't do the math, try it yourself with a friend and three cards.

I was backing you up That Guy. You are right, if you pick a wrong door and switch you will always win and if you pick the right door and switch you will always lose...I wasn't arguing that. What I was saying was that the chances of you picking the right door in the first place is 1/3 while the chances of picking the wrong door the first time is 2/3. Therfore, as you said, the chances are better that you picked a wrong door in the first place so you are more likely to win if you switch.

I think my explanation wasn't quite clear...Rick explains it better in a later message (clearly better than I because you agreed with him and disagreed with me and we both said the same thing). The point of my message was to clear up the fallacy that after a door is revealed the chances are 50/50, which is not true because the odds do not change during the process.

So I didn't sleep through either of my statistics courses :), I'm just not a good teacher.

Kevin

Wait a sec, I just got here. Are saying you will always pick the right door if you switch after Monty reveals one of the doors?

Betsy "I'm learnding!" Where

GeekCode S1.3: HOM+++ FRI++@! The Ramones+++ CBG+++!!!!* BUR> TEE+++ NSync-- f++ n- $+++ 4F05, 5F13 F1981

No, if you picked the right door in the first place and then switch you will pick the wrong door. But if you picked one of the wrong doors in the first place and switch then you will pick the right door. It's simple, if you pick one of the wrong doors in the first place and Monty reveals the other wrong door then the only door left is the right one.

Since the chances that you picked a wrong door in the first place are 2/3 then the chances that you will win by switching is also 2/3.

Kevin

Betsy Where wrote:

Sorry, I misunderstood you. I guess me English is even worse than my math! :-)

"Kevin Madden" wrote in message news:[email]...

and

will

chances of

are

later

with

up

not

a

door

three.

the

this

the

winning

do

The fallacy in your reasoning is this:

Monty must get rid of a loser door.

If he got rid of a random door (well, one of the two you didn't pick), you'd be right; the chances are 50/50. However, Monty is FORCED to pick only one of the doors 2/3 of the time. If you pick a bad door, Monty can't even choose which door to eliminate. He must pick the only other bad one.

Actually, I'll give you the reasoning that was used to teach me.

Chances of you picking the winning door when first asked: 1/3

Think about this statement. Really think about it for a second. These odds will not change, no matter how much you learn about the system in the future. The odds are, and always will be, 1/3.

If Monty eliminates a door, *and we know that one is still a prize*, i.e., theres a 1/1 chance one is still a prize, the other door's chances of having the prize is 1/1-1/3, or 2/3.

Good explanation, Rick. It's amazing how many different ways there are to explain this.

"Rick Nelson" wrote in message news:[email]...

You are more full of shit than an overflowing cesspool.

look, im not full of shit. you are right. my hypothesis was to switch and my experiment proved it. we got 63% good prize when you switch, which adjusted to odds is 2thirds.

but with out switching we got 35% good prize.

each of the odds are doing the experiment 100 times each way.

read a book called "the man who loved only numbers" theres an interesting section on this problem. where it says out right that erdos said it didnt matter if you switched ot not. a lot of famous mathematicians didnt get it, including paul erdos.

if you dont belive me read the book, and educate yourself.

heres the amazon link, or take out of the library.

"ben-plot" wrote in message news:[email]...

http://www.amazon.com/exec/obidos/ASIN/0786884061/o/qid=994116940/sr=2-1/ref =aps_sr_b_1_1/103-6630619-9650258

I apologize. I never realized this was such a difficult problem for even advanced math wizards. It seems so simple to me... and yet I am not an advanced mathematician by any stretch of the imagination.

Okay, I'm getting involved now:) I've looked at all of these links and I'm still having a little trouble, maybe That Guy can clear this up for me. Here's my problem: Once you eliminate one of the doors, that door is removed from the equation. It is no longer x/3, but now x/2 since there are only 2 possibilities left. That third door is no longer a factor since it cannot be chosen. It doesn't count anymore and is therefore an impossible outcome. The equation then becomes a whole new equation with 2 possible outcomes (x/2)rather than three. When you say you have a 2/3 chance of picking the prize door after one door is revealed and after switching, you are still counting that dead door that cannot be chosen. If it's not a possible outcome, it's no longer a factor and cannot be entered into the equation. Oohh..headache.

Wockle

i think everyone should take a break, trim some cheeba off of the plant in your closet and light it up. you should use a gravity bong, this gravity bong is one of the best uses for my knowledge of physics.

I'm with you...except I'm going for a nice glass of ale. I've put my two cents in and this off topic thread has gone on way too long. Those who know that the answer to this is a mathematical certainty and those who think this is some sort of philosophical question that will never be answered have all made their points.

Kevin

ben-plot wrote:

"wockle" wrote in message news:[email]...

one

tell

to

You're basing your equation on the false assumption that Monty chooses a door at random. Monty does not choose a door at random. He chooses only a losing door, and that changes everything.

Also, you're overanalyzing. This is the simplest way to look at it:

Here are all possible scenarios (2) if you always switch:

1. You pick the right door (a one in three chance) and you switch to another door, so you lose.

2. You pick a losing door (a two in three chance.) There are two doors remaining, the winning door and a losing door. Monty can't pick the winning door, so he picks the losing one. The only one left is the winner, so you switch and you win.

Therefore, if you always switch, then whenever you pick a losing door initially, you win, and whenever you pick a winning door initially, you lose. Since there are two losing and one winning, your chances of picking a losing are two in three, so your chances of winning if you always switch are also two in three.

Here's another way to look at it. You picked a door and your chances of it being a winner are one in three. The chances that one of the other doors is a winner is two in three, simply because there are two of them. If you could pick BOTH of the other doors, your chance that ONE of them is a winner is two in three. When Monty eliminates a losing door, the chances that one of the two doors you didn't pick is a winner is STILL two in three, because Monty will always pick a losing door.

I can sum it up in a much easier to understand way: If the player switches, she wins if and only if her first choice is incorrect, an event that has a probability 2/3. If the player never switches, then she wins if and only if her first choice is correct, an event with a probability 1/3. See, your more likely to guess incorrectly the first time so you should switch.

Betsy "glavin!" Where

Yes, that is better. Thanks!

"Betsy Where" wrote in message news:[email]...

switches,

if her

*takes a bow* No applause, just throw money.

Betsy "mmm...hug" Where S1.3: LIS+++ FRI++@! The Ramones (RIP Joey)+++ HIB-- CBG+++!!!!* NSync-- Ron Howard++ f++ n- $+++ 4F05, 5F13 F1981

Keep the door you picked. It's now 50-50

I don't quite understand something about Scenario B. If you pick door Y (a losing door), you claim that Monty will always elimate Z, the other loser. But the original problem states that Monty will reveal one losing door. How do you know that he won't sometimes reveal the door you already picked to be the loser? Is it part of the problem that Monty can't tell you anything about the door you picked?

If you could clear that up this would make sense.

(I'll feild this one)

Yeah, Monty will never reveal the door you picked. Thus, he's actually forced to reveal a certain door 2/3 of the time (which is the same as saying the door he didn't open has the prize 2/3 of the time, since he will never reveal a winning door). Hope that clears it up.

Ok, I got it. I am officially converted. ;) Thanks to both of you who responded.

"Philip Werner" wrote in message news:[email]...

If

down:

You have a very good point. I mistakenly assumed that all were familiar with the rules of the game. The rules are that Monty will never pick the door you have picked, and he will never pick a winning door, and he will always pick one door. And he will never tell you whether you picked the right door or the wrong one.

The knee-jerk "duh, of course" answer is that if there are got two doors, but only one prize, you have a fifty-fity chance of picking the right door. This isn't necessarily true, if the situation is set up such that: the host knows which door has the prize behind it; and the only doors that he opens are the ones that don't have the prize behind them.

Well that's just "on paper." In real life, one should keep in mind that he doesn't always open one of the doors, because that ruins the entertainment value. If it is true that the host ALWAYS opens one of the doors without a prize behind it, then of course it would behoove the contestants to ALWAYS switch to whatever the OTHER door is. Where's the sport in that?

If, switching back to the world of statistics again (hold all comments about Schroedinger's cat, please), it is true in the long run that the odds of the prize being behind door one in comparison to the prize being behind door two, given that the host ALWAYS opens a door when the contestant is right, but only fifty percent of the time when the contestant is wrong.

Cecil Adams offers this variation on, and insight into, the question:

"Suppose our task is to pick the ace of spades from a deck of cards. We select one card. The chance we got the right one is 1 in 52. Now the dealer takes the remaining 51 cards, looks at them, and turns over 50, none of which is the ace of spades. One card remains. Should you pick it? Of course. Why? Because (1) the chances were 51 in 52 that the ace was in the dealer's stack, and (2) the dealer then systematically eliminated all (or most) of the wrong choices. The chances are overwhelming--51 out of 52, in fact--that the single remaining card is the ace of spades."

I can't believe I got involved...

Alee

S1.3 FRI*+++ RWG++ GIL-- MIL++ MAG++ TEE+ LIS* f+++ n+ $ 9F01,7F19,7G02,2F10, 2F22,1F02,4F04 F1980

"Katijaikat" wrote in message news:[email]...

but

which

ones

You're right, but you're talking about something else.

Let me restate the original problem:

"> >there are three doors. behind one is a car, and behind 2 are donkeys.

This isn't about the game show itself and statistics relating to it, it's a simple math problem involving set rules.

The problem you are talking about (real world) is much more complex, and the example you gave below is more complex than I'd care to tackle, but FWIW it appears correct.

about

the

two,

percent

select

the

ace

(1)

the

The

card

Oh, a five THIRteen . . .

Dave

"That Guy" wrote in message news:Nw8%6.63$[email]...

cannon",

about

away

see

when