I couldn’t believe the problems I had explaining the three door problem to a bunch of educated, computer-literate folk. A lot of them couldn’t accept that the opened door was a red herring.
The problem needs to be stated correctly, as it was in the article, or it’s not really a math problem. In real life, Monty didn’t always open a door, which means that whether switching is correct or not becomes an empirical issue (although you should still switch if your ONLY evidence of the likelihood a door will be opened is your own experience).
In case it wasn’t clear in the above, many internet wiseguys who post the problem leave out that Monty is required to open a door. So then THEY are the fools.
The opened door isn’t a red herring, it’s essential. It transforms the choice to picking between your original choice at the original 2-to-1 odds and picking the field. (Of course, this depends on the original disposition being random and Monty being obligated to open a door, as CaptBackslap says.)
It’s a red herring because the problem is precisely equivalent to giving the player a choice of keeping his door, or choosing both other doors (without opening one), which of course makes the answer obvious.
I think a lot of the trouble with getting people to see it is that they have enough knowledge to be dangerous. They don’t even consider that they might be wrong.
It’s easier for people to understand if you describe a version with a million doors instead of three, and then gradually taking away all of the doors except for the one you originally picked and one other. Do you really think that you got the right door out of a million on the first try?
The frustrating thing about Politico et al is that it is very, very easy to think of a critique of Silver that holds up.
To wit:
“Silver, and those who use his numbers, are relying on the innumeracy of Americans to report a matter of statistics as though it were a matter of fact; by ‘predicting’ a 3 in 4 chance of victory for Obama they subtly want Americans to see Obama being the winning side as a fait accompli, lowering the motivation of potential Romney voters and making it more likely that undecideds will break for Obama.”
Bullshit, of course, but a lot closer to a relevant critique (not to mention being more intellectually rigorous) than what they’re peddling.
No, this isn’t much better. “Matter of fact” and “statistics” are not mutually exclusive categories. There can be statistical, as well as probabilistic, facts. It’s a matter of fact that a fair coin flip has a 50% chance of turning up heads.
Of all the things I’ve read and heard over the last two weeks, nothing gives me more optimism for an Obama victory than the entire right-wing world’s pitchforks & torches attack on a guy who is pretty much just aggregating numbers using conventional methods.
This emotional approach is not incompatible with a statistical approach. It can be used to describe the causes and effects of the phenomena that are measured statistically.
The best way to explain the Monty Haul problem is to expand the number of choices. I always use a deck of cards, putting all 52 face down and asking the player to pick the ace of spades. He picks one, I turn over 50 cards showing that they’re not the ace of spades, leaving two face-down cards – the one he picked and the one remaining card I hadn’t turned over. Then I ask him if he wants to swap cards. If he says yes, I look confused and ask why – isn’t it a 50-50 choice? Usually the light goes on and they understand it’s not a 50-50 choice, and that let’s them see that the Monty Haul problem isn’t 50-50 either.
Any AM radio call-in sports show has more reference to empirical evidence and far more presumed understanding of probability, the scientific method, etc. than people who are paid to talk about politics for a living. It constantly amazes me. A lot of sports fans aren’t interested in advanced metrics, or don’t understand them, but they still fundamentally understand what probability *is*.
Nobody thinks that a WAR of 9 means a player is going to hit exactly 9 walk-off home runs in a year. Nobody thinks the bookies are wrong if a team wins but fails to cover the spread. Nobody thinks that because Kevin Durant averages 32 points a game, any time he scores more or less than that it proves that mathematics is fundamentally flawed and/or John Hollinger is a homosexual.
I wish political conversations in this country could be 1/10th as empirically and statistically driven as two drunks at the sports bar arguing the Manning brothers. Nobody’s like “Pythagorean wins are unamerican, real Americans look at PyBUSHian wins” or some crap.
But what really fucking chaps my hide about this Nate Silver backlash is that it seems like you should know, even if you can’t do grade school arithmetic, that to refute somebody’s mathematical argument would require, you know, some math. Like, some squiggly Greek characters and stuff? Maybe a graph or a chart or something?
If this were about global warming or wind power or stem cells or whatever, they’d be sure to dig up something at least pseudoscience-y. But actual math is so far beyond their experience they have nothing to fall back on but booger-flinging rage. (Did you know that Andrew Wiles solved Fermat’s last theorem by showing that Fermat was a wicked big quee-ah who had hands like a little bitch? It’s true, he did!)
As a result, these anti-Silver articles read like somebody who doesn’t speak any French phonetically transliterating something into English, and therefore deciding that French isn’t actually a real language because of that. “Comb oh tally view”? I don’t know what a tally view is, but I’m pretty sure you can’t comb it!!! LOL french is stupid!
But Mitt’s got momentum! Or I thought he did, but now with the new polls, it’s starting to look like maybe Obama’s taken the momentum back. It’s probably because he’s looking so Presidential in the aftermath of the hurricane.
Anyway, anyone who thinks they can assign a probability to who will win this election is just blowing smoke. It’s just a fifty/fifty chance.
I wish political reporting would be like sports reporting in another way: the acceptance of empirical results.
With only a few exceptions who don’t matter, there are no sports pundits who argue for two-way players or the single-wing offense in the NFL, or that having a jump ball after every basket would restore the NBA to its glory.
Yet in 2012 political discourse, we have pundits who promote views as old and useless as those.
Baseball, basketball? Great. But football is, in part, theater (more so than other sports). There is way too much noise and way too few N to say much about anything but the obvious.
And people are utterly irrational about football–it’s about the same as politics, really.
I’ll be honest. I consider myself relatively perceptive and I have no fucking clue WHY the Monty Haul problem comes out the way it does.
I accept that its better to switch, because people have proven empirically that it is so, but I don’t understand WHY. The part of my brain that’s semi-decent at logic problems rebels at that choice. It screams at me “What the fuck does the open door matter? You have two doors. One of them is a goat, the other is a car. You already have a claim on one door that you can switch to the other one if you choose to. Why does how you GOT here matter?”
And it’s never gotten an answer that makes sense to it, in the way that it has gotten answers that make sense as to other counter-intuitive things. It accepts the math, because the math is solid, but doesn’t understand it at all.
If you know Monty Hall is randomly opening a door, then his choice reveals nothing and you have no more information than when you started. There’s no point in switching in this case.
If you know that Monty Hall is opening a door based on his knowledge of where the goat is (and isn’t), then his choice reveals information that you didn’t have when you started. Switching doors is how you act on that information in this case.
Not sure why you think it’s off, Julian. If Monty opens a door without knowing where the car is (not the original problem), you have to take into account the odds he picks the door with the car (1/3), and decide what to do in that case. If you just toss these cases out, what’s left is 50/50.
As I said, it depends on how you deal with the cases in which Monty chooses the door with the car. If the choice remains your door or the remaining unopened door, it’s a wash (i.e if Monty picks the car, you lose, in all other cases it’s 50/50).
But, as mentioned earlier, this is not the problem as presented.
Right, the first example wasn’t the problem as presented, but I suspect it’s the one Murc was confusing for the problem as presented. That’s why I distinguished it from the second example (which was the problem as presented).
But Julian would rather pick the car after Monty shows him the car. If he thinks Monty will ever show him the car, I’ve got a goat to sell him.
I thought it was blindingly clear, but I guess it was not: I am aware that Monty will never show me the car.
Monty will always pick a door with a goat.
Whether or not he knows that the goat is behind the door is beside the point (although, given that the game is over if he picks the car, it seems likely that he knows where the car is).
Because Monty isn’t going to open the door with the car and therefore end the contest. (obviously everybody chooses the door where they know the car is). Monty opens the door with a goat because he knows a goat is behind that door and not a car. Therefore you know that you have a better chance of picking the car if you switch than if you stay. But its the same mathematical theory whether you think he may open the door with the car or if he does not have that option.
FMGuru’s explanation doesn’t make sense to me either. If I didn’t already know the right answer (without knowing why its right) I’d always make the wrong decision, and I’ve never seen a logic chain (as opposed to cold hard math) that makes sense to me otherwise.
Put another way: my brain keeps thinking “The fact that all the wrong choices save one were taken away is irrelevant. There are two cards. One is the Ace of Spades. The other is not. I have already picked one. I may change if I want. If a random stranger were to wander in right now and were faced with these two cards, his choice would be 50/50. Why wouldn’t mine be the same?”
Basically I’ve never been able to grok why that initial pick matters so much. No matter what you pick, all of the wrong choices (but one) are going to be cleared away. So whenever I sit down and try REASON (as opposed to math) I always come up with ‘the initial choice means dick.’
But you are not a random stranger. You already made a pick that had a 1/52 chance of being right, then 50/52 of the wrong answers were taken away so that there is the 1/52 card you made and the 51/52 card. If you were to spin around in circles so you got dizzy and somebody shuffled the cards and asked you to pick one of the two cards, then you would be that person.
Or think of it that the other 50 cards are all combined with the other card that was not taken away. You can either stay with the one card that you first picked, or you can take the other 51 cards all together.
The reason is that you make your pick, and you are effectively offered either your pick (1 in 3 chance of winning) or the field (2 in 3 chance of winning).
Think of it in terms of three Cracker Jack boxes, only one of which contains a toy surprise. You pick one box. I then give you the choice of swapping the box that you chose for the two boxes that you didn’t choose. You can see right away that two boxes give you twice the chance of getting a toy surprise that one box does. Now, after you make your choice, I could dump the contents of the two boxes that you didn’t choose into a bigger box, start grabbing popcorn out of that bigger box and strewing it all over the place, whatever. But it doesn’t change that the choice is between the box that you chose at the original 1 in 3 chance of winning and the field at a 2 in 3 chance of winning.
Cracker Jack is not the same. You’re offering me two boxes for one, true, but you’re not showing me that one of those boxes has no toy. Monty is, and that changes the odds to 1/2.
So how about if I take the big box, divide the contents back between the two smaller boxes, and offer you the two boxes for the one that you originally picked? Or divide the big box between four even smaller boxes? It doesn’t change the fact that the choice is between the original box at 1-in-3 or the field at 2-in-3.
I mean, I don’t need to show you that one of the boxes that you didn’t choose contains only popcorn. There’s only one toy surprise for the three boxes. You chose one box at 1-in-3 odds. It doesn’t matter how I repackage the other two boxes, there’s 2-in-3 odds that the other two boxes contain the toy surprise.
You already know that one of those boxes doesn’t have a toy in it, because the rules of the game are that there are three boxes and only one of them has a prize. Since you haven’t gotten any new information, the odds stay one in three.
Basically I’ve never been able to grok why that initial pick matters so much.
First principles: The odds you initially picked the right card are 1/N, where N is the number of cards. Nothing can change that. If N-2 of the remaining cards are shown not to be the card, and all probabilities add up to one, the odds that the remaining card is the card is 1-(1/N)=(N-1)/N.
The odds you initially picked the right card are 1/N, where N is the number of cards. Nothing can change that.
You see, that’s what I don’t understand. If I was standing there not making a choice, and one of the doors was shown to be a loser, the other two doors would each be, for me, a 1/2 probability. If I had made an original choice of one of those two doors, my odds of being correct will have changed from 1/3 to 1/2, just like the observer’s odds. I can “pick” my original door or the other one, and it does not change my chances.
If I was standing there not making a choice, and one of the doors was shown to be a loser, the other two doors would each be, for me, a 1/2 probability.
That’s right. But that’s a different game.
If I had made an original choice of one of those two doors…
Hindsight and probability don’t mix well (if Schroedinger put a dead cat in the box…). You’re overthinking, SPP.
You go in knowing nothing, and pick a door. That establishes the 1/3 odds. Then Monty opens a door showing a goat. That’s always possible, whether you picked the right door or not – at least one of the remaining doors has one. The odds you picked the right door are still (always!) 1/3. Since there is only one door left, the odds for that one must now be 2/3.
Anyway, don’t waste time worrying about it. Probably never happen!
One perspective: Your initial choice puts a constraint on Monty. Suppose one goat is white, one is black. If you pick the white goat, Monty has to show you the black one; if you pick the black one, he has to show you the white one. In the 2-out-of-3 cases where you picked a goat the first time, you’re forcing Monty to show you where the other goat is (because he never shows you what’s in the door you picked, and he never shows the car).
Contrast: If you make no choice, Monty can choose to either show you the white goat or the black goat — you havent’t constrained him at all.
Try this: Ignore the opening of the door, and pretend instead that Monty offers the choice of sticking with your original pick, or taking BOTH the two doors you didn’t pick. Should you switch?
If a random stranger were to wander in right now and were faced with these two cards, his choice would be 50/50.
It doesn’t matter if the second pick is made by a random stranger, as long as the cards aren’t re-dealt. Card No. 1 still has a 1/52 chance of being the ace of spades, and card no. 2 still has a 1/2 chance.
Card No. 2 has a 51/52 chance of being the ace of spades.
Of course, the random stranger would have to know the history (which card was chosen by the player, which card was left by the dealer), otherwise from her point of view, the odds are 50/50.
(2) the random stranger thinks the odds are 50-50–but they are not.
Odds depend on available information. She may know that one of the cards was picked by the player, but if she doesn’t know which one, they are indistinguishable to her. So in that case, to her, the odds really are 50/50. For someone (Monty) who has all the information, the odds are 0/100.
Say you take Door #1. The chance that your choice is correct is 1/3. That chance doesn’t change when Door #2 is opened to reveal a goat, so the remaining 2/3 chance is now with Door #3.
Nope. I still don’t get it. Why doesn’t my pick go from 1/3 to 1/2 when I see the goat? If it does, then the other unopened door is also 1/2 and it doesn’t matter if I switch or not.
It doesn’t because there were three doors and there are still three doors. Two of them are losers and one of them is a winner. After one of the kausgates is opened, two of the doors are losers and one of them is a winner.
Play it with FMguru’s deck-of-cards version ten times. Switch five of those times. See how many times you win.
Let’s call the initial possibilities Y (the prize is behind your door), L (behind the left unpicked door) and R (behind the right unpicked door). The probability of any of these is 1/3.
If Y, Monty will open the L door 1/2 the time and the R door 1/2 the time.
If R, Monty opens L every time.
If L, Monty opens R every time.
So say Monty opens R. This choice includes 100% of the L cases but only 50% of the Y cases. Therefore L is twice as likely as Y.
If Monty ALWAYS picked R in the Y case, then it would be 50-50 whether to stand or switch. But as he picks R in only half the Y cases, he’s eliminated half the ways in which the prize can be behind your door.
Suppose you think of frequencies instead of probabilities, then. Suppose you have, say, 300 copies of the Monty Hall problem—i.e. 300 stages each with 3 doors. And you set off walking from stage to stage opening 1 door on each stage, so after several tedious and RSI-inducing minutes, you’ve opened 300 hundred doors. And your chance of getting a car is 1/3 each time, so at the end you can see roughly 100 cars and 200 goats behind the open doors, right?
So now Monty Hall goes along the 300 stages opening doors with goats, or actually he could just chalk “GOAT” or “X” on the doors or something.
And you feel that this changes the chances of your original pick from 1/3 to 1/2, so … what? Before, you could see about 100 cars and 200 goats behind the open doors, but now you can see about 150 of each?? Monty Hall chalking a mark on some doors caused roughly 50 goats to physically transmogrify into cars in front of your eyes? If not, in what sense has the probability changed?
The odds of my flipping a coin and getting heads 100 times in a row are astronomically tiny. But if by some miracle I flip 99 heads, the odds of getting that last heads aren’t astronomically tiny; they’re one in two, just like they were one in two for every flip before and will be for every flip after.
Because they don’t have the option of moving the car between your first and second choices, the probability of the car being behind your chosen door (1/3) or either of the other two (2/3) does not change.
The coin flip outcomes are random; Hall’s choice in regard to which door to open is not: it’s constrained by the fact that, if you chose a goat with your initial choice, Hall has no choice but to show you the one remaining door that has a goat behind it. Since you will choose a goat initially two thirds of the time, Hall has no choice which door to open in two thirds of cases, which means that in those cases the car is behind the door Hall can’t open per the rules of the game. Does that help?
Because you make the first choice without knowing what’s behind any of the doors. You have a one in three chance of being right, and a two in three chance of being wrong.
Now a.door you didn’t choose is opened. Your odds of having guessed right the first time are still one in three. You just know that one of the two doors you didn’t choose was wrong – but that was true anyhow. There’s still a two in three chance that you chose the wrong door. That’s why you always change the door.
Actually, the prior choice doesn’t matter as long as (a) all possible initial arrangements are equally likely and (b) in the case where the prize is behind your door, Monty is equally likely to pick any of the remaining doors.
The prior choice matters because it constrains Hall’s choice. The key to this problem is recognizing that Hall’s choice of door to open is constrained both by your initial choice and the rules of the game. If you chose the car originally he can open either other door – but that only happens 1/3 of the time.
2/3 of the time your initial choice will be a goat – and that means that Monty is forced to open the only door left available to him that has a goat behind it. Which means that 2/3 of the time the other door remaining closed will have the car behind it and so you should switch.
Two things make this problem work – the fact that YOU make the original choice to constrain Monty’s choice AND the fact that Monty is also constrained to only pick a door with a goat behind it. If you remove either of those elements the “always switch” rule disappears because the probabilities become independent of each other.
Jesus, Murc and SPP. Thank you. I am a SMART person, I swear to God, but this makes absolutely no sense to me. Thank you for explaining how your/my brain works, because every time I try, I get tied up in logic knots. Whether it’s the goat behind the door, or the Ace of Spades, how the fuck is it not 50/50????
I think it’s easier to understand if you imagine playing a lot of games at once. Say you play 300 games at the same time. In about how many of these games do you pick the goat on your first choice? About 200 times, right? So 100 times you picked the car.
Now Monty acts. In the 200 cases where you picked a goat, he opens the other door with a goat. The remaining door is the car, right?
In the 100 cases where you picked the car, he opens either of the other doors at random. The remaining door is the goat, right?
So in 300 games, the car’s behind the third door about 200 times, and behind your door about 100 times.
Metereologists (used to, anyway) over-report the chances of precipitation. Since every jackass who dropped out of high school but is still capable of discerning rain on the top of his head thinks “they just make it up,” they reasoned that it was better to be wrong in that direction. Nobody will really care that much if you forecast rain and it doesn’t appear, but they’ll be up in arms if you say 30% and a brief shower appears.
Most of these ‘problems’, it seems to me, hinge on verbally eliding a distinction between colloquial and technical meanings of terms, which when explained properly, are perfectly clear.
“If, for example, this particular weather forecasting model predicted a 75 percent chance of rain on 100 separate days over the previous decade, and it rained on 75 of those days, then we can estimate the model’s accuracy in this regard as 100 percent.”
But … that isn’t how you should judge probabilistic predictions of “one-off” events. You should use the Brier score, or any other proper scoring function! A “100 percent accurate” model would be certain of rain on 75 of those days, and certain of sunny skies on the other 25 days.
Instead, you should judge the accumulated “distance” between the probability provided and what actually happens over a time frame. Whichever model has the closest distance wins.
I don’t see the point of pretending all Nate Silver does is average polls. If that was all he did, NYT wouldn’t be paying him to do it because a script can be written to do that in 2 minutes.
Nate Silver averages polls, assesses the weighting of these polls based on his estimation of their accuracy, his estimation of the time-window of validity of these polls, and the questions they ask; he has something called “state fundamentals” scores which assess the underlying political environment of a state apart from polling on a specific election question, and then this information affects his forecasting model. He runs, afair, Monte Carlo simulations of elections, which means he has to assign a confidence level-like thing to each state’s predicted election outcome (I guess this is like the “energy cost” to deviate from standard behaviour in MC terms as I understand it?).
My point is, Nate Silver puts in a lot af added value into 538. That’s clearly a very good thing. I don’t actually know what difference that makes to his predictions as opposed to “brute force averaging all the polls and that’s it” but presumably there is some difference. But obviously you can disagree with his decisions on all of those points.
So the conservative/punditocratic critique of “averaging is morally wrong! durrr!” and Paul’s defense of “well, it’s just averaging, there’s nothing to it” are both wrong to me. A reasonable critique of Silver would be “well, I disagree with his decisions on all of these points because blah blah blah, and since presidential elections are much more rare than weather, we haven’t accumulated sufficient data to decide whether his approach is actually good”. I don’t see anyone saying this, though.
“Innumerate potshots” was clever.
I couldn’t believe the problems I had explaining the three door problem to a bunch of educated, computer-literate folk. A lot of them couldn’t accept that the opened door was a red herring.
The problem needs to be stated correctly, as it was in the article, or it’s not really a math problem. In real life, Monty didn’t always open a door, which means that whether switching is correct or not becomes an empirical issue (although you should still switch if your ONLY evidence of the likelihood a door will be opened is your own experience).
In case it wasn’t clear in the above, many internet wiseguys who post the problem leave out that Monty is required to open a door. So then THEY are the fools.
Does Monty know where the two goats are or is he also picking at random?
Never mind, looks like it was answered between typing my question and actually hitting “submit”.
I really need to get less distracted by shiny moving objects.
Monty knows.
Monty always knows.
It isn’t a red herring. Its new information. People have problems with it because its Bayesian.
The opened door isn’t a red herring, it’s essential. It transforms the choice to picking between your original choice at the original 2-to-1 odds and picking the field. (Of course, this depends on the original disposition being random and Monty being obligated to open a door, as CaptBackslap says.)
It’s a red herring because the problem is precisely equivalent to giving the player a choice of keeping his door, or choosing both other doors (without opening one), which of course makes the answer obvious.
Touché.
That’s an insightful way of explaining it.
I think a lot of the trouble with getting people to see it is that they have enough knowledge to be dangerous. They don’t even consider that they might be wrong.
That’s brilliant, at least in the sense that it makes the nature of the problem (and the logic of the answer) so much more obvious.
Mickey Kaus has the most trouble with this problem, because he can’t figure out why you wouldn’t just switch to the opened door.
Awesome
It’s easier for people to understand if you describe a version with a million doors instead of three, and then gradually taking away all of the doors except for the one you originally picked and one other. Do you really think that you got the right door out of a million on the first try?
Actually, yes, I do.
And – what if I really like goats?
mickey, is that you?
The frustrating thing about Politico et al is that it is very, very easy to think of a critique of Silver that holds up.
To wit:
“Silver, and those who use his numbers, are relying on the innumeracy of Americans to report a matter of statistics as though it were a matter of fact; by ‘predicting’ a 3 in 4 chance of victory for Obama they subtly want Americans to see Obama being the winning side as a fait accompli, lowering the motivation of potential Romney voters and making it more likely that undecideds will break for Obama.”
Bullshit, of course, but a lot closer to a relevant critique (not to mention being more intellectually rigorous) than what they’re peddling.
No, this isn’t much better. “Matter of fact” and “statistics” are not mutually exclusive categories. There can be statistical, as well as probabilistic, facts. It’s a matter of fact that a fair coin flip has a 50% chance of turning up heads.
Of all the things I’ve read and heard over the last two weeks, nothing gives me more optimism for an Obama victory than the entire right-wing world’s pitchforks & torches attack on a guy who is pretty much just aggregating numbers using conventional methods.
Sabermetrics with real sabers.
+1
the monty hall problem even confounded Paul Erdős … instead of goats & cars, i use obama and rmoney …
Nate Silver doesn’t understand the intangibles, like clutchiness or locker room leadership.
How dare he discount the Romney team’s veteran experience?
Hey looking at the two candiates we know which one would be called scrappy right?
To be fair, lots of people underestimated the added value Romney’s hustle would bring to this campaign.
This emotional approach is not incompatible with a statistical approach. It can be used to describe the causes and effects of the phenomena that are measured statistically.
But that is not really informative, so ignore my trolling.
The best way to explain the Monty Haul problem is to expand the number of choices. I always use a deck of cards, putting all 52 face down and asking the player to pick the ace of spades. He picks one, I turn over 50 cards showing that they’re not the ace of spades, leaving two face-down cards – the one he picked and the one remaining card I hadn’t turned over. Then I ask him if he wants to swap cards. If he says yes, I look confused and ask why – isn’t it a 50-50 choice? Usually the light goes on and they understand it’s not a 50-50 choice, and that let’s them see that the Monty Haul problem isn’t 50-50 either.
That is a really nice explanation.
Yeah, it becomes much more obvious with more than three doors.
Any AM radio call-in sports show has more reference to empirical evidence and far more presumed understanding of probability, the scientific method, etc. than people who are paid to talk about politics for a living. It constantly amazes me. A lot of sports fans aren’t interested in advanced metrics, or don’t understand them, but they still fundamentally understand what probability *is*.
Nobody thinks that a WAR of 9 means a player is going to hit exactly 9 walk-off home runs in a year. Nobody thinks the bookies are wrong if a team wins but fails to cover the spread. Nobody thinks that because Kevin Durant averages 32 points a game, any time he scores more or less than that it proves that mathematics is fundamentally flawed and/or John Hollinger is a homosexual.
I wish political conversations in this country could be 1/10th as empirically and statistically driven as two drunks at the sports bar arguing the Manning brothers. Nobody’s like “Pythagorean wins are unamerican, real Americans look at PyBUSHian wins” or some crap.
But what really fucking chaps my hide about this Nate Silver backlash is that it seems like you should know, even if you can’t do grade school arithmetic, that to refute somebody’s mathematical argument would require, you know, some math. Like, some squiggly Greek characters and stuff? Maybe a graph or a chart or something?
If this were about global warming or wind power or stem cells or whatever, they’d be sure to dig up something at least pseudoscience-y. But actual math is so far beyond their experience they have nothing to fall back on but booger-flinging rage. (Did you know that Andrew Wiles solved Fermat’s last theorem by showing that Fermat was a wicked big quee-ah who had hands like a little bitch? It’s true, he did!)
As a result, these anti-Silver articles read like somebody who doesn’t speak any French phonetically transliterating something into English, and therefore deciding that French isn’t actually a real language because of that. “Comb oh tally view”? I don’t know what a tally view is, but I’m pretty sure you can’t comb it!!! LOL french is stupid!
But Mitt’s got momentum! Or I thought he did, but now with the new polls, it’s starting to look like maybe Obama’s taken the momentum back. It’s probably because he’s looking so Presidential in the aftermath of the hurricane.
Anyway, anyone who thinks they can assign a probability to who will win this election is just blowing smoke. It’s just a fifty/fifty chance.
I wish political reporting would be like sports reporting in another way: the acceptance of empirical results.
With only a few exceptions who don’t matter, there are no sports pundits who argue for two-way players or the single-wing offense in the NFL, or that having a jump ball after every basket would restore the NBA to its glory.
Yet in 2012 political discourse, we have pundits who promote views as old and useless as those.
Hell, most reporters can’t even understand empirical data in SCIENCE reporting, let alone politics.
Exclude football.
Baseball, basketball? Great. But football is, in part, theater (more so than other sports). There is way too much noise and way too few N to say much about anything but the obvious.
And people are utterly irrational about football–it’s about the same as politics, really.
The comment above and the responses to it are entirely too generous to sports fans, is all I have to say.
How much should we be reading into the Youtube url for that Scarborough video? (http://www.youtube.com/watch?v=TbKkjm-gheY)
I’ll be honest. I consider myself relatively perceptive and I have no fucking clue WHY the Monty Haul problem comes out the way it does.
I accept that its better to switch, because people have proven empirically that it is so, but I don’t understand WHY. The part of my brain that’s semi-decent at logic problems rebels at that choice. It screams at me “What the fuck does the open door matter? You have two doors. One of them is a goat, the other is a car. You already have a claim on one door that you can switch to the other one if you choose to. Why does how you GOT here matter?”
And it’s never gotten an answer that makes sense to it, in the way that it has gotten answers that make sense as to other counter-intuitive things. It accepts the math, because the math is solid, but doesn’t understand it at all.
If you know Monty Hall is randomly opening a door, then his choice reveals nothing and you have no more information than when you started. There’s no point in switching in this case.
If you know that Monty Hall is opening a door based on his knowledge of where the goat is (and isn’t), then his choice reveals information that you didn’t have when you started. Switching doors is how you act on that information in this case.
close, but you’re way off
Not sure why you think it’s off, Julian. If Monty opens a door without knowing where the car is (not the original problem), you have to take into account the odds he picks the door with the car (1/3), and decide what to do in that case. If you just toss these cases out, what’s left is 50/50.
There are three doors. I point at one of them (my first “choice”). Monty opens one of the other two doors (the two I did NOT pick).
Either Monty opens the door with the car, in which case I pick the door with the car, or he opens the door with the goat, in which case I switch.
Show me how it can possibly matter whether or not Monty knows where the goat is.
As I said, it depends on how you deal with the cases in which Monty chooses the door with the car. If the choice remains your door or the remaining unopened door, it’s a wash (i.e if Monty picks the car, you lose, in all other cases it’s 50/50).
But, as mentioned earlier, this is not the problem as presented.
Right, the first example wasn’t the problem as presented, but I suspect it’s the one Murc was confusing for the problem as presented. That’s why I distinguished it from the second example (which was the problem as presented).
But Julian would rather pick the car after Monty shows him the car. If he thinks Monty will ever show him the car, I’ve got a goat to sell him.
I thought it was blindingly clear, but I guess it was not: I am aware that Monty will never show me the car.
Monty will always pick a door with a goat.
Whether or not he knows that the goat is behind the door is beside the point (although, given that the game is over if he picks the car, it seems likely that he knows where the car is).
Monty will always pick a door with a goat.
Whether or not he knows that the goat is behind the door is beside the point…
You’ve lost me mate. If he doesn’t know where the goats are, how does he always pick a goat door?
Maybe it’s a caper ex machina?
I have no idea what you are talking about. If Monty picks the car? Monty will never pick the car.
Monty will always pick a door with a goat. You are always better of switching. Monty’s knowledge is irrelevant.
Because Monty isn’t going to open the door with the car and therefore end the contest. (obviously everybody chooses the door where they know the car is). Monty opens the door with a goat because he knows a goat is behind that door and not a car. Therefore you know that you have a better chance of picking the car if you switch than if you stay. But its the same mathematical theory whether you think he may open the door with the car or if he does not have that option.
See FMguru’s explanation above.
FMGuru’s explanation doesn’t make sense to me either. If I didn’t already know the right answer (without knowing why its right) I’d always make the wrong decision, and I’ve never seen a logic chain (as opposed to cold hard math) that makes sense to me otherwise.
Put another way: my brain keeps thinking “The fact that all the wrong choices save one were taken away is irrelevant. There are two cards. One is the Ace of Spades. The other is not. I have already picked one. I may change if I want. If a random stranger were to wander in right now and were faced with these two cards, his choice would be 50/50. Why wouldn’t mine be the same?”
Basically I’ve never been able to grok why that initial pick matters so much. No matter what you pick, all of the wrong choices (but one) are going to be cleared away. So whenever I sit down and try REASON (as opposed to math) I always come up with ‘the initial choice means dick.’
But you are not a random stranger. You already made a pick that had a 1/52 chance of being right, then 50/52 of the wrong answers were taken away so that there is the 1/52 card you made and the 51/52 card. If you were to spin around in circles so you got dizzy and somebody shuffled the cards and asked you to pick one of the two cards, then you would be that person.
Or think of it that the other 50 cards are all combined with the other card that was not taken away. You can either stay with the one card that you first picked, or you can take the other 51 cards all together.
The reason is that you make your pick, and you are effectively offered either your pick (1 in 3 chance of winning) or the field (2 in 3 chance of winning).
Think of it in terms of three Cracker Jack boxes, only one of which contains a toy surprise. You pick one box. I then give you the choice of swapping the box that you chose for the two boxes that you didn’t choose. You can see right away that two boxes give you twice the chance of getting a toy surprise that one box does. Now, after you make your choice, I could dump the contents of the two boxes that you didn’t choose into a bigger box, start grabbing popcorn out of that bigger box and strewing it all over the place, whatever. But it doesn’t change that the choice is between the box that you chose at the original 1 in 3 chance of winning and the field at a 2 in 3 chance of winning.
Cracker Jack is not the same. You’re offering me two boxes for one, true, but you’re not showing me that one of those boxes has no toy. Monty is, and that changes the odds to 1/2.
So how about if I take the big box, divide the contents back between the two smaller boxes, and offer you the two boxes for the one that you originally picked? Or divide the big box between four even smaller boxes? It doesn’t change the fact that the choice is between the original box at 1-in-3 or the field at 2-in-3.
I mean, I don’t need to show you that one of the boxes that you didn’t choose contains only popcorn. There’s only one toy surprise for the three boxes. You chose one box at 1-in-3 odds. It doesn’t matter how I repackage the other two boxes, there’s 2-in-3 odds that the other two boxes contain the toy surprise.
Only until I *know* that one box has no toy. Then I’m not choosing two boxes vs. one, but one vs. one.
You already know that one of those boxes doesn’t have a toy in it, because the rules of the game are that there are three boxes and only one of them has a prize. Since you haven’t gotten any new information, the odds stay one in three.
Basically I’ve never been able to grok why that initial pick matters so much.
First principles: The odds you initially picked the right card are 1/N, where N is the number of cards. Nothing can change that. If N-2 of the remaining cards are shown not to be the card, and all probabilities add up to one, the odds that the remaining card is the card is 1-(1/N)=(N-1)/N.
You see, that’s what I don’t understand. If I was standing there not making a choice, and one of the doors was shown to be a loser, the other two doors would each be, for me, a 1/2 probability. If I had made an original choice of one of those two doors, my odds of being correct will have changed from 1/3 to 1/2, just like the observer’s odds. I can “pick” my original door or the other one, and it does not change my chances.
If I was standing there not making a choice, and one of the doors was shown to be a loser, the other two doors would each be, for me, a 1/2 probability.
That’s right. But that’s a different game.
If I had made an original choice of one of those two doors…
Hindsight and probability don’t mix well (if Schroedinger put a dead cat in the box…). You’re overthinking, SPP.
You go in knowing nothing, and pick a door. That establishes the 1/3 odds. Then Monty opens a door showing a goat. That’s always possible, whether you picked the right door or not – at least one of the remaining doors has one. The odds you picked the right door are still (always!) 1/3. Since there is only one door left, the odds for that one must now be 2/3.
Anyway, don’t waste time worrying about it. Probably never happen!
Obviously it’s time for deep research. I will accept that switching is correct, but I still don’t get it. To the Wiki!
One perspective: Your initial choice puts a constraint on Monty. Suppose one goat is white, one is black. If you pick the white goat, Monty has to show you the black one; if you pick the black one, he has to show you the white one. In the 2-out-of-3 cases where you picked a goat the first time, you’re forcing Monty to show you where the other goat is (because he never shows you what’s in the door you picked, and he never shows the car).
Contrast: If you make no choice, Monty can choose to either show you the white goat or the black goat — you havent’t constrained him at all.
Try this: Ignore the opening of the door, and pretend instead that Monty offers the choice of sticking with your original pick, or taking BOTH the two doors you didn’t pick. Should you switch?
If a random stranger were to wander in right now and were faced with these two cards, his choice would be 50/50.
It doesn’t matter if the second pick is made by a random stranger, as long as the cards aren’t re-dealt. Card No. 1 still has a 1/52 chance of being the ace of spades, and card no. 2 still has a 1/2 chance.
Card No. 2 has a 51/52 chance of being the ace of spades.
Of course, the random stranger would have to know the history (which card was chosen by the player, which card was left by the dealer), otherwise from her point of view, the odds are 50/50.
(1) you’re right about card no. 2, of course.
(2) the random stranger thinks the odds are 50-50–but they are not.
(2) the random stranger thinks the odds are 50-50–but they are not.
Odds depend on available information. She may know that one of the cards was picked by the player, but if she doesn’t know which one, they are indistinguishable to her. So in that case, to her, the odds really are 50/50. For someone (Monty) who has all the information, the odds are 0/100.
Say you take Door #1. The chance that your choice is correct is 1/3. That chance doesn’t change when Door #2 is opened to reveal a goat, so the remaining 2/3 chance is now with Door #3.
Nope. I still don’t get it. Why doesn’t my pick go from 1/3 to 1/2 when I see the goat? If it does, then the other unopened door is also 1/2 and it doesn’t matter if I switch or not.
It doesn’t because there were three doors and there are still three doors. Two of them are losers and one of them is a winner. After one of the kausgates is opened, two of the doors are losers and one of them is a winner.
Play it with FMguru’s deck-of-cards version ten times. Switch five of those times. See how many times you win.
Let’s call the initial possibilities Y (the prize is behind your door), L (behind the left unpicked door) and R (behind the right unpicked door). The probability of any of these is 1/3.
If Y, Monty will open the L door 1/2 the time and the R door 1/2 the time.
If R, Monty opens L every time.
If L, Monty opens R every time.
So say Monty opens R. This choice includes 100% of the L cases but only 50% of the Y cases. Therefore L is twice as likely as Y.
If Monty ALWAYS picked R in the Y case, then it would be 50-50 whether to stand or switch. But as he picks R in only half the Y cases, he’s eliminated half the ways in which the prize can be behind your door.
Suppose you think of frequencies instead of probabilities, then. Suppose you have, say, 300 copies of the Monty Hall problem—i.e. 300 stages each with 3 doors. And you set off walking from stage to stage opening 1 door on each stage, so after several tedious and RSI-inducing minutes, you’ve opened 300 hundred doors. And your chance of getting a car is 1/3 each time, so at the end you can see roughly 100 cars and 200 goats behind the open doors, right?
So now Monty Hall goes along the 300 stages opening doors with goats, or actually he could just chalk “GOAT” or “X” on the doors or something.
And you feel that this changes the chances of your original pick from 1/3 to 1/2, so … what? Before, you could see about 100 cars and 200 goats behind the open doors, but now you can see about 150 of each?? Monty Hall chalking a mark on some doors caused roughly 50 goats to physically transmogrify into cars in front of your eyes? If not, in what sense has the probability changed?
Does that help a little bit?
Why does the prior choice matter?
The odds of my flipping a coin and getting heads 100 times in a row are astronomically tiny. But if by some miracle I flip 99 heads, the odds of getting that last heads aren’t astronomically tiny; they’re one in two, just like they were one in two for every flip before and will be for every flip after.
Because they don’t have the option of moving the car between your first and second choices, the probability of the car being behind your chosen door (1/3) or either of the other two (2/3) does not change.
The coin flip outcomes are random; Hall’s choice in regard to which door to open is not: it’s constrained by the fact that, if you chose a goat with your initial choice, Hall has no choice but to show you the one remaining door that has a goat behind it. Since you will choose a goat initially two thirds of the time, Hall has no choice which door to open in two thirds of cases, which means that in those cases the car is behind the door Hall can’t open per the rules of the game. Does that help?
It helped for me. Thanks.
Because you make the first choice without knowing what’s behind any of the doors. You have a one in three chance of being right, and a two in three chance of being wrong.
Now a.door you didn’t choose is opened. Your odds of having guessed right the first time are still one in three. You just know that one of the two doors you didn’t choose was wrong – but that was true anyhow. There’s still a two in three chance that you chose the wrong door. That’s why you always change the door.
Actually, the prior choice doesn’t matter as long as (a) all possible initial arrangements are equally likely and (b) in the case where the prize is behind your door, Monty is equally likely to pick any of the remaining doors.
See my other comment.
The prior choice matters because it constrains Hall’s choice. The key to this problem is recognizing that Hall’s choice of door to open is constrained both by your initial choice and the rules of the game. If you chose the car originally he can open either other door – but that only happens 1/3 of the time.
2/3 of the time your initial choice will be a goat – and that means that Monty is forced to open the only door left available to him that has a goat behind it. Which means that 2/3 of the time the other door remaining closed will have the car behind it and so you should switch.
Two things make this problem work – the fact that YOU make the original choice to constrain Monty’s choice AND the fact that Monty is also constrained to only pick a door with a goat behind it. If you remove either of those elements the “always switch” rule disappears because the probabilities become independent of each other.
Jesus, Murc and SPP. Thank you. I am a SMART person, I swear to God, but this makes absolutely no sense to me. Thank you for explaining how your/my brain works, because every time I try, I get tied up in logic knots. Whether it’s the goat behind the door, or the Ace of Spades, how the fuck is it not 50/50????
I think it’s easier to understand if you imagine playing a lot of games at once. Say you play 300 games at the same time. In about how many of these games do you pick the goat on your first choice? About 200 times, right? So 100 times you picked the car.
Now Monty acts. In the 200 cases where you picked a goat, he opens the other door with a goat. The remaining door is the car, right?
In the 100 cases where you picked the car, he opens either of the other doors at random. The remaining door is the goat, right?
So in 300 games, the car’s behind the third door about 200 times, and behind your door about 100 times.
A hat-tip to Paul from the happy mutants at Boing Boing:
http://boingboing.net/2012/10/31/what-nate-silver-is-actually-t.html
On the chances it will rain: If it’s 40%, I’m taking me brolly. I’m guessing Joe Scarborough only takes his if it’s 100%.
Metereologists (used to, anyway) over-report the chances of precipitation. Since every jackass who dropped out of high school but is still capable of discerning rain on the top of his head thinks “they just make it up,” they reasoned that it was better to be wrong in that direction. Nobody will really care that much if you forecast rain and it doesn’t appear, but they’ll be up in arms if you say 30% and a brief shower appears.
Tell the average person that .999… equals 1. The response usually falls just short of burning you at the stake as a witch, but not by much.
I love that one.
10X – 9X = X. Ergo, DAMN YOU TO HELL.
Oh god, not that discussion.
Don’t make me bring up how the chances of a girl’s sibling being a boy change depending on whether you know which one is older.
TOO LATE.
Most of these ‘problems’, it seems to me, hinge on verbally eliding a distinction between colloquial and technical meanings of terms, which when explained properly, are perfectly clear.
In these cases, it’s not so much that as confusion about what counts as new information.
“If, for example, this particular weather forecasting model predicted a 75 percent chance of rain on 100 separate days over the previous decade, and it rained on 75 of those days, then we can estimate the model’s accuracy in this regard as 100 percent.”
But … that isn’t how you should judge probabilistic predictions of “one-off” events. You should use the Brier score, or any other proper scoring function! A “100 percent accurate” model would be certain of rain on 75 of those days, and certain of sunny skies on the other 25 days.
Instead, you should judge the accumulated “distance” between the probability provided and what actually happens over a time frame. Whichever model has the closest distance wins.
I don’t see the point of pretending all Nate Silver does is average polls. If that was all he did, NYT wouldn’t be paying him to do it because a script can be written to do that in 2 minutes.
Nate Silver averages polls, assesses the weighting of these polls based on his estimation of their accuracy, his estimation of the time-window of validity of these polls, and the questions they ask; he has something called “state fundamentals” scores which assess the underlying political environment of a state apart from polling on a specific election question, and then this information affects his forecasting model. He runs, afair, Monte Carlo simulations of elections, which means he has to assign a confidence level-like thing to each state’s predicted election outcome (I guess this is like the “energy cost” to deviate from standard behaviour in MC terms as I understand it?).
My point is, Nate Silver puts in a lot af added value into 538. That’s clearly a very good thing. I don’t actually know what difference that makes to his predictions as opposed to “brute force averaging all the polls and that’s it” but presumably there is some difference. But obviously you can disagree with his decisions on all of those points.
So the conservative/punditocratic critique of “averaging is morally wrong! durrr!” and Paul’s defense of “well, it’s just averaging, there’s nothing to it” are both wrong to me. A reasonable critique of Silver would be “well, I disagree with his decisions on all of these points because blah blah blah, and since presidential elections are much more rare than weather, we haven’t accumulated sufficient data to decide whether his approach is actually good”. I don’t see anyone saying this, though.