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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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5658 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
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.
 
2012-11-09 03:46:07 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.


It's easy. Start with a million boxes. Take away 999,998 of them, showing the ones you took away were wrong. Human intuition understands that if you pick 1 in a 1,000,000, you are unlikely to have picked the right one. So it makes a lot of sense to switch to the other one (which is almost certainly holding it, since you know the others didn't).

It only becomes non-intuitive when we leave it the low numbers.
 
2012-11-09 03:53:59 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.


Link  People are terrible at knowing they don't know something.
 
2012-11-09 04:07:47 PM

Dogmatrix: Let's consult the experts: Is Nate Silver a witch?


The only way to make that site funnier would be to give a Nate-Silver-esque number for how likely he is a witch.

Otherwise, lulz.
 
2012-11-09 04:09:59 PM

boredgrad: 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.


Morons are terrible at knowing they don't know something. Intelligent people are painfully aware of their own ignorance.
 
2012-11-09 04:16:05 PM

FitzShivering: 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.

It's easy. Start with a million boxes. Take away 999,998 of them, showing the ones you took away were wrong. Human intuition understands that if you pick 1 in a 1,000,000, you are unlikely to have picked the right one. So it makes a lot of sense to switch to the other one (which is almost certainly holding it, since you know the others didn't).

It only becomes non-intuitive when we leave it the low numbers.


I am sure an extreme case would help some people, it won't solve the main issue. A lot of people will still look at that problem in the end and say, "There are two there, whichever I pick has a 50/50 shot at being right."

People simply have a lot of trouble understanding probabilities and statistics (I admit I have trouble doing the stuff myself), which is what brings the enormous backlash against them as seen with the reaction to prediction models such as FiveThiryEight the past few months or for years from "traditionalists" in baseball.
 
2012-11-09 04:25:03 PM

dywed88:

It only becomes non-intuitive when we leave it the low numbers.

I am sure an extreme case would help some people, it won't solve the main issue. A lot of people will still look at that problem in the end and say, "There are two there, whichever I pick has a 50/50 shot at being right."



That would be the intuitive answer, though, which is the point. It doesn't make people dumb, it just makes them unfamiliar with probabilities, or as I like to call them, "normal."

from "traditionalists" in baseball.

Yeah? Fark you and your "designated hitter," you damn hippy.
 
2012-11-09 04:28:11 PM

dywed88: FitzShivering: 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.

It's easy. Start with a million boxes. Take away 999,998 of them, showing the ones you took away were wrong. Human intuition understands that if you pick 1 in a 1,000,000, you are unlikely to have picked the right one. So it makes a lot of sense to switch to the other one (which is almost certainly holding it, since you know the others didn't).

It only becomes non-intuitive when we leave it the low numbers.

I am sure an extreme case would help some people, it won't solve the main issue. A lot of people will still look at that problem in the end and say, "There are two there, whichever I pick has a 50/50 shot at being right."

People simply have a lot of trouble understanding probabilities and statistics (I admit I have trouble doing the stuff myself), which is what brings the enormous backlash against them as seen with the reaction to prediction models such as FiveThiryEight the past few months or for years from "traditionalists" in baseball.


I've discussed the Monty Hall problems with hundreds of people. Universally, people get it wrong when it's three, and argue to the death. Present it as a million, and they universally get it immediately.

But I agree with you 100% in general about peoples' normal troubles with probabilities and statistics.

/I ranted about the Monty Hall problem for 4 hours the first time it was presented to me
//was actually sort of right since the person posed the problem incorrectly, but still, that's mostly a cover for how outraged my brain was at the non-intuitiveness of it
 
2012-11-09 04:34:23 PM

Rent Party: That would be the intuitive answer, though, which is the point. It doesn't make people dumb, it just makes them unfamiliar with probabilities, or as I like to call them, "normal."


Read this statement carefully, for it is the death of our future if we cannot change it.
 
2012-11-09 04:40:05 PM

Khellendros: Rent Party: That would be the intuitive answer, though, which is the point. It doesn't make people dumb, it just makes them unfamiliar with probabilities, or as I like to call them, "normal."

Read this statement carefully, for it is the death of our future if we cannot change it.


We have managed to muddle through most of history with most people being bad at math. I doubt very much that this portends anything other than we will manage to muddle through the rest of it.
 
2012-11-09 04:47:11 PM

FitzShivering: A lot of people will still look at that problem in the end and say, "There are two there, whichever I pick has a 50/50 shot at being right."


Forgive me if I'm quoting the wrong person. Also forgive me for my ignorance, but why is this incorrect?

You were shown 3 boxes.
You pick one, which is irrelevant (like other players' cards in a hand of poker - they're just as random as the next card in the deck) since you can't see it, and its contents are the whole point of the exercise. Moving on...
One is taken away.
You're left with a choice between 2 boxes - the one you first chose and the third one. How is this not 50/50?

Or am I misunderstanding, and most people would switch, thinking they now had a 2/3 shot (which someone else may have to explain to me. That is a very silly leap of logic)?
 
2012-11-09 05:00:12 PM

Dr Dreidel: FitzShivering: A lot of people will still look at that problem in the end and say, "There are two there, whichever I pick has a 50/50 shot at being right."

Forgive me if I'm quoting the wrong person. Also forgive me for my ignorance, but why is this incorrect?

You were shown 3 boxes.
You pick one, which is irrelevant (like other players' cards in a hand of poker - they're just as random as the next card in the deck) since you can't see it, and its contents are the whole point of the exercise. Moving on...
One is taken away.
You're left with a choice between 2 boxes - the one you first chose and the third one. How is this not 50/50?

Or am I misunderstanding, and most people would switch, thinking they now had a 2/3 shot (which someone else may have to explain to me. That is a very silly leap of logic)?


Because when you first picked a box, the odds were that the money was behind one of the boxes you didn't pick. You had a 33.3% chance of getting right, and a 66.6% chance of getting it wrong.

Now remove one of the two boxes. By hanging onto the box you originally picked, you are *still* going with the 33.3% shot. The odd were that the money was in one of the other two boxes, and now one of them has been removed. You take that box, and you have a 66.6% chance of getting the right one.

I like the idea of explaining that with a million boxes. It becomes intuitive then.
 
2012-11-09 05:08:35 PM

Dr Dreidel: Or am I misunderstanding, and most people would switch, thinking they now had a 2/3 shot (which someone else may have to explain to me. That is a very silly leap of logic)?


A little from column A a little from Column B

You are shown 3 boxes - One wins, Two lose
You pick one.
From the two you did not choose, the host shows you one he knows to be a loser.
You are now given the option to choose to keep yours or pick the third,

Now, the probability side of things:

When you pick the first box, there is a 1 in 3 chance that it was the winner. That means there is a 2 in 3 chance the winner was not selected.
If you picked a winner (remember this happens 1 in 3 times), then there are two loser boxes and it doesn't matter which is removed, you lose by switching.
However if you picked a loser (which of pretty straightforward to understand occurred 2 of 3 times) the host must remove the other loser and the remaining box must be the winner. Switching wins.

Because the host is showing you a loser you basically have the option of picking one box or two. And I think everyone understands picking two gives you better odds of winning.

Most people actually choose to stick with their first because the probabilities are immediately apparent and people feel committed to their choice.
 
2012-11-09 05:09:27 PM

Dr Dreidel: FitzShivering: A lot of people will still look at that problem in the end and say, "There are two there, whichever I pick has a 50/50 shot at being right."

Forgive me if I'm quoting the wrong person. Also forgive me for my ignorance, but why is this incorrect?

You were shown 3 boxes.
You pick one, which is irrelevant (like other players' cards in a hand of poker - they're just as random as the next card in the deck) since you can't see it, and its contents are the whole point of the exercise. Moving on...
One is taken away.
You're left with a choice between 2 boxes - the one you first chose and the third one. How is this not 50/50?

Or am I misunderstanding, and most people would switch, thinking they now had a 2/3 shot (which someone else may have to explain to me. That is a very silly leap of logic)?


Just because you have two choices, and the outcomes are random, doesn't mean that the odds are 50/50. In this case, you had a 33% chance of picking correctly. That doesn't change just because you are shown one empty box.

See a few posts up regarding 100000 boxes. It's more intuitive.
 
2012-11-09 05:12:32 PM

Rent Party: Dr Dreidel: FitzShivering: A lot of people will still look at that problem in the end and say, "There are two there, whichever I pick has a 50/50 shot at being right."

Forgive me if I'm quoting the wrong person. Also forgive me for my ignorance, but why is this incorrect?

You were shown 3 boxes.
You pick one, which is irrelevant (like other players' cards in a hand of poker - they're just as random as the next card in the deck) since you can't see it, and its contents are the whole point of the exercise. Moving on...
One is taken away.
You're left with a choice between 2 boxes - the one you first chose and the third one. How is this not 50/50?

Or am I misunderstanding, and most people would switch, thinking they now had a 2/3 shot (which someone else may have to explain to me. That is a very silly leap of logic)?

Because when you first picked a box, the odds were that the money was behind one of the boxes you didn't pick. You had a 33.3% chance of getting right, and a 66.6% chance of getting it wrong.


But the only way to know if you were right or wrong is to open it, right?

Now remove one of the two boxes. By hanging onto the box you originally picked, you are *still* going with the 33.3% shot. The odd were that the money was in one of the other two boxes, and now one of them has been removed. You take that box, and you have a 66.6% chance of getting the right one.

I like the idea of explaining that with a million boxes. It becomes intuitive then.


I could rephrase my question as - "If you know he's going to remove an empty box after you make your choice, shouldn't you know it's 50/50 (the box you choose, plus the one he doesn't open) going in, rather than thinking of it as a 1-in-3 shot?"

The box you chose could have been the right one, but you know he's always going to open an empty one regardless. You have a 50/50 shot.

// I am good with computer and teh maffs
// how am I not getting this?
 
2012-11-09 05:15:35 PM

iSancho2k: See a few posts up regarding 100000 boxes. It's more intuitive.


Not to me. That example does nothing - unless, of course, I'm still missing something.

dywed88: And I think everyone understands picking two gives you better odds of winning.


But you only get to open one. How does it matter if the right one passed through your hands, but you switched it out?
 
2012-11-09 05:25:10 PM

Dr Dreidel:
Because when you first picked a box, the odds were that the money was behind one of the boxes you didn't pick. You had a 33.3% chance of getting right, and a 66.6% chance of getting it wrong.

But the only way to know if you were right or wrong is to open it, right?



Well, that would settle the matter to 100% certainty, but we are here just playing the odds.


I could rephrase my question as - "If you know he's going to remove an empty box after you make your choice, shouldn't you know it's 50/50 (the box you choose, plus the one he doesn't open) going in, rather than thinking of it as a 1-in-3 shot?"


No. You are picking one of three, so the odd of getting it right are 1:3, even if you know that a leftover box is going away. The odds of the right box being left on the table are 2:3. That is the stat that matters. It is more likely that you initially picked wrong than it is that you initially picked right.

In this scenario, the odds *never* become 50/50.


The box you chose could have been the right one, but you know he's always going to open an empty one regardless. You have a 50/50 shot.


Because you don't have a 50/50 shot if it one box goes away. The odds were 2:3 that one of the boxes on the table contained the money, and that doesn't change just because one of them went away. What does go away is the another bad choice, and that's the non-intuitive aspect of things.

The box you picked was going to be wrong most of the time. The boxes on the table were going to be right most of the time. One of the boxes on the table just went away, but the odds of the right box being on the table didn't change at all. It makes sense to pick that one instead of the one in your hands.
 
2012-11-09 05:28:20 PM

Dr Dreidel: iSancho2k: See a few posts up regarding 100000 boxes. It's more intuitive.

Not to me. That example does nothing - unless, of course, I'm still missing something.


Lets do the million examples then.

There are one meeeelion boxes on a table, one of which contains the prize.

You select one box. The odds of you selecting the right box are incredibly small.

I now remove 999,998 of the boxes, leaving one on the table.

Do you change out for the leftover box, or do you stick with your million to one shot?
 
2012-11-09 05:31:59 PM
The answer to the Monty Hall problem -- which I haven't seen mentioned, but forgive me I missed it -- is "You don't really know whether to switch or not, because you don't know the rules that Monty is using."

Scenario 1: The One The People Saying To Switch Have In Mind

In this scenario, you pick one of the three boxes. You have a 1/3 chance of being right.
Monty (or the producer) knows the box with the prize, and will always reveal and empty box.
If you chose the right box (a 1/3 chance), Monty reveals one of the two empty boxes (it doesn't matter which). If you switch, you lose.
If you chose the wrong box (a 2/3 chance), Monty reveals the only empty box you didn't pick. If you switch, you win.
Therefore, since switching is right if your initial choice was wrong, and your initial choice was probably wrong, you should switch.

Scenario 2: Clueless Monty

In this scenario, you pick one of the three boxes. You have a 1/3 chance of being right.
Monty has no idea what box has the prize, and will reveal one of the two boxes you didn't pick, choosing at random.
Three things can happen.
* 1. You chose the right box, and Monty reveals one of the two empty boxes (it doesn't matter which). The chance of this case is 1/3 (your chance of choosing the right box). If you switch, you lose.
* 2. You chose a wrong box, and Monty reveals the other empty box. The chance of this case is 2/3 (your chance of choosing a wrong box), times 1/2 (Monty's chance of choosing the other empty box), or 1/3. If you switch, you win.
* 3. You chose a wrong box, and Monty reveals the box with the prize. The chance of this case is 2/3 (your chance of choosing a wrong box), times 1/2 (Monty's chance of choosing the right box), or 1/3. In this case, Monty says "So sorry, you lost," and sends you home with Turtle Wax, and you never arrive at the state described in the problem.
So in this scenario, if you do get to point where Monty offers the choice, it means that you're either in case 1 (in which case you lose if you switch) or in case 2 (in which case you win if you switch). The two cases are equally likely, so your odds are even -- do whatever makes you happy.

Scenario 3: Evil Monty

In this case, the show is over its prize budget, and Monty is out to shaft you. He knows what box has the prize, and will use that against you.
If you choose the wrong box, Monty will open the box you chose, reveal nothing, open the other box to reveal the prize, say "So sorry, you lost," and send you home.
If you choose the right box, Monty will reveal one of the two empty boxes, and offer you the opportunity to switch. If you switch, you lose.
So in this scenario, you should always stick with your choice.

Since the original problem didn't say what rules Monty is using, we don't know which of these scenarios (or any number of other possible scenarios) is true, so we don't know the right answer. Some people tend to subconsciously assume that it's Scenario 1, and others that it's Scenario 2; that, combined with the fact that we're bad at probability, leads to lots of arguments.
 
2012-11-09 05:36:19 PM

Rent Party: The box you picked was going to be wrong most of the time. The boxes on the table were going to be right most of the time. One of the boxes on the table just went away, but the odds of the right box being on the table didn't change at all. It makes sense to pick that one instead of the one in your hands.


In that case, aren't you faced with a 50/50 choice? He's removed one, now you have a choice between 2: the one you initially had (and just because the odds of it being right were 1/3 before, doesn't mean you have any special new info about it now), and the one he didn't remove.

Rent Party: Do you change out for the leftover box, or do you stick with your million to one shot?


Even so, the box you first picked is every bit as likely - in both simulations - as any other one to have the ONE HUNDRED JILLION FAFILLION SHABADOOBADEE-illion...yen inside. Picking another one doesn't change that.

It's not 50/50 (or 1/2), though. It'd be 1/999,998. For both the box you're holding, and each one in the gigantic stack next to you.
 
2012-11-09 05:38:08 PM

KickahaOta: The answer to the Monty Hall problem -- which I haven't seen mentioned, but forgive me I missed it -- is "You don't really know whether to switch or not, because you don't know the rules that Monty is using."

Scenario 1: The One The People Saying To Switch Have In Mind


Scenario 2: Clueless Monty

Scenario 3: Evli Monty


Scenario 3 isn't really the same, though, as if you select wrong (which you will most of the time) you have no opportunity to switch.
 
2012-11-09 05:44:36 PM

Dr Dreidel:
In that case, aren't you faced with a 50/50 choice? He's removed one, now you have a choice between 2: the one you initially had (and just because the odds of it being right were 1/3 before, doesn't mean you have any special new info about it now), and the one he didn't remove.


You are almost there. You don't need any new information to change the odds. All you need to do is know what the odds are.

You are left with a choice of two, but it is not a choice between two equally likely outcomes. Outcome 1 - the prize being in your hand is a 1:3 chance. Outcome 2- The prize being on the table is a 2:3 chance. Removing one of the boxes on the table doesn't change that. The odds of the prize being on the table are still 2:3, even if there is only one box there. That is why you switch boxes. Because you were most likely wrong when you picked yours, and still are even though one of the other's has been taken away.



Rent Party: Do you change out for the leftover box, or do you stick with your million to one shot?

Even so, the box you first picked is every bit as likely - in both simulations - as any other one to have the ONE HUNDRED JILLION FAFILLION SHABADOOBADEE-illion...yen inside. Picking another one doesn't change that.

It's not 50/50 (or 1/2), though. It'd be 1/999,998. For both the box you're holding, and each one in the gigantic stack next to you.


You're getting there. It isn't 50/50. It is "odds of me picking right originally
If I remove all of the boxes on the table except one, that gives you higher odds than the one you have in your hand.
 
2012-11-09 05:50:47 PM

kidgenius: 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%


"In 138 games in which you have switched, you have won 90 times"

90/138 = 65%

Just thought I'd run it through the other way to help support the point.
 
2012-11-09 05:53:33 PM

Rent Party: You're getting there.


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
 
2012-11-09 05:55:51 PM

Rent Party: Scenario 3: Evli Monty

Scenario 3 isn't really the same, though, as if you select wrong (which you will most of the time) you have no opportunity to switch.


That's the same for Scenario 2 as well. But the problem just says "You have a chance to switch"; it doesn't say "You'll have had a chance to switch in all plays of this game." That's an unspoken assumption, and unspoken assumptions are the bane of probability problems, because they're not necessarily true. The author of TFA made just that sort of mistake by making the assumption that the outcomes of different elections are independent.
 
2012-11-09 06:10:04 PM
Obviously, Silver was being (sunglasses) conservative. (Yeaaaaaaaaaaaaaaaaa)
 
2012-11-09 06:15:43 PM
Dr Dreidel

Another way to look at it:
asking you to switch is like asking you if you want to stick with your initial one-in-a-million pick or if you prefer to have the other 999,999 boxes.
There are _always_ at least 999,998 wrong boxes among the 999,999 you didn't pick.
That someone opens those guaranteed-to-be-empty 999,998 boxes before asking you if you want to stick with your choice doesn't change the fact that he's essentially asking you if you prefer having your initially selected box or the other 999,999. 

It's like a game of roulette where you place a bet on the number 17 and the casino then asks you: "would you prefer to play a game where you win 36 times your bet if the ball lands on 17 or would you like a game where you win 36 times your bet if the ball lands on any number but 17?".
 
2012-11-09 06:24:11 PM

The Voice of Doom: It's like a game of roulette where you place a bet on the number 17 and the casino then asks you: "would you prefer to play a game where you win 36 times your bet if the ball lands on 17 or would you like a game where you win 36 times your bet if the ball lands on any number but 17?".


Um, you don't get to open the rest of the boxes, do you? (Unless it's the 1-in-3 scenario. Then, you're left with the box he took away and the box you don't choose - technically, you opened "the remaining box.") I don't know at which point you're responding (sorry), but the box you're still holding has the same 1-in-whatever chance as any other in the second run.

So stick with it, pick another - even if you do a little dance, your odds are no better or worse. That's what I'm saying.

// please tell me I'm not still wrong
 
2012-11-09 06:28:04 PM

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 06:32:04 PM
And if that doesn't work, get out a pencil and paper and just do it by exhaustive search. Since there are only 3 doors, it is not difficult to enumerate all of the possibilities - there are only 18. (3 prize locations * 3 initial guess locations * 2 playstyles (switching or not-switching).

You will immediately see that if you don't switch, you win exactly 3 of the 9 games. If you do switch, you win exactly 6 of the 9 games.
 
2012-11-09 06:38:12 PM

China White Tea: 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.


I get that explanation - really, I do.

Except...the laws of physics still apply, and there are still 2 boxes left. Both had a 33% chance of being right initially. The flaw I find (and again, this may be my tiny man-brain too busy focusing on football instead of the squigglies what look like letters but ain't) is the part where "This value is still 66[%]!".

...is it? Like I said, you still have 2 boxes left for the second run, and I don't think it's fair to the lesbian sister of biology that you lump Boxes 2 and 3 together in the second run. Box 2 has been removed, and while it and 3's collective percentage was 66 in the first run, 3 isn't in the second run. I may as well counter that Box 1 has a 50% shot and 2 + 3 together have 50% - you know Monty's going to take a box away regardless of which one you choose.

// I joke, but I really want to understand this one
// my mind is a raging torrent, etc
// am I too smart in jumping to the second run before the first is done?
 
2012-11-09 06:41:03 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 kick your ass, take the $1m and run.

I love poorly worded polling questions.
 
2012-11-09 06:52:59 PM

Dr Dreidel: China White Tea: 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.

I get that explanation - really, I do.

Except...the laws of physics still apply, and there are still 2 boxes left. Both had a 33% chance of being right initially. The flaw I find (and again, this may be my tiny man-brain too busy focusing on football instead of the squigglies what look like letters but ain't) is the part where "This value is still 66[%]!".

...is it?


Yes. It is. Again, just jot out the exhaustive distribution of possible games and you will see that if you switch, you win 2/3, if you don't, you win 1/3.
 
2012-11-09 06:57:57 PM
Alternatively: If you DO switch, the ONLY time you will lose the game is if your FIRST pick was correct.

EVERY other time you win. Since you understand that your initial choice had a 33% chance of being correct, what does that leave us with?
 
2012-11-09 06:58:53 PM
But I want the goat, Monty.
 
2012-11-09 07:47:33 PM

China White Tea: Again, just jot out the exhaustive distribution of possible games and you will see that if you switch, you win 2/3, if you don't, you win 1/3.


This may be hours after you're gone, but dammit, I'm gonna do it. This is real-time.

("your" pick in the first run is starred, in the second is bold; I'll strike the one Monty pulled for the second run)
1* 2 3
1* 2 3
1 2* 3
1 2* 3

1* 2 3
1* 2 3
1
2 3*
1 2 3*

1 2* 3
1 2* 3
1 2 3*
1 2 3*

It doesn't matter which is the "right" pick, it looks like (unless I forgot something, but I've spent more time on this than pretty much anything else today. I hope I didn't fark this up royal. Now how did I fark this up royal?) you switch (lines 2, 3, 6 and 7) half the time, and you stay put (1, 4, 5, 8) half the time.

If Box 1 is the correct choice, scenario 3 doesn't apply (we assume they're not removing the winner). You win (1, 3, 5, 7) half the time, and you lose (2, 4, 6, 8) half the time. The same is true for the other real combinations.

So where did I fark up royal? 

// I know I did, and someone's gonna make me look like a fool
 
2012-11-09 07:57:07 PM
China White Tea
Alternatively: If you DO switch, the ONLY time you will lose the game is if your FIRST pick was correct.


That's why I said that the choice boils down to "your box or all of the others":
you don't chose between two boxes, you chose between two sets of boxes:
one set contains your initial pick, the other set contains all the other boxes.
With n boxes your set with just a single element has a chance of 1:n that it contains the winner while the other set with its n-1 boxes has a chance of (n-1):n of containing the winner.
You know that at least all but one of the boxes in the other set with its (n-1):n chance must be empty, i.e. that there are at least n-2 empty boxes in that set with its (n-1):n winning odds.

Now some person - who knows which boxes are empty(!) - comes along and opens n-2 empty boxes in that set.
Since the guy does NOT open the boxes in that set at random, this does fark all for the odds of the set.
The odds of that set with its 1 unopened and n-2 opened, empty boxes to contain the winner are still (n-1):n .
Explicitly pointing out some of the empty boxes in that set doesn't change the odds..

Only thing is:
now that there's only one unopened box left in that set.
Which means the odds of (n-1):n of the winner being in that set "transfer" to that unopened box:
You know there are (n-1):n odds of the winner being in that set and you also know that all the other boxes in that set are empty.
 
2012-11-09 08:04:46 PM

Dr Dreidel: Box 1 has a 50% shot and 2 + 3 together have 50% - you know Monty's going to take a box away regardless of which one you choose.


But his choice is not independent of the outcome -- he only eliminates losing outcomes, not a random outcome. Therefore Monty's action removes some uncertainty from the game compared to the initial state.
 
2012-11-09 08:13:38 PM

Dr Dreidel: So where did I fark up royal?


The critical component: the host cannot eliminate the winner. He knowingly reveals a loser

The order of the boxes don't matter, so lets say you pick a box and that is put in the first slot, this cuts it down to 3 possible permutations. Then Monty eliminates a loser. Winner is in bold, eliminated is struck out, your choice is 1 and switch leaves you with the remaining 2 or 3.

If 1 is the winner

1 2 3
Win if you keep

If 2 is the winner

1 2 3
Lose if keep

If 3 is the winner

1 2 3
Lose if keep
 
2012-11-09 08:14:57 PM
The Wikipedia article on this has come very clear reasoning:

http://en.wikipedia.org/wiki/Monty_Hall_problem#Simple_Solutions
 
2012-11-09 08:18:48 PM

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."


The thing of it is, it's very easy to check on it easily, the formula is public. Silver didn't hide any of it, several of the state by state factors are actually national economic indices. It's obscenely easy if you understand stats and aren't trying to bend over backwards to call an outstanding model into question because you believe the r^2 is simply too high.
 
2012-11-09 08:21:05 PM
If you have a deck of cards, you can easily test it out. Let the Ace of Spades be the winner. Pull one random card out of the deck. Go through the deck and eliminate the first 50 cards that aren't the Ace of Spades. If the Ace of Spades is the remaining card in the deck switch wins, else don't switch wins. Do this a number of times and see how often you would win by switching.

I purposely use the 52 card deck because the more extreme variant results in a more extreme and obvious results, you can do the same by using just the Ace of Spades and 2 other cards. to show the 3 box example on its own.

There is actually a Mythbusters episode where they manual demonstrate this problem.
 
2012-11-09 08:41:56 PM
I'd like to see a statistical breakdown of Silver's model's accuracy. I think you could get away with applying Bayes Theorem, describing the probability that the model is correct given the outcome as a function of the probability of the outcome given the model is correct. You could apply verification stats from the last three elections to determine the priors.

I like that the model predicted FL as a 50/50 toss-up. Not only does the model predict the easy states, it predicts the hard states and effectively predicts how hard it's going to be to call those states. His system is simple but light-years ahead of the next guy.
 
2012-11-09 09:32:25 PM

LouDobbsAwaaaay: I'd like to see a statistical breakdown of Silver's model's accuracy. I think you could get away with applying Bayes Theorem, describing the probability that the model is correct given the outcome as a function of the probability of the outcome given the model is correct. You could apply verification stats from the last three elections to determine the priors.

I like that the model predicted FL as a 50/50 toss-up. Not only does the model predict the easy states, it predicts the hard states and effectively predicts how hard it's going to be to call those states. His system is simple but light-years ahead of the next guy.


Wouldn't you have to know the degree of dependence in his state-by-state probabilities? I guess you could get a rough estimate by looking at the overall election percentages if you were good at math.
 
2012-11-09 10:24:16 PM

LouDobbsAwaaaay: I'd like to see a statistical breakdown of Silver's model's accuracy. I think you could get away with applying Bayes Theorem, describing the probability that the model is correct given the outcome as a function of the probability of the outcome given the model is correct. You could apply verification stats from the last three elections to determine the priors.

I like that the model predicted FL as a 50/50 toss-up. Not only does the model predict the easy states, it predicts the hard states and effectively predicts how hard it's going to be to call those states. His system is simple but light-years ahead of the next guy.


My best advice, watch his blog and the PEC over the coming weeks.

Wang at the PEC already did some analysis of his and Nate Silver's results. His preliminary conclusion was that Nate Silver's use of "fundamentals" skewed his predictions and made them less accurate than that at the PEC. Although the PEC had Florida leaning Romney 60-80%, but he had higher odds for both parties in most states (and something like 98% chance of Obama winning). In the Senate the trend continued with PEC generally having higher percentages, however he got Montana and North Dakota correct.
 
2012-11-09 11:13:56 PM
The Mony Hall problem is easier to understand with 4+ boxes.

Take a series of n boxes. One holds the money. You choose one at random. Your odds of having selected the correct box is 1/n. If you repeat this experiment n times, the expected outcome of having chosen the correct box should be exactly 1 of those times, and the incorrect box (n-1) times, right off the bat. (you can prove this probability by simply ending the game here.)

The host then stands in front of one of the other n-1 boxes, this is noe the host box. All of the remaining n-2 boxes are then revealed showing them as empty. You are then offered a choice of keeping your original box or trading for the host's.

Now keep in mind, the game rules prohibit the host from showing a full box, and the host knows which box that is. Therefore, if you did not choose the winning box at the start, the host has no choice, per the game rules, but to go stand in front of the actual winning box. To do it oherwise would be revealing the winning box, which would violate the game rules.

If you repeat this experiment enough times, therefore, the following probability table is generated:

1/n - correct box is chosen at start. Host's box is wrong. (25% chance for 4 boxes this state is true)

(n-1)/n - an incorrect box is chosen at start. The host's box must now be correct. (75% chance for 4 boxes that this state is true)

Therefore, you are always more likely to be in the "good to switch" state than in the "bad to switch" state. Thus, one should always switch. QED.
 
2012-11-10 12:32:43 AM
Sorry, the page you were looking for in this blog does not exist.



Looks like the blogger changed his mind...
 
2012-11-10 02:12:26 PM

bemused outsider: Sorry, the page you were looking for in this blog does not exist.



Looks like the blogger changed his mind...


He saw the comments had turned into a bunch of arguing over some decades-old game show that nobody understood the premise behind and killed the article in dismay.

I mean, all this talk of boxes being opened or closed and nary a cat in sight. How is a phycisist supposed to relate to that?

The Monty Hall example should be re-imagined as two dead cats and one live cat. If you don't find the live cat, it dies of radiation poisoning.

Now both internet people and physicists have something relateable in the problem. Too late to save his blog though :(
 
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