I agree that puzzles alone shouldn’t e.g. determine whether you’re a good fit for a job, though. That’s one of the more annoying parts of software interviewing.
Eg., rather than asking about the risks of A,,B,C given various probabilities etc. ask them to make bets in a highly familiar environment with a resource that has uniform marginal utility to them... so eg., "suppose you were at home and your friend does..., how much of your time on a sunday would you bet to do... "
You find that when "the very same question" is asked in highly familiar terms people get it right.
Then this investigator-academic should ask: what features of their own puzzle induce the kind of mistakes they see?
In my experience its often that people are far less socially incompetent than questioners, so put "interpretation & trust" priors on terms/presentations of questions that mean they don't parse the presentation into the problem the investigator has in mind.
People who write puzzles tend to be the most bureaucratic sort of literalists who have profoundly eccentric modes of interpretation
The problem that most people are solving is: "what do i say to make this person/question go away"
This is it.
Never attribute to incompetence that which can be readily attributed to apathy.
No, because the point is that they always hinge on an arbitrary distinction to give one answer. But if you made a different arbitrary distinction you'd get a different answer. And the arbitrary distinctions are, well, arbitrary. They reflect neither truth nor capability. Just whether you can read the questioner's mind as to what arbitrary and intentionally unstated assumptions they are making.
No thanks.
That said, if someone can't fathom the most widely used symbolic languages humans use (math, logic, language, etc) they probably do have a cognitive deficit of some sort when compared to those who can.
If you are a primitive farmer abstract thinking isn't really useful to you. Everything you deal with in your life, except religion, can almost entirely be dealt with absolutes with little in the way of abstractions.
If it rains at the right time then you can have a good harvest. If the weather is bad then it sucks. If there is animals threatening your crops or herd you need to take steps to deal with them.
There is a lot of logic in dealing with these things. You have to know the seasons, know the stars, know the dirt, etc. You have to understand the life cycle and manipulate the behavior and biology of plants and animals at the right stages in their lives. Things have a logical sequence and there are direct consequences that are predictable from events and your actions.
Where as in modern society you have been conditioned to think in terms of hypothetical and abstractions through being exposed to testing your entire life.
You first need to know how test questions work before you are able to answer them accurately.
For a person who isn't exposed to this then the whole affair of asking hypotheticals and assuming imaginary situations with specific rules that don't actually apply to the present reality is very confusing.
They don't even understand the question. So, of course, they are going to suck at answering them.
And ultimately that is all IQ testing measures.. your ability to take tests.
I agree, but would also say that you should be capable of learning to understand those questions. For example, If you can't speak English, you'll be bad at reading books in English. If you were never taught math, you'll be bad at math. Similarly, If you never learned to reason you'll be bad at solving logic puzzles. It's almost tautological.
However, if a person is incapable of learning to do one of those things, despite the majority of the world being fully capable of doing it, they probably have a cognitive deficit.
> And ultimately that is all IQ testing measures.. your ability to take tests.
I disagree. I think it measures how well you've learned to reason, though I do agree that reasoning is a learned skill for most people.
To speak in your analogy, people walk around with different maps of the same territory and realizing this is the self-awareness mjburgess is talking about.
- the local Russell hater
I used that phrasing to drive home the idea that logic is not some inherent aspect of nature, or even fundamental to the way humans perceive the world.
"All" bring a common colloquial term doesn't have a strict set theory definition here. It is reasonable many people think zero hats is means the lie is in this very first word.
A lot of people will consider "all" to implicitly mean 1 or more, while I think strict logicians will map colloquial all to 0 or more.
All mat imply colloquially 2 or more as well, as why bother say "all" if you had one hat in the truthful sense
"My hats" contrasts with the "has a hat" because having a hat in your possession that you could have borrowed does not confer ownership that the word "my" can imply.
So great, a three letter word and a two letter word and we are knee deep in ambiguity.
They could be wearing the hat to try to publicly locate the true owner who might say "hey I lost that hat at x".
"Are green"... Green as in vegetable? Green as in the specific wavelength defined as green and not lime or some other named shade? Completely green dyed being undermined by a black spot or a pattern on the hat?
Imo zero hats of ownership is a viable lie to the statement, as is having one red-green hat.
Does that actually happen in academia? It seems to mostly be a social media thing.
Suppose you're writing a paper what do you write: option A) Average People Cannot Understand Probaility!?!?!, option B) Inexperienced test takers with unfamiliar notation fail to grasp meaning of a novel question; option C) survey participants on technical questions often do not adopt a literal interpretation of question; D) etc. etc.
In general researchers are extremely loath to, or poor at, recognising there's 101 alternative explanations for any research result using human participants and 99.99% of the time just publish the paper that says the experiment evidences their preferred conclusion.
The contrapositive is a rule that says that "A => B" is the same as "not B => not A". This is very confusing to people, and few can follow verbally why it works.
But here is a fun experiment. People are presented with a selection of envelopes, all face down, and are asked to verify the fact that, "All unstamped envelopes are small." They immediately begin turning over the large envelopes, then have trouble explaining their (correct) reasoning!
Here is a correct implication process for their actions.
"All unstamped envelopes are small." => "unstamped envelope => small envelope" => "not small envelope => not unstamped" => "large envelope => stamped"
At which point it is easier to just check the large envelopes!
It really isn't. The problem is usually given in this form:
> Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice? [1]
The only correct ansewr to this question is "I don't know." People who answer yes are just taking way too many standardized tests.
The green hat problem hinges on subjective interpretations of the meaning of both "liar" and the different ways in which the liar's sentence may be false. It may be false because the liar owns many hats, none of which are green. Or they own many hats, only some of which are green. Or they own no hats. These are all reasonable interpretations of how the sentence might be false, and the answers presented are not necessarily mutually exclusive.
If Monty knows the door with the prize and is aiming for the game to continue, then you should switch. (This is the usual argument.)
If Monty doesn't know where the prize is, then you learned nothing. (Monty's result was luck, and he can't impart information that he doesn't have.)
If Monty knows where the prize is and wants you to lose, absolutely don't switch. (Monty will only drag the game out as a way to try to make you lose.)
The reasoning behind these statements is completely solid, and there are no hidden assumptions being snuck in.
You can't just sneak in the assumptions. You have to state them somewhere.
You specifically need information that we haven't been given about the counterfactuals. What might Monty have done in other scenarios that we have not yet observed? We don't know. And we're not actually told. That makes that an implicit assumption that wasn't specified.
Edit: Furthermore, I don't think that author's solution to the Crawl problem is correct. When the host eliminates a door, either you will get information that says you should switch, and you'll win 100% of the time; or you won't get information, and you should still always switch and win 2/3 of the time.
That's the missing assumption. I would say assuming that people are perfectly random falls into the "standardized test" category.
> you won't get information, and you should still always switch and win 2/3 of the time.
You always get some information, the set of possible results becomes narrower, so saying the probabilities don't change is not sufficient. Not a good idea to discuss the problem in informal language though.
That was when I realized many people just memorize the Monty Hall's solution without understanding it, a.k.a. "standardized tests".
Edit: from your other reply I see your beef is more with the wording.
Edit edit: although I disagree with your point even then. But it seems to lead to rather fruitless argument, so let's leave it here.
Yes, it's also correct that you don't "know" the result, and you might prefer goats to cars (even then you should probably sell the car and buy several goats), but there's a reasonable enough interpretation of "advantage" that you shouldn't dismiss the problem outright.
But anyway I'll link the relevant section from Wikipedia:
https://en.wikipedia.org/wiki/Monty_Hall_problem#Other_host_...
> The version of the Monty Hall problem published in Parade in 1990 did not specifically state that the host would always open another door, or always offer a choice to switch, or even never open the door revealing the car. However, Savant made it clear in her second follow-up column that the intended host's behavior could only be what led to the 2/3 probability she gave as her original answer. (emphasis mine)
My point was that when people ask this question, they often word it like the very 1990 version did, lefting out this critical statement (which Savant considered needed as well, therefore she clarified in the follow-up column), making the question ambiguous.
(Although Savant also said "Very few (out of people who said is 2/3 is wrong) raised questions about ambiguity"... so perhaps people are actually just bad at probablity...?)
The problem is that you're presented with two envelopes with 2 real numbers inside. You randomly select one, then look, and try to guess if you got the larger number. It doesn't seem like you can do better than even, but you can!
Unfortunately everyone hates the answer. Which is that you make up a random number and pretend it is the other one. Your odds of being right are
50% + (probability of choosing between the numbers) / 2
Which can always be strictly bigger than 50%. (Though possibly by only a little amount.)
Special case. If those numbers and yours were all independently randomly chosen from the same distribution, you'll be right 2/3 of the time.
The question leaves that distribution completely hidden, and your answer smuggles it back in. That feels less like a counter-intuitive math/stats question and more like a badly worded gotcha.
We can't have an uniform distribution over all real numbers, so it's quite pointless to discuss if looking into the envelope gives any new information, cause we don't even know the distribution yet.
There are no hidden conditions. It is just a shocking result that we don't expect.
If your solution is the same as this article's, it's plain wrong. Even the natural number case is plain strong.
It's very easy to demostrate as well: consider a trivia case where the distribution is just {P(1)=1/3, P(2)=1/3, P(3)=1/3} and you see 2 in the first envelope. There is no strategy to get a better chance than 50%. Therefore, any strategy that gives a better chance than 50% must implicitly make an assumption over the initial distribution (and therefore excludes a distribution like {P(1)=1/3, P(2)=1/3, P(3)=1/3})
Actually the article is even "wronger" than this, because "started A" and "switched" aren't independent and one can't simply use the product of their probability. The above example is a quick way to demonstrate it's not a general strategy without assumption to get >50% winning chance. Similarily, one can just use {P(1)=1/3, P(2)=1/3, P(3)=1/3} (this is a valid distribution over real numbers!) to demonstrate the real number strategy isn't general.
Again, for both natural number and real number case, the discussion over strategies is only meaningful is we know something about the distribution.
Interestingly, this article is wrong more or less in the same way as believing switching does give you more expected value in the original "twice money in another envelope" variation.
Edit: For people who are interested in the switching strategy, check Randomized Switching in the Two-Envelope Problem (2009). Spoiler: full of discussion over the initial distribution.
Seeing 2 is only one of the many possible cases. You haven’t calculated the total probability.
I don't actually care how you convince yourself. But the explanation is right. If your random number is outside of the range, you've got even odds. If it is inside of the range, you've got 100% odds. As long as there is a positive probability of being between, you've got strictly better than even, by half of the probability of being between.
Many, many distributions guarantee positive odds of being in between. The one I chose for my program was:
(log(rand) * (flip_coin() ? 10 : -10 ))
The strategy is straightforward and bulletproof (if you allow a random generator of real numbers, otherwise you may keep tossing coins indefinitely): keep tossing coins until you get tails. If the number you saw is less than the number of heads you got, you don't switch.
For the simplest case assume that one envelope always contains 1 and another always contains 2. You choose one envelope randomly, so in 50% of cases you get 1, which you switch in 50% of cases. And in 50% of cases you get 2, which you switch in 25% of cases. Hence, you pick the higher number in 62.5% of cases. The same works with any numbers N, M; or any complex distributions; or even real numbers with a bit more complicated strategy. You don't have to know whether you are between two values in advance, you just have to guess.
In other words, I'm merely trying to be informative.
I thought you meant the strategy can make the winning chance always >50% even after the player opens the first envelope, which isn't possible.
However you actually meant the strategy can make the expected winning chance >50% before the player opens the first envelope, for any well-defined distribution of real number, even the distribution is not known to the player, which now I realize is true.
(I haven't thought through some edge case like Cantor distribution, but now I incline to it's true not just for "many distributions". Of course for a discrete distributions, we need to specifiy the two envelopes can't have the same number. Besides that, it seems to hold true for any distribution?)
But before you pick, your odds were still bigger than 50%. Just not by much.
It's not saying that after the player see the number in the first envelope, the strategy guarantees a >50% outcome.
It's saying that give any distribution, over all possible outcomes, >50% times the strategy will end up pick the larger number. You can say this >50% is the expected winning chance before the player see the number in the first envelope.
I'd say this is "intuitve" because, if your strategy can guarantee "when the player see a large number in the first envelope, he's less likely to switch than if he saw a small number", it would be better than blindly switching by coin toss. So intuitively such a strategy exists.
The only "trick" here is that since the player doesn't know the initial distribution, they can't tell "how large counts as large?" therefore they needs something that preserves some property over the whole real number line. That's why the strategy involves sampling from a another distribution whose PDF is non-zero everywhere.
Let's play this game exactly once.
You choose two unequal real numbers. I don't know what they are, and I don't know the distribution from which you choose them. You write them down and put them in separate envelopes.
I'm allowed to choose one envelope and open it to see the number inside, and my job is then to say which envelope holds the larger number.
I claim I have a strategy now which lets me win strictly more than 50% of the time. My strategy is this.
I choose a real number R at random from a distribution that has dense support. In other words, for any two reals, L and U with L<U, P(L<R<U) > 0. This is easy to do ... one method is to list the rationals, positive and negative, then roll a die, discarding numbers until you get a 6.
Now I flip a coin and thereby choose an envelope at random. I proceed by assuming my chosen number is between your two numbers. There is a non-zero chance this is true ... call it e. So e>0.
If I'm wrong then my choice is 50% ... probability is 1-e.
If I'm right then my choice is 100%. ... probability is e.
Combined, my chance of being right is 0.5(1-e) + e = 0.5+e/2, which is strictly greater than 50%.
You can make it as small as you like, and if we play the game repeatedly then you can make it approach 50%. But as it stands, with a one-off game, I can win with a probability that depends on your chosen numbers, but which is strictly bigger than 50%.
I am a mathematician, so please bear with me when I try to explain how this can work.
The rational numbers are countable, and that means that I can write a list of them. There are several ways of doing this, but personally I like the Calkin-Wilf tree[0]. That only gives the positive ones, but we can include zero and the negative ones by interleaving them.
So, whatever interval you choose, there are infinitely more reals outside the interval as inside (by that I mean that you can fit an infinite number of copies of that interval up to infinity). So the probability e is not >0, it is effectively 0.
One you have chosen the two numbers, L and U, I note that there are rational numbers in between. Choose one of those numbers, call it M.
M is in my list above. Now I roll a die, discarding numbers from the list until I get a 6. There is a non-zero probability that the number retained is M, so there is a non-zero probability that my chosen number is between L and U. So e is definitely non-zero.
The second problem is, what does it mean to choose a real at random?
It doesn't have to be uniformly at random -- that's the mistake nearly everyone makes -- and the above process does it perfectly well. It only ever chooses a rational number, but that's OK. It's still a real number, it's still a random number, and for any non-empty interval, there is a non-zero chance the chosen number is inside.
... as a human living in the finite universe there are limitations to your choice.
Yes, but that is accounted for in the explicit description of how to choose the number.
Any number you can write using all the atoms in the universe is infinitely outnumbered by all numbers that you can't.
Again, this is accounted for by the fact that we are not choosing uniformly at random.
So ... yes, choose from a Gaussian, but then you have to tell me exactly how.
That's tricky.
Yes, it's established there isn't "uniform distribution over all real numbers" without violating axiom of probability. You're 100% correct on this.
But it doesn't make Colin's solution wrong, because e > 0 for any* well-defined distribution.
> Which is a different problem than stated originally
There are two ways to inteprete the original problem:
A. The numbers are truly randomly picked over all real numbers.
B. The numbers are picked from a well-defined distribution which is unknown to the player.
Since A. is invalid mathematically speaking (without changing the commonly accepted definition of probability), it's reasonable to only consider B., in which case, Colin's solution is correct.
I made a more intuitive explantion on why a strategy better than coin toss exists here: https://news.ycombinator.com/item?id=42372972
*: More strictly, any distribution that guarantees the probability that the two numbers in envelope are the same = 0.
This is one reason that probability theorists have learned to be more careful in stating the double money version to rule out the solution that I gave. But my version has been in the literature since, I believe the 1950s. Under, as I first encountered it, the same exact name. (I encountered it from Dr Laurie Snell at Dartmouth College in the mid 1990s.)
You need to (at least) add this statement to the original question:
> The host must open a door in any situation, and you know this rule.
Because "host" in daily language is a human being with agency to choose whether to not open a door. Without this statement, the original problem is actually a game theory problem with two players. The host might be playing a strategy where he only opens the door when he knows you picked the car at the first place, or any other strategy.
The above statement removes the ambiguity and that's what the original author meant (but failed to put in words).
(If you're interesting in this, the wikipedia page linked above actually contains a quite extensive discussion in history over it)
https://web.mit.edu/rsi/www/2013/files/MiniSamples/MontyHall...
In the case of a logician and the properties of the elements of the empty set the frame of mind of the examiner is probaly going to be about using algebraic logical connectives.
For another nice example I can quote [0] via [1]
> Luria: All bears are white where there is always snow. In Novaya Zemlya there is always snow. What color are the bears there?
> Peasant: I have seen only black bears and I do not talk of what I have not seen.
> Luria: What what do my words imply?
> Peasant: If a person has not been there he can not say anything on the basis of words. If a man was 60 or 80 and had seen a white bear there and told me about it, he could be believed.
This is a more extreme case, but in my opinion it is the same phenomenon of being asked to take external things as true and work on them.
[0] https://languagelog.ldc.upenn.edu/nll/?p=481
[1] https://www.astralcodexten.com/p/somewhat-contra-marcus-on-a...
I've rarely encountered a case where it is isnt an extreme lack of self-awareness in the questioner -- eg., being extremely overfit to language/notation/etc. localised to their own area of expertise.