In their book Merchants of Doubt that I reviewed very favorably here, authors Naomi Oreskes and Erik M. Conway make the case that those people and business interests that oppose the scientific consensus that goes against their business and ideological interests (like the perils of smoking tobacco, second-hand smoke, acid-rain, the ozone hole, and climate change) base their opposition strategy on exploiting the way people make decisions.
The theory of rational decision-making says that most people, when faced with doubt or the choice between competing theories that they are not in a position to evaluate for themselves and with no compelling reason to favor any specific one, tend to favor the status quo and taking no action or choose the option that is easier or less expensive or less discomfiting. So if you can make people feel that there is a great deal of doubt or uncertainty on some issue, they will tend to favor the status quo. Hence they use their vast resources to launch a relentless media campaign to try and give the impression that there is a great deal of doubt on an issue when there is actually a scientific consensus on what needs to be done.
An example of how people make decisions in favor of the status quo can be seen in the Monty Hall problem that I discussed in an earlier post. If you are not sure what this famous problem is, see here and/or watch this video.
Although the logic is very clear that one should switch to the other door when given the choice, most people who are not aware of the statistics, do not do so and stick with their first choice, and in my earlier post. I speculate as to why this might be so.
But the point is that people usually do not start neutral on some issue that requires action but have a preference, even if they are unaware of it. Getting them to take action requires overcoming this emotional inertia. To do so requires existing conditions to be so bad that people are willing to overcome the inertia and make the change, or to cloak the change in a way that makes it look like it is actually a reversion to the status quo
This is why the Brexit vote in favor of leaving the EU, although on the surface somewhat close with 51.9% voting to leave while 48.9% voted to stay, was actually quite significant because people were voting to change the status quo in a major way, showing that the winning slogan ‘Take Back Control’ was very effectivet. It was the word ‘back’ that I think was significant. It implied that by leaving the EU, the UK was reverting to the status quo, going back to what was normal and familiar, rather than changing things. Meanwhile, those who felt that things were very bad for them personally and blamed membership in the EU for it, could vote for leaving too. For them, the desire for change overcame the inertia. So the Leave campaign had it both ways.
We have seen that conservative politicians in the US also appeal to a return to a mythical past. The challenge for progressive politicians is, like the Leave campaign, try to create a dual message, that change is necessary and good but also that the change will restore the ‘natural order’ of things. What would that ‘natural order’ look like? The argument should be that the natural order is for justice and equality and that the Republican policies of the past few decades have take us far away from that desirable state.
While it’s a certainty that every politician is attempting to manipulate the public in some way or other, it is nevertheless genuinely disturbing to see how the contemporary Right has made such a success of using recent(ish) insights from social science and psychological research to do so in such an effective manner.
A (hypothetical) politician who is trying to maintain some kind of ethical principles, trying to be more-or-less honest with the public, and trying to act in the way they believe is genuinely best for their nation (and the world at large) is always going to lose out to an amoral liar who is willing to say and do whatever it takes to achieve their agenda.
It’s honestly frightening.
On an entirely unrelated note (hollow laugh), back to Westminster tomorrow to noisily encourage MPs to pass the legislation to prevent a No Deal Brexit -- and ideally bring this mandateless government of liars, chancers and fools down also. Wish I’d been there to berate Johnson this evening.
Zooming animation warning for the video.
I don’t know if the video explains the problem. But it’s no surprise that people get the “wrong” answer when presented with an incomplete explanation and trying to solve the “wrong” problem. I know I did.
As for Brexit, the original referendum didn’t distinguish a negotiated exit from a no-deal exit, and was apparently spoiled by disinformation and one assassination. So it’s not clear what answer that 52% gave.
Not the best explanation, IMO. Much easier to recognize that Monty opening one of the doors is precisely the same as asking the contestant if they realize that at least one of the unchosen doors has a zonk behind it. Of course it does. So opening a door is a complete red herring, and the offer amounts to “keep your choice or choose both other doors”, without opening anything.
I think I’m too stupid to get the Deal game (I’m the worst at math and odds). 🙂 If I pick 1 door out of 3 doors, my odds are 1/3 of hitting the jackpot, right? It seems to me that if the host removes a door, then the original 1-of-3 game is over and a new game with 2 doors is now what’s being played, and I can pick either door, with 1/2 odds of guessing right.
I don’t get why you even add up the odds of the other doos. But can’t you just as easily look at it that the opened door bequeathed its 1/3 odds over to the door I picked rather than the one I didn’t, so that the odds of my first choice being the prize are now 2/3 while the remaining door is still 1/3? Why do these consolidated odds only go to the door I didn’t initially pick?
Actual practice may prove switching to be the right move more often than not, but I can’t grasp the reasoning of why that should be so.
Ridana @4: Suppose, if instead of opening a door, I just ask you whether one of the doors you didn’t choose has a zonk behind it. Because there is only one door with a car behind it, you know the answer is “yes”. Then I say, without opening any doors, that you can keep your choice, or choose both other doors. You’d choose both other doors, right?
That is exactly the same as opening one of the unchosen doors and showing a zonk behind it. It gives you no information other than that the host knows what’s behind the doors.
Ridana,
If it is any consolation, this problem is one that arouses much heated discussion and analysis, and even eminent mathematicians have refused to accept the idea that switching is better. But computer simulations have shown that the probability of winning if you switch is 2/3.
You can read much more about the problem and the controversy and the various arguments here.
Mano @6:
The only possible case for which the odds are the same for keeping your original door and switching, is if the host doesn’t know where the prize is, and picked a door with a zonk behind it by sheer chance. If the game is infinitely repeatable, that is a nonsensical assumption.
If there are eminent mathematicians who refuse to accept the idea that switching is better even if the host knows where the prize is, they should have their eminence surgically removed.
@7, I’m guessing such mathematicians were only confused initially. I remember taking it a while to sink in back when I first encountered that problem.
At this point I’ve thought about it enough that it really is as simple as you have the option to pick one door or two others. Which would you rather go with?
Erdos really required a computer sim to be convinced? My respect for him just dropped several notches.
unitooo @ 1: …back to Westminster tomorrow to noisily encourage MPs to pass the legislation to prevent a No Deal Brexit
Out of curiosity, what would that legislation be? What do you think Parliament can do to stop a No Deal?
Are you expecting them to cancel Brexit? That seems unlikely.
Are you expecting them to accept the negotiated deal they’ve refused three times already? That seems unlikely.
What other option do you see?
Ridana @ 4: If I pick 1 door out of 3 doors, my odds are 1/3 of hitting the jackpot, right? It seems to me that if the host removes a door, then the original 1-of-3 game is over and a new game with 2 doors is now what’s being played, and I can pick either door, with 1/2 odds of guessing right.
No. You are discarding important information; the odds you first picked from. That is IMPORTANT, it is data that is built upon by adding knowledge about the doors you DID NOT pick. You do NOT restart anything, you are gaining information that if used improves your chances. This should become clearer if you forget the three door example and concentrate on the 100 door example.
Rob Grigjanis @ 7: If there are eminent mathematicians who refuse to accept the idea that switching is better even if the host knows where the prize is, they should have their eminence surgically removed.
You are mistaken. The hosts knowledge is irrelevant -- what they add to YOUR knowledge by showing you the zonk is what matters. People having a problem with this invariably over-look the knowledge you had from the start: the odds on offer. That is important information which is added to when a door you didn’t choose is opened.
And being one of the people who wrote one of the simulations that runs this puzzle millions of times I affirm Mano’s point that it demonstrates results exactly as explained in this film -- which using the 100 Door example usually convinces people of.
Here’s a simulator people can run for themselves to test the problem; https://www.mathwarehouse.com/monty-hall-simulation-online/
Rob Grigjanis @ 7
Ooops, I think I misunderstood what I quoted of you -- getting your meaning jumbled up in my mind (serves me right for rushing over breakfeast).
As you seem to have actually written that the hosts knowledge does NOT matter, you’re right. Sorry.
The odds depend on Monty Hall’s behavior.
If he generally picks the best remaining door, then it’s better not to switch.
If he picks at random, then it’s no better or worse to switch.
If he generally picks the worst remaining door, then it’s better to switch.
The math problem assumes that Monty Hall picks the worst remaining door, but descriptions of it often fail to explain this part. So people unfamiliar with the problem may assume he picks the best remaining door.
fentex @11: I’m not sure you did understand. The host’s knowledge is crucial. If he plays the game repeatedly, but doesn’t know where the prize is, he will inevitably, one third of the time, open the door with the car behind it. It’s implicit in the presentation of the problem that he always picks a door with a zonk behind it. So he must know where the prize is.
If the host knows where the prize is, and the contestant picks the right door, the host can choose either of the two remaining doors to open. If the contestant picks a wrong door (one with a zonk), the host simply picks the remaining door which has a zonk behind it. Either way, showing the zonk teaches you nothing. Switching is always the right move.
No, not nothing.
If the game is understood from the beginning, then you already know that you should switch doors after Monty opens another. (Like you said, this is because Monty certainly knows to open one that reveals a goat/zonk/whatever — not the big prize — which is useful information to the player.)
In other words, the fact that it’s best to switch doors isn’t what you need to learn. However, you don’t know which of the other doors will be the one for you, until after Monty has made his choice.
If there are three doors (A, B, C) and you picked A, then you have not yet been taught (by Monty, when he shows you which door he’s opening) that Monty chose B or that he chose C. It’s not as if you already know from the beginning that your strategy is [A + C]. You only learn that in the event that Monty chooses to open door B. Otherwise, it’ll be [A + B], because he certainly won’t open A (your choice) and he certainly won’t open a door with the big prize.
This may be clearer in the 100-door example from the Numberphile video: upon choosing door #1 at the start, you would not have known to switch to door #37 specifically. You won’t know to do that, until Monty opened all of the other 98 doors.
So, you do have to learn (or be taught) what your choice will be when it’s your turn to switch, based on the door(s) that Monty has decided to show you.
Obs, the right choice is to grab Monty by the throat and threaten to go medieval on them unless they fess up.
(Or, what Rob wrote above)
The interesting thing about this problem is not its trivial solution but the tangle people get into trying to solve it. It is one of five that I selected for people to try. Here is another:
Paradoxical Daughters
Q1 Jane has two children. One is a daughter. What is the probability that Jane has two daughters?
A1 To answer the problem we must look at all the possible ways in which Jane can have two children. These are as follows:
son – son
daughter – son
son –daughter
daughter – daughter
We know that Jane has at least one daughter, so we can ignore the son – son option. This leaves the last three options. Only one of these involves a second daughter, therefore the probability that she has two daughters is 1 in 3.
Q2 Jane has two children. One is a daughter named Emma. What is the probability that Jane has two daughters?
A2 In the second problem a daughter is uniquely identified by her name, Emma. The options now become:
Emma – son
son – Emma
daughter – Emma
Emma – daughter
Thus the probability that Jane has 2 daughters is now 1 in 2.
It seems paradoxical that the only difference between Q1 and Q2 is the addition of the words: named Emma. Yet the probability that Jane has two daughters has changed drastically. Is this just because we know the name of one daughter?
I will put the others and their solutions up on my blog:
https://agrumpyoldphysicstechnician.wordpress.com/
in due course.
Marja Erwin @ 13: The odds depend on Monty Hall’s behavior.
If he generally picks the best remaining door, then it’s better not to switch.
The ENTIRE point is that he shows you where a zonk is. This DOES NOT require he knew where it was as the only other possiiblity is that he shows your where the car is -- an outcome that doesn’t matter in the context of the question, you’ve already won.
The question is entirely unconcerned with a outcome where the host simply gives you the car.
Rob @ 13:Either way, showing the zonk teaches you nothing. Switching is always the right move.
Yes, switching is always the right move -- but that’s because being shown the zonk DOES TEACH you something, otherwise you’d have no information to inform your decision.
There’s somerthing going on with peoples insistence that Monty KNOWS where the zonk is, that is some kind of misunderstanding. The idea he knows what he’s doing is important has nothing to do with your decision, it’s incidental to events that a predicate of the question is that he shows you a zonk. Why he does it is immaterial -- we all know he did it because he knows wherew the zonks are, but that’s got nothing to do with the extra information he gifts.
It’s the FACT he shows you a zonk that’s important. The problem can be run forever with Monty making a random choice -- but as all instances in which he shows you the car are dropped (because you’ve just won and have no need to cotinue) the are irrelevant opt question of whether you ought change your choice.
Fentex @10
This bill: https://twitter.com/hilarybennmp/status/1168560598650621953?ref_src=twsrc%5Etfw
This isn’t about Brexit specifically (at least, that’s not why I’m protesting) -- this is about government abuse of power. Michael Gove has refused to confirm that this government will obey the rule of law and respect parliamentary sovereignty, and that’s not an uunreasonable interpretation of what Johnson said yesterday evening also. Many of those criticising or scorning those of us protesting would themselves be portesting were a Corbyn government trying the same tactics to, for example, force through economic reform. Here’s the thing: I’d be protesting against that as well. Even though I’d probably agree with most of the policies, government abuse of power (and refusing to accept the rule of law, FFS) is a Very Bad Thing no matter who is doing it or why. If those who sneer at us truly gave a damn about the UK, if they were capable of seeing past their own self-interest, they’d be standing alongside us, saying “Not this way. Not in my name.” The fact that they aren’t speaks volumes.
Personally, I think Parliament has proven itself incapable of resolving Brexit. Here’s what I think is the best way forward: firstly, get the Benn bill passed. Secondly, a referendum (yep) offering a choice between the deal negotiated by the May government, or cancelling Article 50 and remaining in the EU. Parliament has proven itself unable to resolve this, so a fresh mandate is required. Thirdly and finally, a general election.
For those who are enthusiastic about leaving without a deal, it’s worth remembering that after leaving the EU, the UK will still need to reach a trade deal with the EU. What do you expect the EU will demand in return? Might it be, oh I don’t know, something rather like the agreement reached with May’s government, backstop included?
@16, Alanuk, I think what makes that daughter problem confusing is language, and not anything about probability per se. I wonder if that’s where most people run into problems understanding probability (so long as they understand the key to truly understanding any probability problem is having complete vision of the sample space).
Question 1) How many options in the sample space with one daughter have two daughters?
Question 2) The parents obtain 1 daughter. Now, what is the likelihood another child is a daughter?
I feel like that wording makes things more clear as to how to go about thinking about these problems.
@16, by the way, Alanuk, your answer to problem two would, of course, be incorrect if you lived in a society where all daughters of that generation are named Emma (or even if Jane had a rule that she named all daughters Emma). Your case really relies upon being able to uniquely identify a specific child.
cr @14: You choose A, Monty opens B. The choice then given by Monty is not ‘keep A, or switch to either of B or C’. If it were, you’d be right. The choice is ‘keep’ or ‘switch to other unopened door’. You could play it blindfolded and not know which door he opened. You’d still say ‘switch’.
fentex @18:
You’re imagining a scenario which isn’t presented in any version of the problem I’ve seen. The woman in the video says “The door that Monty opened would always have behind it a zonk”.
If he’s choosing at random, and instances in which he reveals the car are edited out, how do you know that the contestant wins in that case? It could equally be the case that revealing the car requires a “do-over” until a zonk is revealed. In other words, you’re assuming something about “behind-the-scenes rules of the game”.
If a do-over is required, the odds are the same for keeping and switching. So it would matter whether Monty knew where the car is.
We tend to think that the odds no matter what door we choose are 1/3, so how can switching to another door change our odds, as each door has 1/3 chance of being the right one, doesn’t it? But that is not the right way to think about this.
We have 3 doors, A, B, & C. Each has 1/3 chance of being the right door. When we choose one door, say A, that means the odds of it being correct are 1/3. But that means that the odds of the correct door being B or C are 2/3 in total. So when Monty eliminates one of those doors for us (say, B), that means the odds of B or C being the correct door are STILL 2/3 in total. With B eliminated as a correct choice for us, that means the odds of C being the right one are 2/3, because the odds of B or C being correct are still 2/3 in total, just as they were when we chose door A.
I wonder whether the reason that people still with their original choice is really about sticking with the status quo, or about trust? My gut reaction in that situation, under pressure and without a chance to think it through, would be “They are trying to trick me!”, not “Oh, they are trying to help me win”. So on gut reaction to that, I would stay.
Sam N @20:
No, it only seems that language is confusing things. 🙂
The difference between Q1 and Q2 is distinguishability. If you ask how many ways you can put two identical objects into two slots, the answer is one. If you ask how many ways you can put a blue object and a red object into two slots, the answer is two. So, ‘daughter-daughter’ only counts as one possibility, but when the daughters are distinguishable (Emma and not-Emma), there are two.
But I don’t have to play it blindfolded, and I don’t have say precisely what you said. I could also say “I can see that you chose B. So now I want to pick door C.” That’s a clear, valid answer which they would accept.
So, I did learn something when Monty revealed the goat behind door B, because I wouldn’t have known to give my answer in that form (“door C”) until then. At first, I did not know that the sequence of events would be choosing A, waiting for Monty’s choice, then choosing C. Then, I did know. Like I said, that’s not nothing.
I did already know the abstract/generic type of strategy that I’d use. But the actual, specific choice I make really is informed by whatever Monty does, even if your response above is a way to not express it in those terms. If someone’s confused about the epistemology of this scenario, I don’t think it’s me, and what you said before may not be very helpful for them.
Monty also didn’t know in advance that I would choose A. He had to patiently wait until I made it, to learn what I chose, so that he could then make an appropriate choice with B (assuming it has a goat) or C (if it has a goat). We both learned something, and that affected what we did.
Thanks everyone for trying to help. I think the only way for me to come at this is with the 100 doors and thinking of the odds against me. Of 100 doors I have a really low chance of picking the right door. After Monty eliminates all but one, then the prize is either behind my choice or the only one out of 99 he chose not to reveal, and since I was really unlikely to guess right the first time, it’s probably the one he left. Notice how I sneakily avoided using any numerical odds? >:D You think my math phobia is strong, try putting a $ in front of a number and you’ll find me cowering in the corner.
As for the daughter problem, I’m not seeing why gender order matters, since it wasn’t part of the question (she could have had twins or picked them up Baybays-R-Us at the mall). Daughter-son is the same untargeted outcome as son-daughter, i.e., not two daughters. They are not different ways of “having” kids.
But I’m probabilitied out now. This was enough to last me for at least a year. -_-
Ridana @27:
In these kinds of problems, there are almost always implicit assumptions. In this case, they are; Jane gave birth to the kids; if Jane has a kid, it’s equally likely to be (assigned) boy or girl; NO TWINS. The assumptions should really be spelled out, but if not, the least boring thing to do is assume the simplest/most common scenario. You could spend all day thinking up complications (two Emmas, twins, Monty’s just guessing, the dice are rigged, etc), but (for me anyway) that’s not very interesting.
Gender order matters because probabilities depend on the number and type of distinct outcomes. For example, if you toss a coin three times, what is the probability you’ll get two heads and one tail? To get the probability, you have to count the distinct outcomes, each of which has the same probability;
h-h-h
h-h-t
h-t-h
t-h-h
t-t-h
t-h-t
h-t-t
t-t-t
So there are eight distinct possible outcomes, each equally likely. Three of those have two heads and one tail. So the probability of getting two heads and one tail is 3/8.
The proper solution to the Monty Hall problem hinges on a detail that seems irrelevant but is actually extremely important: That the host always picks a non-prize door. This is the classic Monty Hall problem. In this, you have a 2/3 chance of winning if you switch. I find this incredibly easy to understand, especially if you generalize the problem to a large amount of doors.
This detail doesn’t seem like it should matter, but it does. The related Monty Fall problem has the host not knowing which door has the prize, and picking at random. In this game, you have only a 50% chance of winning if you switch.
Why the difference? Because it is possible to completely lose the game if the host reveals the prize, basically not even giving you a chance to choose. The chance of that happening is 1/3. (2/3 chance that you won’t pick the prize, then 1/2 chance the host will pick the prize door).
Why is it different between Monty Hall and Monty Fall? Because every situation in Monty Fall where the host reveals the prize would have given you a guaranteed switch win in the original Monty Hall game.
If you don’t believe me (but are willing to trust my random spreadsheet from the internet), I made this simulation in Excel 3 years ago because while I had no trouble understanding Monty Hall, the Monty Fall problem is what really screwed with my head. https://drive.google.com/open?id=0B80Hh3Ka3Ob_T1IwWTNzVkhFZ2M
Essentially, in Monty Fall there is a 1/3rd chance that you pick the prize door, then there is a 100% chance the host opens a non prize door. There is a 2/3 chance of picking a non-prize door, then a 50% each of the host picking either the prize or no-rpize door to reveal. So there are 3 equally likely outcomes to the Monty Fall problem, only 2 of them let you even continue to the switching choice, and of those two outcomes one has you picking the prize door and the other doesn’t. Stay or switch, your chance of winning is 50%.
Ridana:
Right. It doesn’t matter here, and the math is not hard. Assuming boys are equally likely as girls, the answer is just 50% in both cases. (Only two genuine possibilities, and it’s 1/2 of them.)
One could ask a different type of question, for example, about the chances that Emma (or the unnamed child in Q1), who’s known to be a daughter and one of only two children, is the first/oldest of two daughters or the second/younger one. Of course, it may be that the other child is a son (ranked/ordered however you want), not a daughter. Here, you need to count these permutations as distinct possibilities.
But in the questions from #16, the point is to count unordered sets, where different orderings are regarded as equivalent (which is not to say “identical”). What’s being counted then is not what we were counting above, so there is no such distinction being made (for the purposes of the question). It’s just irrelevant information which you can safely toss out, apparently put in there to trick you … much like naming the daughter “Emma” the second time, as if that made any difference at all.
Do you disagree with any of that Rob? The questions stipulate, with probability 1, that the unnamed child (from Q1) or Emma (from Q2) is a daughter. We don’t have to calculate how likely it is that she’s a daughter, because that was given to us. There’s just the one mystery child to worry about, and you may assume that it’s (roughly) 50/50.
cr @30: If you’re saying that the answer in Q1 is 1/2, then I do disagree. Different orderings are not equivalent.
Consider 100,000 randomly picked families with two kids. About 50,000 will have eldest kid a boy. 25,000 of those will have second kid a boy. Discount them. So, 25,000 with elder a boy, younger a girl. 25,000 with elder a girl, younger a boy. 25,000 with both girls. So, 1/3 with two girls. This is the scenario in Q1.
Now, with the same 75,000 families with at least one girl, choose only those with a girl called Emma, assuming (say) 1/100 girls are called Emma. That will give 500 families which have a boy and a girl (250 + 250), plus a number from the two-girl families. Of those 25,000 families, 250 will have the elder girl called Emma, 250 will have the younger one called Emma. So 500 from them. Thus, of the 1,000 families with a daughter called Emma, 1/2 have two daughters.
No, it’s not Q1. The last bit suggests you’ve misinterpreted something, since you talk as if there is something new that you’re setting up for the next paragaph. I’ll quote it again:
So, we do have a scenario in Q1 with at least one girl, with no talk of younger or older, their order, or what have you. The only difference in the first question is that the girl you were told about wasn’t given a name, just called “a daughter.” I’ll quote Q2, again for convenience:
It’s the same problem, except for the phrase “named Emma.” If I’m the one who’s misinterpreting or misunderstanding something … okay, it happens. I’m very sure this time, but I’ve been fooled in the past, so I’ll leave the possibility open.
Anyway, we both know that’s not how the math works: giving her a name won’t do that.
I’m convinced you don’t know how the math works. I’ve explained it about as simply as it is possible (for me, anyway) to do so, doing all the counting for you. Maybe someone else has the patience and energy to pick up the baton…?
@32, specifying a daughter makes all of the difference, making the probability 1/2. Without specification it is 1/3. This really can be a bit mind bending. It does rely on the assumption that the name Emma specifies a unique individual (which is typical convention). But comment 16 drew out the complete sample spaces for you.
Question 1: you can have son-son, daughter-son, son-daughter, daughter-daughter. Of the four possibilities, only 3 have at least 1 daughter. Of those 3, only one has 2 daughters.
Question 2: you can have Emma-son, son-Emma, Emma-daughter, daughter -- Emma.
Do you see how the specification in the second case gives you two ways to have 2 daughters now?
Question 2 in my opinion is the easier one to grasp. A couple has a daughter, at this point the probability of sex of the other child is independent of that child.
By having a generic daughter that could be in either position, you no longer have daughter-daughter in two unique ways. It is a bit mind bending due to the very small change in wording and how that influences the conditional nature of the probability problem.
I’m not sure how to make this more clear. I think that is sort of up to you to demonstrate to yourself.
I think the surveying a population, like Rob explains in 31 can be a good way to think about it if the straight up brute force of looking at the sample spaces isn’t working for you…
I’m convinced you cut off the end of my sentence after the colon, because you obviously did. Most people try to at least read a full sentence before they zone out.
Suppose that the daughter specified in Q1 does have a name, just like the one in Q2 does, and what’s going on is that you just don’t happen to know what it is. Does that matter? Tell me whether Q1 and Q2 should be analyzed differently, with different math, etc. And if you think so, try to tell me why that is.
I quoted them for you. Perhaps read them again and tell me what you see? I see literally the same question printed twice, with the exact same statements to set up the exact same scenario, except for the extra bit of information giving a name in Q2. You took them to be substantively different mathematical questions. I’ve studied a lot of math, but I’m definitely not sophisticated enough to detect a such difference. I think it’s because there isn’t one. And I think that if you’re seeing one, it’s some kind of confusion or hallucination, because I’ve read it far too many times now and am losing patience with this.
To be clear, your calculations themselves are not the issue, so don’t bother — no problems with them, when removed from context or taken in isolation.
Instead, what’s baffling to me is that you would be claiming to derive two different answers to the same question. Either they’re different questions (according to you, somehow, which you’ve failed to explain) or else you’d have a plain contradiction, because it can’t be that 1/2 = 1/3. I don’t really care which one it is, so take your pick.
Seriously, it’s like saying that when Monty opens up door B, the goat’s name is Frederic.
Or *waves hands* in a “different” scenario, the goat’s name is never uttered, whether or not it happens to be Frederic.
I do not see the relevance.
That’s the thing. And this is why I said the problem isn’t the calculation, the problem is our language. That is what is tripping you up. That naming of a daughter is doing something drastic to the mathematics here. That is the entire point of this exercise.
By naming a daughter, by specifying an identity, you have completely changed the sample space. This is no longer a generic category: daughter. This is now a specific individual, and now the question is really what is the probability that the other child is a daughter?
That is drastically different from the first question, which does not specify an individual. In that first case the problem is basically down to making two coin flips. Of the cases where at least one is heads, what is the probability of two heads.
The second question is given that we have a coin that is heads, what is the chance the other coin flip is tails. This is from the massive amount of information that comes along with naming a child--it has specified one of the coin flips, changing the very underlying nature of the question.
I don’t know how to make this any more clear to you, other than you are focusing far too much on the tiny changes in the words of the sentences and not reflecting on how massively they change the perspective the problem takes on. Specifying an individual daughter changes the very perspective the problem takes on. That is why there is such a difference in the sample space compared to treating daughter merely as a simple category.
Bob @ 23; You’re imagining a scenario which isn’t presented in any version of the problem I’ve seen. The woman in the video says “The door that Monty opened would always have behind it a zonk
That’s true, but I was only trying to point out that it doesn’t matter what Monty knows -- sure, he knows where the zonks are and picks one to show you but it doesn’t matter that he knows, it only matters that he does so (which given as a predicate of the question).
I’m trying ot find a way to make it clear that while it’s true for the question to take the shape it does Monty will have known where the zonks are it doesn’t matter to the contestant whether he did or not -- for the reasoning on whether to switch or not to hold true it only matters you be shown where a zonk is behind a door you did not choose.
The question could be “After you choose your door, another is opened at random and reveals a zonk -- should you change you choice?”.
Yes, because it converts your 1/3 odds to 2/3 odds. That your first choice was, in a statistical sense 1/3 car, 2/3 zonk matters.
With the Monty Hall problem it has always been a puzzle to me why anyone would want to have an expensive, polluting, noisy automobile rather than a nice friendly goat.
But it *does* matter what Monty Hall knows and habitually does.
If Monty Hall knows and habitally chooses the best remaining prize-- (for example, you may *not* be able to switch to the one he shows)-- and this time he shows a bad prize, then you have no reason to switch.
If the explanation amounts to “3 doors” and “this time he shows a bad prize,” why wouldn’t people reach this interpretation?
cr @35:
The difference lies in this, and I’ve bolded the important sentence:
Suppose the ratio of girls called Emma in a random sampling of unrelated girls is 1/M. So, in a group of N₁ families with one boy, one girl, the number with a girl called Emma is N₁/M.
For two sisters, the probability that one of them is called Emma is 2/M. So, in a group of N₂ families with two girls, the number with a girl called Emma is 2N₂/M.
If you can convince yourself that the bolded sentence is true, that’s most of the work. Either way, that’s it from me.
I used to watch the Monty Hall Let’s Make a Deal.
It was almost never the case that there was only the offer to switch doors.
After the zonk was revealed, Monty would almost always offer another choice.
Something behind a curtain or in a box or an envelope of cash, etc.
What does that do for the odds?
Me @41:
I think this is only a very good approximation, not exact.
For example, if 1/2 of all girls were called Emma (so M=2), this would say that all families with two girls have one called Emma, which is nonsense unless there is only one other name to choose. It’s only exact if all names occur with the same frequency, but it is a very good approximation if all frequencies are much less than 1.