Stupid things people thought about Finance

[wannabefree]

Originally Posted by Mathew Green

Not money-related, but typical of the dumb stuff that gets passed off as wisdom nowadays: On the TV show 'Num3ers' the character who is suposed to be a total mathmatical genius poses a question to his class:

"I have three doors. There is money behind one of them and a goat behind the other two. Pick a door." Two students pick two different doors, and he opens one to show that it is a goat. Then he asks to the other student if they want to exchange their door for the one neither of them picked and if doing so would make any difference in the odds of them winning. The class concensus is that it would make no difference, but the 'professor' explains that it would because when they first chose that door the odds of choosing the correct door were only one-in-three, but now that there are only two doors left the odds of picking the right door are 50-50... "so if you switch doors now, the odds are higher that your new choice will be right!"

The logic of that scene is so bad that even a gradeschool kid ought to see through it, but aparently neither the scriptwriters, not the actors, nor the "mathmatical consultant" listed in the credits at the end of the show realized that having a 50-50 chance of the money being behind the unpicked door means there is also a 50-50 chance that it's behind the door they already chose!

Hey Matthew you might want to check out this website http://en.wikipedia.org/wiki/Monty_Hall_problem

This is actually a pretty famous math problem it's called the Monty Hall paradox. It was originally posed by marilyn vos savant (aka the smartest person in the world) in a column and she had many famous mathematicians who wrote her and told her how stupid she is. She made them seem foolish.

[asmom]

Originally Posted by Mathew Green

A group of us were discussing what we'd do if we won the lottery ($40-million). One co-worker who had never made more than $20k/year in his life said that he'd use the $40-million to buy rental houses so he'd have enough income to quit working.

Oh dear, that's hilarious!

[Jacklad]

Originally Posted by wannabefree

Hey Matthew you might want to check out this website http://en.wikipedia.org/wiki/Monty_Hall_problem

This is actually a pretty famous math problem it's called the Monty Hall paradox. It was originally posed by marilyn vos savant (aka the smartest person in the world) in a column and she had many famous mathematicians who wrote her and told her how stupid she is. She made them seem foolish.

People get this one wrong because they think in terms of probability, assuming a random distribution. But that doesn't apply to this problem because it doesn't deal with "pure" random probability. The secret to the trick is that Monty always knows where the "prize" is and will never reveal it. It is his influence on the problem that causes the scenario where it is better to switch.

Take Monty out of the problem and it works the way people think it should.

Jackie

[Mathew Green]

Originally Posted by wannabefree

This is actually a pretty famous math problem it's called the Monty Hall paradox. It was originally posed by marilyn vos savant (aka the smartest person in the world) in a column and she had many famous mathematicians who wrote her and told her how stupid she is. She made them seem foolish.

This is a simple probability problem with only one correct answer and, given the problem as stated, marilyn vos savant's answer is wrong. I know that a lot of people, including some highly educated people, believe switching doors would improve your odds of winning, but IMHO it just ain't so.

Let's look at the problem from a different angle:

1. Mary picks door #1
2. Susie picks door #3
3. Susie's door has a goat behind it.
4. At this point Bob walks into the room with no knowledge of what happened before he got there.
5. The professor says, "I have two doors. Pick one."
6. Bob picks door #2

Does Bob have a better chance of being right than Mary?

If Bob had picked door #1, would he have inhierited Mary's one-in-three odds?

No. Every chance to make a decision (or to change a decision) has it's own set of odds based on conditions as they exist at the moment. A flipped coin doesn't remember how many times it's come up heads, and the odds of a random choice being right can't be effected by the odds that existed during previous choices. IOW: If you have a 50-50 chance of being right, you also have a 50-50 chance of being wrong, and nothing that has happened previously changes that simple mathmatical fact.

Or look at it this way: Since the professor wanted to make a point, he probably let Susie pick her door first, then asked for a volunteer who disagreed with Susie's choice. IOW he asked for someone who would make a choice between door #1 and door #2. That sets Mary's initial odds at 50-50, not the one-in-three odds she thought she was getting.

Here's another angle:

1. Mary picks door #1
2. Susie picks door #3
3. The professor opens door #2 and we see it has a goat behind it.
4. At this point there are two doors left, each girl has picked one, and we know the money has to be behind one of those doors.

Would either girl improve her odds by switching doors? No. If either girl could improve her odds by switching to the oposite door, the same would be equally true for the other girl; and obviously it's impossible for both girls to improve their odds by simply exchanging doors with each other.

Since I'm not a mathematician, and I don't even play one on TV , I'm not really qualified to argue this point. But I do know hocus pocus disguised as mathmatical proof when I see it. The correct answer is that switching doors at that point would have zero effect on your odds of winning the money.

[Jacklad]

Originally Posted by Mathew Green

Since I'm not a mathematician, and I don't even play one on TV , I'm not really qualified to argue this point. But I do know hocus pocus disguised as mathmatical proof when I see it. The correct answer is that switching doors at that point would have zero effect on your odds of winning the money.

Did you read the Wikipedia entry? The odds are 2/3 that you will be successful if you switch. You have a 1/3 chance of choosing the car, which means that 2/3 of the time the car will be in the remainder. That probability doesn't change, regardless of Monty showing you the goat.

Case one: you pick the car. Monty shows a goat.
Case two: you pick goat #1. Monty shows you goat #2.
Case three: you pick goat #2. Monty shows you goat #3.

In both case two and three (i.e. two thirds of the time) you are better off to switch.

You can even take Monty out of it altogether. You pick a door, and I'll give you a choice between your single door, or the other two (that's right, your door versus two doors). Would you switch? Of course, and you'd see easily that 2/3 of the time it would get you the car. Monty showing you the goat doesn't change that, it just distracts you.

Jackie

[Mathew Green]

Originally Posted by Jacklad

Take Monty out of the problem and it works the way people think it should.

You are correct that an active dealer would influence the outcome; but the problem (in it's pure form) assumes that Monty is totally honest. That is to say, he always reveals a goat and he always gives the remaining contestant a chance to switch to the other door. Since his prior knowledge of which door has the money doesn't change his behaviour, his presence has no effect on the outcome of the problem. If he were replaced by a set of preprinted instructions, the outcome would be the same.

My remarks in this thread apply only to the puzzle in it's pure form with a passive dealer, because that is the way it was presented in the tv show. In the real world "professor numbers" would have asked the class for a show of hands to see how many liked each door, and then called two students who had chosen a 'goat' up front. In other words he would cheat to insure the experiment coming out the way he wanted it to. But then it's no longer a real experiment.

[Mathew Green]

Originally Posted by Jacklad

Did you read the Wikipedia entry?

Yes.

Originally Posted by Jacklad

The odds are 2/3 that you will be successful if you switch.

I disagree. And according to the Wikipedia entry, ten tousand readers, including several hundred mathematics professors wrote to marilyn vos savant to tell her switching doors wouldn't change the odds. That's good enough for me. It is, at most, an open question; and I believe strongly that the "2/3" advocates are wrong.

[Jacklad]

Originally Posted by Mathew Green

I disagree. And according to the Wikipedia entry, tousands of mathematicians wrote to marilyn vos savant to tell her switching doors wouldn't change the odds. That's good enough for me. It is, at most, an open question; and I believe strongly that the "2/3" advocates are wrong.

Read it again - particularly the part where you're dealing with a hundred doors. Then read the history.

I prefer the original Bertrand's Box construction, as it gets rid of the ambiguity of Monty, what he knows and how honest he is or isn't. http://www.cenius.net/refer/articles...oxparadox.html

And if you're up for some reading, http://ist-socrates.berkeley.edu/~fi...8/kyburg_3.pdf

Last but not least, try http://home.att.net/~numericana/answer/counting.htm. Dr. Michon is pretty readable for a "pure" mathematician.

Jackie

[Mathew Green]

Jacklad,
If you really want to convince me, explain how switching would improve the odds of either girl in my third example:

1. Mary picks door #1
2. Susie picks door #3
3. The professor opens door #2 and we see it has a goat behind it.

At this point there are two doors left, each girl has picked one, and we know the money has to be behind one of those doors. Would either girl improve her odds by switching doors? No. If either girl could improve her odds by switching to the oposite door, the same would be equally true for the other girl; and it's obviously impossible for both girls to improve their odds by simply exchanging doors with each other.

IMO what is true under that circumstance is equally true in all cases.

[PrincessPerky]

k let me get this straight..you have three options, one prize, two losers (goats)

basic probability, three options, one right = 1/3

always shown a loser, you now have 2 doors, one right = 1/2

suppose you picked door one (never mind what it was origionally called, it is now 1 out of 2 options, coins don't care if they were tails last time or heads, there is still two options)

anyway you pick door number one, then you switch or stay, and win or lose..write it all out pretty picture and all (yes I drew one)

make sure you start with 'switch or stay' then right or wrong......

And it all works out to you have 4 'paths' to the car out of 8 total paths..that is 50/50 chance.......even if you decide to care what the door was origionally called, follow that path it is still a long convoluted way of saying 50/50......of course you can now draw the pretty picture for door number two, it looks the same..leaving you with 8 paths out of 16 total..which is a fancy way of saying 1/2..or 50/50.

switch or stay you have a 50/50 chance of being right. (or wrong) weather it was your first pick or second you still have a 50/50 chance..which is nicer than the 1/3, but no need to change. less you get 'that feeling..'


Out of 100 it is even easier to see, suppose you have 100 options, one winner 99 losers, your host shows you one loser, you now have 99 options, one winner...odds went up but they did so regardless of you staying or going.......you are merely the pointer to a path, each individual path still has the same chance of being right.......

Bertrands box is different, because it involves the probability of a box, in a box. (which box, combined with which coin)

Now three people pay 10 bucks each for dinner...the waiter says you overpaid by 5 bucks, you have to split it, but no one has change, so you each get a buck and the waiter gets a 2$ tip...so you all paid 9 bucks each, which is 27, and the waiter got 2$....27 plus 2 = 29, whered your extra buck go?

the solution is to stop listening to your host (me in this case) the math is simple..

[Jacklad]

Originally Posted by Mathew Green

Jacklad,
If you really want to convince me, explain how switching would improve the odds of either girl in my third example:

1. Mary picks door #1
2. Susie picks door #3
3. The professor opens door #2 and we see it has a goat behind it.

At this point there are two doors left, each girl has picked one, and we know the money has to be behind one of those doors. Would either girl improve her odds by switching doors? No. If either girl could improve her odds by switching to the oposite door, the same would be equally true for the other girl; and it's obviously impossible for both girls to improve their odds by simply exchanging doors with each other.

IMO what is true under that circumstance is equally true in all cases.

Think of it this way - I am Mary. I make my choice. What do I know? I know that I have a 1/3 chance of having the car. But I also know that there is a 2/3 chance that I don't. I also know with 100% certainty that at least one of the non-chosen ones is a goat. When that goat is revealed, it doesn't change anything that I already know - there is still a 2/3 chance that I didn't choose the car, so it's in my best interest to switch. Make more sense that way?

What makes this confusing is that the same logic applies to Susie, and it's in her best interest to switch as well. That's not impossible, because a 2/3 chance of success still implies a 1/3 chance of failure. And, of course, that's what confuses people into thinking it's a 1/2 chance and the only reason some versions of the puzzle introduce a second player. (The classic version, and the original Bertrand's Box, avoid that.)

Jackie

[Jacklad]

Originally Posted by PrincessPerky

k let me get this straight..you have three options, one prize, two losers (goats)

basic probability, three options, one right = 1/3

always shown a loser, you now have 2 doors, one right = 1/2

Nope. That's where people get it wrong. Your original probability doesn't change.

There's a 1/3 chance that you picked the car. There's a 2/3 chance that you didn't.

Originally Posted by PrincessPerky

anyway you pick door number one, then you switch or stay, and win or lose..write it all out pretty picture and all (yes I drew one)

make sure you start with 'switch or stay' then right or wrong......

There's only three cases:

Case 1 - I choose the goat. Door 2 has a goat. Door 3 has the car. A goat is revealed. I switch, and that's the right choice.
Case 2 - I choose the goat. Door 2 has a car. Door 3 has a goat. A goat is revealed. I switch, and that's the right choice.
Case three - I choose the car. Door 2 has a goat. Door 3 has a goat. A goat is revealed. I switch, and that's the wrong choice.

Basically, all the doors, cars, goats and even Monty are just designed to confuse the basic question - would you rather stay with a 1/3 chance of being right or switch to a 2/3 chance of being right. The "trick" is that they will never reveal the car.

Consider three sealed envelopes. You choose one, and one is ripped up. Switch or not? Well, in this case it doesn't matter - because the sealed envelope could have contained the car, you are left choosing between one 1/3 chance and another 1/3 chance. But in the Monty Hall problem, that's not the case, because you know the car will never be revealed.

Originally Posted by PrincessPerky

Out of 100 it is even easier to see, suppose you have 100 options, one winner 99 losers, your host shows you one loser, you now have 99 options, one winner...odds went up but they did so regardless of you staying or going.......you are merely the pointer to a path, each individual path still has the same chance of being right.......

No - your original odds (1/100) *never* change, regardless of how many wrong choices are revealed. Your choice here is between your 1/100 chance of being right and the 99/100 chance of being wrong. I know what I'm choosing.

Originally Posted by PrincessPerky

Bertrands box is different, because it involves the probability of a box, in a box. (which box, combined with which coin)

Betrand's Box is actually the original, and designed to illustrate the confusion between original choice and subsequant knowledge. See http://home.att.net/~numericana/answer/counting.htm for an explanation of why the confusion arises. Another variant is the "Three Prisoners" (aka Prisoner's Dilemma): http://mathworld.wolfram.com/PrisonersDilemma.html

All these puzzles rely on the fact that subsequent knowledge does not alter the original odds.

Originally Posted by PrincessPerky

Now three people pay 10 bucks each for dinner...the waiter says you overpaid by 5 bucks, you have to split it, but no one has change, so you each get a buck and the waiter gets a 2$ tip...so you all paid 9 bucks each, which is 27, and the waiter got 2$....27 plus 2 = 29, whered your extra buck go?

the solution is to stop listening to your host (me in this case) the math is simple..

It is. We paid $25 for the room, got $3 back and the waiter got $2. 25+3+2 = $30, It's a classic of arithmethical misdirection, while Bertrand's Box and it's descendant problems are classic of probability misdirection.

The bigger question, though, is why so many people play the lottery.

Jackie

[PrincessPerky]

Case 1 - I choose the loser. Door 2 has a goat. Door 3 has the car. A goat is revealed. I switch, and that's the right choice.
Case 2 - I choose the goat. Door 2 has a car. Door 3 has a goat. A goat is revealed. I switch, and that's the right choice.
Case three - I choose the car. Door 2 has a goat. Door 3 has a goat. A goat is revealed. I switch, and that's the wrong choice.

K I don't get the choices? all the extra stuff confuses me (hence why I supplied the money one, all the talk confuses people)

path one: chose loser1, reveal loser, stay, lose
path two: chose loser1, reveal loser, switch, win

path three: chose loser2, reveal loser, stay, lose
Path four: chose loser2, reveal loser, switch, win

path five: chose winner, reveal loser, stay, win
path six: choose winner, reveal loser, switch, lose

AH HAH! I get it, those paths are all the options, but if you only look at switch, out of three switch paths, 2 are wins! (there are 6 paths though, and half win)

because you had a 2/3 chance of losing the first time, so it is a box in a box....

If the switch was random, it would be 50/50 chance though (like if a computer made the choice to switch or stay)

K I get it..no time to look at 1/100 right now, but I get the 1/3 one.

[Jacklad]

Originally Posted by PrincessPerky

AH HAH! I get it, those paths are all the options, but if you only look at switch, out of three switch paths, 2 are wins! (there are 6 paths though, and half win)

because you had a 2/3 chance of losing the first time, so it is a box in a box....

Yay! Unfortunately, I can't give you the car. Sadder still, I can't even give you the goat.

Of course, if I had the choice between a car and a goat, my kids (pun intended) would be voting for the goat. They are awfully cute .

Jackie

[wannabefree]

wow these forums are sweet do I get credit for this thread hijack

[amberfocus]

Ah, I remember this problem from a high school class.

It may seem counterintuitive at first, but switching doors *is* better.

I can explain it in a slightly different way.

Say you run through the scenario a large number of times. Every single time, you pick a door, see the goat in one of the other doors, and you STICK with your choice.

If the odds of winning are truly 50/50, then you should win the car 50% of the time.

But think about it. If you pick a door, and always stick with it, you'll win 33.3% of the time.

After the revelation of the goat, there are only two doors left, and there is a 100% chance that the car is behind one of them. If picking and STICKING yields a 33.3% chance of winning, then SWITCHING to the other door would necessarily yield a 100% - 33.3% = 66.7% chance of winning.

Thus, it's better to switch.

I love these brain twisters.

~mimi

[sweeps]

Originally Posted by Jacklad

The bigger question, though, is why so many people play the lottery.

Maybe this is why. Link

[Great to be Debt Free]

Head is swimming... goats...cars...goats driving cars... Monty Hall behind doors...swimming...swimming...Mrs. Walkers math class...her crooked bony finger...screechy voice...probabilities...sounds like a goat...Monty Hall with a goatee...Monty Hall witha goatee driving a car from behind doors... AAAAAAUUUUGH!

[Jacklad]

Originally Posted by Sweepsplayer

Maybe this is why. Link

But they don't do stories on the millions of people who lost. Or the people who spend the grocery money on lottery tickets. Or stories comparing people who throw $10 at the lottery every week for twenty years to those who saved $10 week over the same period.

Lotteries are a state sponsored tax on the stupid.

Jackie

[krayziebone33]

Originally Posted by Broken Arrow

Well, I used to think that the bank interest rates were applied monthly. Imagine my disappointment when I realized that it was annual.

Same here.