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(Some Guy)   That witch Nate Silver was wrong because he was too right   (physicsbuzz.physicscentral.com) divider line 97
    More: Obvious, Nate Silver, weather predictions  
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5661 clicks; posted to Geek » on 09 Nov 2012 at 11:29 AM (2 years ago)   |  Favorite    |   share:  Share on Twitter share via Email Share on Facebook   more»



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2012-11-09 09:38:08 AM  
This guy is making the mistake of thinking the state-by-state probabilities were independent and not covariant,
 
2012-11-09 09:51:16 AM  
This guy is crying out for someone to pay attention to him.

Somebody send his mommy the link.
 
2012-11-09 09:51:25 AM  
(Where '*' is the symbol for multiplication.)

Thank you, Steve.
 
2012-11-09 10:25:32 AM  
so, because he pitched a one his shut out instead of a perfect game, Nate Silver's a giant loser?
 
2012-11-09 11:32:54 AM  

DamnYankees: This guy is making the mistake of thinking the state-by-state probabilities were independent and not covariant,


His comments on that:

"Some of the bright people here at the American Center for Physics have suggested that I'm overlooking the possibility that individual states' elections can affect each other or may be mutually affected by some outside influence in a way to make them move together. That might be true, but it implies that Silver should sometimes call the entire race perfectly (as he seems to have done again this year) and at other times miss on almost every battleground state, but he would be unlikely to miss by a little. If this is the case, any one state's supposed "Chances of winning" seem to me to be meaningless, and Silver should only be able to predict how blocks of states move together. That would make his model a pretty blunt instrument, despite the fact that most of his fans act like it's the equivalent of a statistical scalpel. In fact, Silver points out in the description of his methodology that such interactions among states are accounted for in the model, which means they're included in the calculations that produce the model's output, including the chances of winning he posted for each race. So this is a pretty plausible explanation, although I don't see how I could check it easily."
 
2012-11-09 11:37:32 AM  

Ambitwistor: DamnYankees: This guy is making the mistake of thinking the state-by-state probabilities were independent and not covariant,

His comments on that:

"Some of the bright people here at the American Center for Physics have suggested that I'm overlooking the possibility that individual states' elections can affect each other or may be mutually affected by some outside influence in a way to make them move together. That might be true, but it implies that Silver should sometimes call the entire race perfectly (as he seems to have done again this year) and at other times miss on almost every battleground state, but he would be unlikely to miss by a little. If this is the case, any one state's supposed "Chances of winning" seem to me to be meaningless, and Silver should only be able to predict how blocks of states move together. That would make his model a pretty blunt instrument, despite the fact that most of his fans act like it's the equivalent of a statistical scalpel. In fact, Silver points out in the description of his methodology that such interactions among states are accounted for in the model, which means they're included in the calculations that produce the model's output, including the chances of winning he posted for each race. So this is a pretty plausible explanation, although I don't see how I could check it easily."


So basically, he answered his own question and he doesn't even know it?
 
2012-11-09 11:39:02 AM  
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 11:40:03 AM  
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 11:47:28 AM  

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 11:48:56 AM  

Ambitwistor: Silver points out in the description of his methodology that such interactions among states are accounted for in the model, which means they're included in the calculations that produce the model's output, including the chances of winning he posted for each race. So this is a pretty plausible explanation, although I don't see how I could check it easily


I don't understand what he's asking Silver to do. If 2 probabilities are covariant, they are covariant. Is it possible to come up with a non-covariant probability? I don't know how you'd do that.
 
2012-11-09 11:53:11 AM  
sparks.brushfireoffreedom.org
 
2012-11-09 11:56:07 AM  
I'm not familiar with the particulars of Silver's model but does it account for the idea of prior probabilities? For instance, if poll after poll shows us that it is reasonable to assume that Obama has a 3% advantage in a state then the odds of him losing the state would be much less if we updated the calculations to account for the idea that the null hypothesis is no longer a 50-50 split, but instead a 53-47 split. A random sample from this larger population on election day would have a much lower chance of producing a Romney win. Not accounting for such things could mean that Silver is actually underestimating the odds.
 
2012-11-09 12:08:29 PM  
This is what happens when some dope on a physics site thinks he knows statistics. You're out of your element, bud.
 
2012-11-09 12:09:49 PM  
From the comments

This is a physics blog right? I would have thought fundamental probability theory is required.

encrypted-tbn2.gstatic.com
 
2012-11-09 12:13:33 PM  
Not sure what the issue is with this article--it's more of a "what-if" mental exercise than a "SILVER IS A WITCH" rant. He acknowledges dependence between states as "a pretty plausible explanation" and doesn't spend much time on it simply because, I would assume, it's just not as interesting as the other things he discusses.
 
2012-11-09 12:23:07 PM  
This should actually be an easy question to answer. My understanding is he runs a kind of Monte Carlo simulation of the election, essentially a thousands of simulated elections, and his confidence levels are based on the number of occurances for each outcome, both for each state and the overall election. It should be a matter of measuring the occurance of every state turning the color it did.

This is also treading dangerously close to the lottery fallacy.
 
2012-11-09 12:28:12 PM  
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
 
2012-11-09 12:28:42 PM  
Let's consult the experts: Is Nate Silver a witch?
 
2012-11-09 12:32:31 PM  

DamnYankees: I don't understand what he's asking Silver to do. If 2 probabilities are covariant, they are covariant. Is it possible to come up with a non-covariant probability? I don't know how you'd do that.


Well, there are tests for the mis-specification of a covariant model. Basically, if the model is wrong in two places that are expected to covary, don't penalize it as much (and vice versa).
 
2012-11-09 12:35:57 PM  

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


i.ytimg.com

I'll take the box!
 
2012-11-09 12:37:50 PM  

Donnchadha: I'll take the box!


Stupid!! You so stupid!!!
 
2012-11-09 12:38:14 PM  
Silver DID underestimate the likelihood of Obama winning, intentionally so. He factored in a percentage chance that the polls were just totally wrong, which wouldn't really change the projections, just tamp down the probability from what they're saying outright. But it turns out the polls weren't all just completely wrong.
 
2012-11-09 12:44:34 PM  
Sam Wang at PEC has some analysis of the performance of his, Silver's, and Simon Jackman's predictions. He claims his predictions were the best calibrated probabilistically, according to the Brier score.
 
2012-11-09 12:51:14 PM  
I do think there is a decent chance that Nate Silver's model is right leaning or too conservative with its predictions. Pretty much all the numbers I have seen not in line with his prediction are to the left. However, I certainly don't have enough information or knowledge to do a proper analysis and say for sure, it just wouldn't surprise me.

But, just to note, based on simulations ran by him, Nate Silver gave the actual outcome of the EC at about 20.5% chance, better than any other outcome (with just Florida flipped at 16.5% and flipping North Carolina at 13%). The only other one over a 5% chance was Obama taking North Carolina and Nebraska 2nd at 5.5%. Under these circumstances, while an unlikely result in and of itself, it is over 4 times more likely than most other possibilities. If Nate got, say, Virginia wrong, it would make some people feel good that he wasn't perfect, but he ends up with an even less likely final result occurring.

Yes, based on the numbers he posted, it was unlikely that he would come up with the perfect electoral map, but as any statistician would say, unlikely does not mean impossible. However it does require a follow-up investigation

I would like to see a more thorough analysis comparing actual results to his projections of the popular vote by state. See if there was a significant trend there or not.
 
2012-11-09 12:51:47 PM  

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?
 
2012-11-09 01:07:53 PM  
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 01:51:48 PM  

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 01:53:07 PM  

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


You always change to the other box.

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.
 
2012-11-09 01:54:36 PM  

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.
 
2012-11-09 01:55:43 PM  

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 01:55:56 PM  

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 01:56:55 PM  

justtray: You always change to the other box.

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.


Both remaining boxes are equally likely to hold the money.
 
2012-11-09 01:57:19 PM  

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 01:57:54 PM  

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 02:00:24 PM  

burndtdan: justtray: You always change to the other box.

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.

Both remaining boxes are equally likely to hold the money.


kidgenius: 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 02:04:25 PM  
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 02:05:32 PM  

dywed88: I do think there is a decent chance that Nate Silver's model is right leaning or too conservative with its predictions. Pretty much all the numbers I have seen not in line with his prediction are to the left. However, I certainly don't have enough information or knowledge to do a proper analysis and say for sure, it just wouldn't surprise me.


I'm sure Silver's model could be further tweaked and improved.

Perhaps he should give less weight to external factors such ("state fundamentals," economic indicators, etc.) when a greater volume of polling data is available.
 
2012-11-09 02:05:38 PM  

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 02:06:02 PM  
tl;dr : Nate's one of the Popular Girls and I'm going to show why he's a poopyhead by bludgeoning you with my Mighty Brain.
 
2012-11-09 02:07:50 PM  
The probability that it rained every day in this hypothetical week is

P= 0.846 * 0.794 * 0.843*0.906*0.797*0.967*0.503


Says who?
 
2012-11-09 02:08:03 PM  

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.


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.
 
2012-11-09 02:09:01 PM  

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 02:09:47 PM  

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 02:11:42 PM  

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?


The mathematical trick lies in the option to switch -- not to re-pick a random box.

In the three door puzzle, you only lose after switching if you originally picked the prize and you win after switching if you originally picked an empty box. Since the odds of picking an empty box the first time is 2/3 (or 66%), then your odds of winning after switching is 2/3 (or 66%).

In your variant, the odds of picking the correct box initially is 50/50 because there are only two boxes to choose from. The added third box does nothing to the odds of the original choice.
 
2012-11-09 02:15:16 PM  

kidgenius: 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.


Two options was the "alternate" and in that scenario you are correct. I hadn't read enough upstream to realize that was to what your response was.

This is what I was referring to.

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?
 
2012-11-09 02:23:20 PM  

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?



You putting an open empty box next to the other one does nothing... why would that even be part of a game?
 
2012-11-09 02:34:47 PM  

Rent Party: This is what I was referring to.

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?


Classic Monty Hall problem - always switch.
 
2012-11-09 02:45:19 PM  
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 03:33:10 PM  
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.
 
2012-11-09 03:35:33 PM  

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.


That's the point of it, though. It's simple enough to present to people that it is easily understood, even though the answer isn't intuitive.

It is subject to truthieness, because it appears simple.
 
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