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(Some Guy)   That witch Nate Silver was wrong because he was too right   (physicsbuzz.physicscentral.com) divider line 20
    More: Obvious, Nate Silver, weather predictions  
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5660 clicks; posted to Geek » on 09 Nov 2012 at 11:29 AM (1 year ago)   |  Favorite    |   share:  Share on Twitter share via Email Share on Facebook   more»



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2012-11-09 09:38:08 AM  
3 votes:
This guy is making the mistake of thinking the state-by-state probabilities were independent and not covariant,
2012-11-09 02:45:19 PM  
2 votes:
I find this to be the best explanation for those confused:

The choice to switch is really a bet on whether or not you chose correctly at the start. You only had a 33% chance of being correct initially. Therefore, 66% of the time you will pick wrong at the start and switching is the correct answer:

Correct first pick (33% of the time) - switching results in a loss.
Incorrect first pick (66% of the time) - switching results in a win.
2012-11-09 11:39:02 AM  
2 votes:
How nice to see a non-statistician fail so badly at statistics, its as if Nate Silver knows something after studying statistics that this guy doesn't know from not studying statistics.
2012-11-09 06:28:04 PM  
1 votes:

Dr Dreidel: Is it that the simulation is actually done with replacement, and I'm thinking without?

I see it as a new puzzle the second time - 1 or 2? Or: 1, 2, 3, 4...999,997 or 999,998?

I dunno, it's not like the second box changed, either - you just know the one he took away wasn't it.

// or was it?
// was he just farking with us the whole time?!?
// I always knew Monty Hall was a bastard
// I've been refreshing like a maniac - this is killing me
// I'm like: but really, which one of us is the moran? Is it really me, again?
// HAS THE WHOLE WORLD GONE CRAZY?!? AM I THE ONLY ONE WHO GIVES A SHIAT ABOUT THE PROBABILITIES?!?!?!?
// slashy record


Here, maybe this awesome mspaint illustration will help:

www.imgjoe.com 

In this example, you have chosen box one. It doesn't matter where the prize is, really - there were three boxes to start with, so there is a 33% chance your box was correct, a 66% chance it was in one of the other two boxes. So, let's draw a virtual box around the other two boxes and give it a 66% chance.

Monty then eliminates one empty, unchosen box - this, by definition, has to reside within our virtual box that has a 66% chance of holding the prize.

Having eliminated a KNOWN empty box, the one remaining box does not have its probability adjusted. There was a 66% chance of the prize residing in our virtual box before we eliminated the known bad box, and there is STILL a 66% chance of the prize residing inside of our virtual box.
2012-11-09 03:53:59 PM  
1 votes:

China White Tea: I get that the Monty Hall problem is difficult for some people to understand, but I wish they could just accept that they don't understand it instead of adamantly derping about how it's 50/50. At least be smart enough to know what you don't know.


Link  People are terrible at knowing they don't know something.
2012-11-09 02:09:47 PM  
1 votes:

Rent Party:
It is counter-intuitive, but it does matter.

When you make the original choice, your odds of picking the million dollars are 1/3, Statistically you are more likely to be holding an empty box than you are the one with the money. Inversely, 2 out of three times, the money is going to still be behind one of the other doors.

When one of the other options is eliminated, you know which one it is.

This is a "Three Prisoners" problem.


No, not in his example. In his example, you only had 2 choices, not 3. So your chance was 50/50 to begin with. A new box is introduced, shown to be empty, and thus you can't swithc to it has zero effect on the probability.

In the true monty hall problem, you switch. in the alternate one, it won't matter.
2012-11-09 02:09:01 PM  
1 votes:

kertus: Next up?

You have three boxes. One contains $1m
You choose a box
I open one of the remaining boxes and show you it's empty
I offer you a choice - keep your box or take the remaining box
What do you do? Why?

I love poorly understood probability problems


Ha! That was my first thought as well.

Marilyn Vos Savant has a nice write up on this here but my favorite part is the hate mail at the bottom
2012-11-09 02:05:38 PM  
1 votes:

justtray: O [O O]
33% 66%
O [O X]
33% 66% - still

I guess that's the visual way of looking at it. Why its '66%' always confused me.


And if anybody doubts it, they can go play:
http://www.stat.sc.edu/~west/javahtml/LetsMakeaDeal.html

"In 138 games where you have not switched, you have won 48 times"

48/138 = 34.7%
2012-11-09 02:04:25 PM  
1 votes:
O [O O]
33% 66%
O [O X]
33% 66% - still

I guess that's the visual way of looking at it. Why its '66%' always confused me.
2012-11-09 01:57:54 PM  
1 votes:

burndtdan:
Both remaining boxes are equally likely to hold the money.


No, they aren't. One box is 66% of holding it, and the one you chose only has a 33% chance of holding it.
2012-11-09 01:57:19 PM  
1 votes:

kidgenius: justtray: When you originally chose, you had a 33% chance of being correct. When one, incorrect option is taken away, the other remaining option has a 50% chance of holding the prize.

No, the other option has a 66% chance of being correct, not 50%. If it was 50% then switching/not switching doesn't matter.


Yeah, sorry, I get confused on the semantics of it. The point is still correct; that you have a higher chance of winning by switching boxes because they can't take away the winning box.
2012-11-09 01:55:56 PM  
1 votes:

kidgenius: Tyrone Slothrop:
Here's an alternate of that:

You have two boxes. One contains $1m
You choose a box
I put a 3rd, open empty box, next to the other two.
I offer you a choice - keep your box or take the remaining (closed) box
Do you make a different choice than in the 1st situation? Why?

It won't matter. it's 50/50 to switch. I could make either choice and I have an equal chance of being right/wrong.


A lot of people will say you switch boxes because you'd be making a choice based on 50/50 odds, but miss that by choosing not to switch you are also making a choice based on 50/50 odds even if you originally selected it with worse odds.
2012-11-09 01:55:43 PM  
1 votes:

Tyrone Slothrop: kertus: Next up?

You have three boxes. One contains $1m
You choose a box
I open one of the remaining boxes and show you it's empty
I offer you a choice - keep your box or take the remaining box
What do you do? Why?

I love poorly understood probability problems

Here's an alternate of that:

You have two boxes. One contains $1m
You choose a box
I put a 3rd, open empty box, next to the other two.
I offer you a choice - keep your box or take the remaining (closed) box
Do you make a different choice than in the 1st situation? Why?


Doesn't matter, it doesn't change your probability. In this scenario, you still have a 50/50 shot either way.
2012-11-09 01:51:48 PM  
1 votes:

Tyrone Slothrop:
Here's an alternate of that:

You have two boxes. One contains $1m
You choose a box
I put a 3rd, open empty box, next to the other two.
I offer you a choice - keep your box or take the remaining (closed) box
Do you make a different choice than in the 1st situation? Why?


It won't matter. it's 50/50 to switch. I could make either choice and I have an equal chance of being right/wrong.
2012-11-09 01:07:53 PM  
1 votes:
Do you make a different choice than in the 1st situation? Why?
Doesn't matter. I had a 50% chance of picking right and you did not do anything affect my choice and therefore the probabilities. So both have equal probability.

The key to the Monty Hall problem is that the host opens one he knows is a loser. Since there is a 1 in 3 chance you picked the winner, there is a 2 in 3 chance that the winner was in a remaining door. And with a loser removed from the unchosen ones there is a 2/3 chance that the winner is left there (the odds that you didn't pick the winner in the first place).

In that case the host affected the choices and therefore affected the probabilities
2012-11-09 12:28:42 PM  
1 votes:
Let's consult the experts: Is Nate Silver a witch?
2012-11-09 11:53:11 AM  
1 votes:
sparks.brushfireoffreedom.org
2012-11-09 11:47:28 AM  
1 votes:

justtray: Rain is discrete. It either rains or it doesn't.


Is it weird that I understood your words more than I did the ACP guy's? I still don't know what he's banging on about, but your explanation makes me think he didn't think his cunning rebuttal all the way through.
2012-11-09 11:40:03 AM  
1 votes:
Rain is discrete. It either rains or it doesn't.

Who wins a state is based on the number of peopel that vote for either candidate. Whoever has more wins. The polls estimate the number of voters for each side, within a % margin of error.

This analogy is so full of fail. A better analogy would be, "How MUCH is it going to rain in a specific area?" If it rains over 2 inches, then correct, less incorrect.

Using that analogy you would find the weatherman's predictions much like Nate's, would be very accurate.

Working backwards, Florida had the closest race, nearly 50/50. But the total difference of winning and losing was a tiny percentage of the total votes. To make the assertion that Nate should have simply said the chance of Obama winning was a near certainty (100%) shows an entire lack of regard for common sense. There are any number of factors that could have tipped the scale the other way.

So again, it's not "did it rain." While that is semantically correct, we don't decide if it rained based on an amount of rainfall, which is in turn, how we decide electoral results for states.
2012-11-09 09:51:16 AM  
1 votes:
This guy is crying out for someone to pay attention to him.

Somebody send his mommy the link.
 
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