oh no you didn't....

[andylawcc]

http://www.businessinsider.com/the-most-controversial-math-problems-2013-3#

#2

 

[IronWing]

A site that requires the reader to stare at an ad for 15 secs before moving to content? Forget that.

 

[andylawcc]

eh? i didn't have to....

 

[DesiPower]

another puzzle... meh... thread title didn't make much sense though...

 

[Pheran]

That was an excellent set of slides, thanks.

 

[KeithTalent]

That was an excellent set of slides, thanks.

Why did I find this post so funny?

KT

 

[Pheran]

Why did I find this post so funny?

KT

I'm not sure. Is my sheer geekiness due to liking math slides amusing?

 

[Rakehellion]

I don't understand the Monty Hall problem.

 

[SKORPI0]

http://en.wikipedia.org/wiki/Monty_Hall_problem

 

[Rakehellion]

http://en.wikipedia.org/wiki/Monty_Hall_problem

Yeah. I think that's wrong and it only works by confusing the question.

If you have two doors, your odds can never be greater or worse than 50%. Just like if there were only two doors to begin with.

 

[andylawcc]

Question #2 is talking about "0.999999999999999 = 1" ?

 

[ShadowOfMyself]

The birthday one... Dafuq? How does that make any sense?

If you have 23 people and there are 365 possible birthdays, how can the chance of two people sharing a birthday be 50%? When I was in school I always had 20+ people classes and there was NEVER anyone with the same birthday as me, and from what I remember no one had the same birthday to begin with

Sounds like bullshit to me

 

[Rakehellion]

The birthday one... Dafuq? How does that make any sense?

If you have 23 people and there are 365 possible birthdays, how can the chance of two people sharing a birthday be 50%? When I was in school I always had 20+ people classes and there was NEVER anyone with the same birthday as me, and from what I remember no one had the same birthday to begin with

Sounds like bullshit to me

Anecdotal evidence is not math.

 

[rcpratt]

The birthday one... Dafuq? How does that make any sense?

If you have 23 people and there are 365 possible birthdays, how can the chance of two people sharing a birthday be 50%? When I was in school I always had 20+ people classes and there was NEVER anyone with the same birthday as me, and from what I remember no one had the same birthday to begin with

Sounds like bullshit to me

Didn't click the link, but I'm sure it means 50% chance of the entire class having one or more "birthday pair."

 

[Jeeebus]

I find the Monty Hall 'Problem' fairly idiotic, so I stopped reading beyond there. Anyone who argues the chances of winning are greater than 50/50 once the other doors are revealed needs to be punched in the prostate.

 

[KeithTalent]

I'm not sure. Is my sheer geekiness due to liking math slides amusing?

No idea, but it was amusing. ^_^

I don't understand the Monty Hall problem.

Is the simplest, clearest explanation.

KT

 

[brianmanahan]

 

[Rakehellion]

No idea, but it was amusing. ^_^

Is the simplest, clearest explanation.

KT

That's wrong. Your odds were 1/3, until you were shown the door with the goat.

If you do the problem in reverse with two doors, one with a car and one with a goat, your odds are 50%. If they show you another goat, your odds are still 50%. It only works by confusing the problem.

Once you're shown the undesirable doors, they're eliminated from the equation, thus it's irrelevant how many there were.

 

[EliteRetard]

The birthday one... Dafuq? How does that make any sense?

If you have 23 people and there are 365 possible birthdays, how can the chance of two people sharing a birthday be 50%? When I was in school I always had 20+ people classes and there was NEVER anyone with the same birthday as me, and from what I remember no one had the same birthday to begin with

Sounds like bullshit to me

You're probably thinking 365 options in a hat and everybody picking one...in which case the chances are zero that anybody will have the same. But if they pick one and then put their result back in the bag the chances are greater that somebody will indeed pick one that was already picked the more people you have picking.

 

[Pheran]

Yeah. I think that's wrong and it only works by confusing the question.

If you have two doors, your odds can never be greater or worse than 50%. Just like if there were only two doors to begin with.

I find the Monty Hall 'Problem' fairly idiotic, so I stopped reading beyond there. Anyone who argues the chances of winning are greater than 50/50 once the other doors are revealed needs to be punched in the prostate.

Both of you are failing at math. This is why this problem is controversial - because the correct answer is counterintuitive.

 

[kranky]

That's wrong. Your odds were 1/3, until you were shown the door with the goat.

If you do the problem in reverse with two doors, one with a car and one with a goat, your odds are 50%. If they show you another goat, your odds are still 50%. It only works by confusing the problem.

Once you're shown the undesirable doors, they're eliminated from the equation, thus it's irrelevant how many there were.

How about if you pick one of the doors, and I get the other two?
Now we show you one of my doors is a goat (which, of course, one of my doors had to be, so this changes nothing.)
If it's 50/50 like you say, then your odds of winning would be the same whether you switch or not. So you don't get to switch.
Do you think we'll both win half the time, or will I win twice as often?

 

[Jeeebus]

Both of you are failing at math. This is why this problem is controversial - because the correct answer is counterintuitive.

The thought of me traveling to Pittsburgh to hit you is counterintuitive. Please do me a solid and punch yourself in the prostate.

 

[Pheran]

That's wrong. Your odds were 1/3, until you were shown the door with the goat.

If you do the problem in reverse with two doors, one with a car and one with a goat, your odds are 50%. If they show you another goat, your odds are still 50%. It only works by confusing the problem.

Once you're shown the undesirable doors, they're eliminated from the equation, thus it's irrelevant how many there were.

Wrong. Look at the following slide that has 50 doors, then see if you still feel the same way.

 

[Pheran]

The thought of me traveling to Pittsburgh to hit you is counterintuitive. Please do me a solid and punch yourself in the prostate.

You can hate on me all you like, it won't change the right answer. Both of you should take a look at the 50 door example, but I'll put it in text.

There are 50 doors, behind them are 49 goats and 1 new car. You choose one door, then I open 48 doors with goats, leaving one unopened door. You still want to keep that first door you chose, which has a 2% chance of being the correct door?

 

[Jeeebus]

You can hate on me all you like, it won't change the right answer. Both of you should take a look at the 50 door example, but I'll put it in text.

There are 50 doors, behind them are 49 goats and 1 new car. You choose one door, then I open 48 doors with goats, leaving one unopened door. You still want to keep that first door you chose, which has a 2% chance of being the correct door?

Yes.

And it doesn't matter if we're doing this 1,000 times, my answer is always going to be yes and my chance of winning is always going to be 50%.

There could be 10,000 options and the odds would always be the same. Because every time we do this exercise, you're going to show me 9,998 doors that are goats. Meaning no matter which door I pick, it's going to come down to my door or one other door.

Your logic is flawed, and your underarms smell too.