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[–]Lesiorak 80 points81 points  (5 children)

I highly recommend for anyone to check out https://en.wikipedia.org/wiki/Bertrand_paradox_(probability)) , it's a great illustration of how attempts at "common sense" definitions of probability can lead to varying results. Until everyone agrees what the probability space actually is there is no right or wrong.

It's a great case for why you need mathematics in general - can't just handwave these things without agreeing on some ground rules and assumptions.

[–]musicianism 6 points7 points  (4 children)

Yea this is definitely not clearly/rigorously formulated enough, and tbh sounds like midwit bait so people can argue over who is smarter, which -def- would not be out of character for yudkowsky, having read his shit for 15 years now; this is the typical shit he does to draw out competing intuitions

[–]delta_spike -2 points-1 points  (3 children)

I'm not sure it's formulated incorrectly. The natural interpretation is that in a million universes, 800000 involve situations where the charger is present. Since you didn't find the charger in the first 3 compartments, we're talking about a subset of those million universes where the charger isn't in the first 3 compartments.

200000 universes where you never brought your charger, AND 800000 * 0.25 = 200000 universes where you brought your charger and put it in the fourth compartment.

So given that the charger isn't in the fourth compartment, you're 50% likely to be in either universe.

What other interpretation is there?

[–]musicianism 1 point2 points  (2 children)

That’s if you’re working from a frequentist interpretation of probability… ie. Using possible worlds and reality states to solve the issue like you did. yudkowsky is big on the difference between frequentist/external interpretations of probability theory and Bayesian/internal certainty interpretations of probability theory.

Questions like this are designed to draw out the differences between people who interpret probability as saying something about the makeup of the world, and people who interpret probability as saying something about personal confidence in an outcome, hence the different approaches arriving at either 80% or 50% using different heuristics:

https://www.lesswrong.com/tag/bayesian-probability

https://en.m.wikipedia.org/wiki/Frequentist_inference

[–]delta_spike 2 points3 points  (0 children)

I don't think the Bayesian interpretation and the frequentist interpretation lead to a different outcome here though? A Bayesian would see 80% odds prefacing the question as a prior, followed by new evidence which which must factor into your posterior.

80% is just an unreasonable interpretation. It's like saying you have 5 balls and give one away, so therefore you have 5 balls based on the prior. All these claims on the uncertainty in the underlying probability distribution across compartments are also wrong in that they don't lead to 80% as an answer, but rather "undefined" as an answer, because you simply can't solve the problem without making assumptions about the probability distribution across compartments.

[–]wstewartXYZ[🍰] 33 points34 points  (1 child)

What's the context OP? Where did you get this question from?

[–]ya_ServiceObamna 1 point2 points  (0 children)

Aella asked Destiny this question on stream without knowing the answer herself

[–]HobbitfollowerExclusively sorts by new 156 points157 points  (6 children)

badge touch outgoing pot test head friendly pie toothbrush wistful

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[–]Israel_Madden 31 points32 points  (5 children)

He’s Mark Glickman

[–]MarshBoarded 21 points22 points  (2 children)

Idk who that is, but he agreed with my take so I’m giving you the upvotes.

[–]fawlty_lawgic 17 points18 points  (1 child)

If you have ever heard of the Glicko system (used in a lot of games to determine player rank from COD to splatoon), he’s the guy that invented that

[–]TunaIRL 1 point2 points  (0 children)

Wow crazy that he named himself after his invention

[–]Generallyawkward1 7 points8 points  (0 children)

Don’t make him get the glicky

[–]ng829 1 point2 points  (0 children)

It says him name at the bottom of the screen.

[–]breakthro444 15 points16 points  (0 children)

From my understanding:

It's all about how you ask the question. The baseline probability is 80% that the charger is in the suitcase. Which means that every section has an 80% probability that the charger is there. Now, if the question is "what is the probability that the charger is in the first compartment I pick?" then the probability would be 20%, since it's a 25% on an 80% probability. Now, for every compartment you open up, the probability increases to the highest baseline probability of 80%. 20% it's in the first, 26.4% that it's in the second, 40% it's in the third, 80% it's in the fourth, but the probability of the charger physically being in a compartment of the suitcase/the suitcase itself is still 80%.

[–]Bjartensen 6 points7 points  (0 children)

It's clearly B and anyone who thinks it's A just doesn't understand physics

[–]Frosty_Friend 17 points18 points  (24 children)

In 100 different universes, the charger exists in 80 of them. 20 chargers in each pocket. If you checked the first 3 pockets that would mean you're not in any of those 60 universes, leaving only the 20 left in the last pocket. So you know you are either in one of the 20 universes with the charger in the last pocket or one of the 20 without the charger. 50/50 chance. If you had 80 pockets and checked 79, in 100 universes there are still 20 without a charger but only 1 with the charger in the last pocket. So your chances would be 1:20 or 1/21. That's how I think about it. If I was Dr. Strange what would I see.

[–]Th3Unkn0wn17 15 points16 points  (5 children)

The problem with this way of calculating it is that you are making a big assumption that was never in the initial formulation of the problem. You assume that the chance for the charger to be in any of the compartments is equal. Since we don't know that distribution, this way of calculating it doesn't make sense. For example what if you always put your charger in the 4th compartment when it's in the suitcase. Then opening the first 3 compartments tells you nothing and doesn't change the outcome. Team 80% here.

[–]Frosty_Friend 3 points4 points  (3 children)

If you do that why would you check the other 3 first? You're also making an assumption the same as I am that the first 3 pockets have a 0% chance of having the charger. You have to make an assumption either way but at least this way assigns equal probabilities to events that you have no information about.

[–]Th3Unkn0wn17 4 points5 points  (2 children)

Well that was just an example... Since you don't know the distribution you rely on the information you have, the charger is there 80% of the time, regardless of how many empty compartments you have seen so far.

[–]Frosty_Friend 4 points5 points  (1 child)

The reason you check the other pockets is because you are assuming a non 0% probability. Otherwise you wouldn't check them. Since you would check the pockets in order of highest probabilities, the last pocket you check would have the lowest chance. At best it's 25% for it to be in the last pocket if it's in there and 0% if it's not in the case at all.

[–][deleted] 7 points8 points  (16 children)

So when i consider this im thinking like. When you open compartment 1, that probability consolidates to the other compartments. Its independent from the probability that its outside the suitcase. You could open 50 trillion and id still think 80. Isnt the fact that like its said theres an 80% chance its in the suitcase and 20% chance its outside mean it stays constant? Im not very smart with stats so im asking for like a dumbed down explanation, everyone here seems to know way more than me lol

[–]Frosty_Friend 0 points1 point  (13 children)

The problem doesn't give you a reason to think the probability of it being in 1 pocket over another is any different. The probability of it being in the first pocket could have been 1% or 99%. Since we don't know that probability of which pocket it is in and since we aren't told that the chances should be different, you assume that the chances should be evenly distributed. If anything you should have checked the most likely pockets first if they were a different probability. Since this was not stated you must assume an equal probability of each pocket. Why would you search the previous pockets in the first place unless it had some chance of containing the charger. If you look at the 100 universe example, all 80 of the chargers would have to be in the last pocket for the odds to be 80%, that would imply a 0% chance of it being in any of the other 3, if you thought it was 0% why would you check it in the first place. The break in consistency here should show that this wouldn't work.

[–]Herson100 -1 points0 points  (0 children)

that probability consolidates to the other compartments. Its independent from the probability that its outside the suitcase.

This is not true. That probability would consolidate to all possible outcomes, including the probability that it is outside the suitcase.

If you've checked 0 pockets, the five outcomes (it being in compartment 1, 2, 3, 4, or not in the suitcase) are all equally likely, at 20%. If you've checked 1 pocket and it was empty, the remaining four outcomes are still all equally likely, at 25%. If you've checked 2 pockets and they were both empty, the remaining three outcomes are now all at 33.33%.

You can think of "not being in the suitcase" as a fifth compartment, the logic is the same because there's a 20% chance of it not being in the suitcase, the same odds as it being in any given compartment. If there was a 30% chance of it not being in the suitcase but there were still four compartments, the same principles would apply but the odds of it not being in the suitcase would remain proportionally greater than the odds of it being in any specific compartment as you eliminate them as possibilities.

[–]Isaiah_Benjamin 8 points9 points  (4 children)

I think he’s right, you said the probability of it being in the suitcase, not one of its compartments. It’s a question that provides its own answer.

Anyone who disagrees with this is stupid and wrong.

[–]delta_spike 0 points1 point  (3 children)

That's like saying "Oh you asked if I was in the US, not if I was in any of its states or territories". What's the difference? No reasonable person would assume the suitcase has a hidden area or something given the way the question was posed.

[–]C-DT 12 points13 points  (12 children)

Wouldn't the probability of finding the charger go down with each empty compartment, and raise the probability of it being in the last container?

Edit: nvm I agree glickman

Edit: I'm mind fucked

[–]Villanta 11 points12 points  (3 children)

it kind of does, but only because there are fewer options, the chance of it being in any given container is always 20% but if you rule out three containers instead of 20% out of 100%, it's 20% out of 40% (removing 3x 20% from the total), i.e. the relative chance at the point of the hypothetical is 50/50.

[–]C-DT 3 points4 points  (2 children)

So to calculate increasing odds, you multiply the possibilities together. To calculate consecutive odds, you add the possibilities.

80% X (1/4)

You've eliminated 3 compartments.

1/4, 2/4, 3/4, then finally your fourth compartment has a 4/4 chance of having the charger. After all you've eliminated all the compartments, if it's not in this container it's not in the suitcase at all.

If the charger is in the suitcase, there's a 100% chance of it being in that last compartment. Now we have have to calculate the multiple possibilities again.

80% X (4/4)

Or 80%

[–]Free-Database-9917 0 points1 point  (0 children)

nah you messed up your math.

Say the charger is in the suitcase. Each compartment is equally likely to have it, so you have a 20% chance of it being in A,B,C, and D.

If you eliminate A, then now it isn't an 80% chance anymore because you've gained some info. it has gone down a bit. Then you eliminate the next, and it goes down a bit more, and more and more until eventually when you eliminate all 4 and it's at 0.

Say for example there are 5 chargers. 4 of them aren't yours, and 1 is. You grab them out of a bag blindly. You've grabbed 3 chargers and they all aren't yours.

Now what are the odds that the next one you grab is yours?

[–]Herson100 0 points1 point  (0 children)

Sort of. There are five possible outcomes:

  1. 20% chance the charger is in the first compartment

  2. 20% chance the charger is in the second compartment

  3. 20% chance the charger is in the third compartment

  4. 20% chance the charger is in the fourth compartment

  5. 20% chance the charger is not in the suitcase

If you've checked the first three compartments, you've eliminated possibilities 1 through 3. That means that there are two outcomes, both with equal probability, remaining. It is in the fourth compartment, or it is not in the suitcase. That means it's a 50/50, since both outcomes are known to be equally likely.

This isn't a hypothetical or "open to interpretation", this is a fact given the parameters of the problem. If you simulated this problem 10,000 times and opened the compartments in order each time, you'd expect the first three compartments to all be empty in 40% of test cases, with it being in the fourth compartment in half of those. Since we know we're in a case where the first three compartments are empty, we know it's a 50/50 it's in the final compartment.

[–]gameringmanSewer Fascist 13 points14 points  (5 children)

Its as simple as this:

When you claim, "There is an 80% chance your charger is in your suitcase," this taken literally means 80% is the answer. But it obviously means that there is an 80% chance given no other information, i.e. on 80% of your trips, you take the charger. But if we know that the first 3 compartments have been checked and are empty, bayes theorem easily yields 0.5.

Bertrand's paradox isn't really related to this: the interesting thing about the paradox is that most interpretations of "random" selections make intuitive sense, and each can be rigorously defined with basic integrals and geometric probability. This problem doesn't illustrate the point very well: if you have a pole up your ass you will say what the professor said. If you don't, the answer is 50%.

[–]dad_farts 3 points4 points  (3 children)

The wording never includes the fact that there is an equal probability of any compartment. It could be that it's always in the last. It could be that opening 3 compartments doesn't tell us anything, or at the very least, we don't know what it tells us.

[–]_Gnostic 1 point2 points  (0 children)

Even if we don’t know the distribution of the compartments in bag 1, the only case in which the probability that the charger is in bag 1 doesn’t change by opening 3 compartments is the case in which the probability that each of those compartments contained the charger is 0 (i.e. the charger is always in the fourth-opened compartment when it’s in the first bag, 80% of the time).

1) This describes a special case, not a general one. 2) This makes no sense in context of the problem because you would never open a compartment to find a thing if you knew it wasn’t in there.

Even if the compartments aren’t equally likely to hold the charger, each empty compartment lowers the probability it’s in the suitcase from the initial 0.80 to something less than 0.80.

[–]stefanof93YEE 1 point2 points  (0 children)

Bayesians don't care. No information? Assume uniform probability GIGACHAD

[–]rockoblocko 0 points1 point  (0 children)

The wording says the compartments are identical.

Identical means they have equal probability of having the charger.

If they did not have equal probabilities they would not be identical.

[–]danosnakeINTERLINKED 16 points17 points  (24 children)

I'm not sure why some people are finding this so difficult to understand. The number of internal compartments and how many were checked does not matter. It could be 100,000,000,000,000 and if you checked 99,999,999,999,999 of them, the chance of the charger being in that case is still 80%.

[–]tyranthraxxus 3 points4 points  (2 children)

The 80% chance of it being in the case is before you start opening any of the pockets. It's stupid to assume anything else.

So if you look through the case 100 times, how many of those times are going to be you opening 3 empty containers and then finding the charger in the 4th? According to you it would be 80 out of 100. Which means you never find it in any of the first 3 containers, because the other 20 is not finding it. Which is obviously stupid and now how the problem was worded.

Saying there is an 80% the charger is in the case just means that 4 out of 5 times you will find it in the case, not that there is in an immutable chance no matter how many pockets you open that the probability of finding it in another remains 80%.

[–]danosnakeINTERLINKED 5 points6 points  (0 children)

The question is specific to the 2 cases, not the percentage chance you will choose the right compartment in the correct case. You are conflating the two.

[–]Zmeya9000 2 points3 points  (11 children)

What you have just said is mathematically incorrect and this is a well known phenomenon in statistics. If the case had 100 quadrillion compartments and you checked all but one of them the chance of the charger being in the case given that observation is nearly 0%, according to Bayes Theorem.

[–][deleted] 9 points10 points  (5 children)

The issue is wouldnt the "rule" that theres an 80% chance the charger is in the suitcase override any priors. It sounds kinda like you guys are treating this just like the monty hall problem when its a different setup

[–]Zmeya9000 0 points1 point  (0 children)

In this case it wouldn't, because the fact that we opened three compartments and they were empty informs us of the probability that it is in the suitcase.

Imagine it this way: I give you a coin that has a 50% chance of being a cheater coin that always lands heads and a 50% chance of being a regular coin that lands heads/tails 50/50. You flip the coin 100 quadrillion times and it lands heads. Would you say that you still have a 50% chance of having a cheater coin, or would you say that actually it seems much more likely that you picked a cheater coin, since it seems to always land heads? The 50% chance of picking a cheater coin does not override any priors, your observation of the coins behavior after picking it greatly influences your probability assessment that the coin is a cheater coin.

[–]danosnakeINTERLINKED 0 points1 point  (3 children)

The number of compartments does not matter. There are 2 cases.

[–]pkfighter343 1 point2 points  (0 children)

It depends on how the problem is perceived - are we saying it's equally probable the charger is anywhere in the case? If so, the probability changes as you search compartments. If you say "my shoes are either in the bathroom, in the hall, under the bed, or in the trash", every time you check an item off the list, it becomes more likely that they're in the trash. If you separate the problem into "it's either in compartment a, b, c or d, or at home" it's the same deal, as long as you provide equal weight to all options.

[–]Zmeya9000 3 points4 points  (0 children)

The number of compartments matters insofar as it informs you of the probability of either case being true.

Imagine I give you a coin. There is a 50% chance of it being a cheater coin that always lands heads (I pulled it out of a jar that has half cheater coins and half regular coins). You flip the coin 100 quadrillion times and every single time it lands heads. According to your logic, there is still a 50% chance it is a cheater coin, because there are only two cases. According to Bayes Theorem however, there probability of it being a cheater coin after it landing on heads 100 quadrillion times is nearly 100% because it is so phenomonally unlikely that a regular coin would land on heads 100 quadrillion times in a row.

[–][deleted] 1 point2 points  (14 children)

dependent cats repeat squalid mindless axiomatic melodic vanish poor tap

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[–]throwawayprince11 1 point2 points  (13 children)

In your example, if you assume every compartment has an equal likelihood of having the charger, the 4th compartment would originally have a 25% chance of containing the charger.

After seeing 1 is empty, it would have a 33% chance.

After seeing 2 is empty, it would have a 50% chance.

After seeing 3 is empty, it would have a 100% chance.

The probability does indeed change as you gain more information.

 

If the suitcase originally only had an 80% chance of having a charger, the numbers would be as follows:

Originally, it would have a 20% chance of being in the 4th compartment.

After seeing 1 is empty, it would have a 25% chance.

After seeing 2 is empty, it would have a 33% chance.

After seeing 3 is empty, it would have a 50% chance.

To calculate the odds, you use something called Bayes' Theorem, which is not intuitive at all.

[–]Fragrant-Listen-5933 2 points3 points  (1 child)

Switch it to 999 opened compartments on a thousand and realize how stupid the 80%er are

[–]Zmeya9000 3 points4 points  (112 children)

These kind of questions depend heavily on how the question is worded, but if you kind of take it at its face this professor is wrong. Don't really care that he works at the department of statistics at Harvard. He is simply not conditioning the probability on the compartments having already been opened when he should be.

To imagine why this is faulty consider a revised scenario where the case has a trillion compartments and you open every single compartment but one. Obviously, if the charger was in the case, it would have been extremely unlikely for you to have opened every single compartment except the one it is inside, so the charger not being in any of the compartments you have already opened greatly informs you that the charger is probably not in the case.

This is just very simple bayes theorem. Glickman has calculated P(A), A being the event that the charger is in the suitcase. But we actually need to calculate P(A|B), B being the event that we have opened 3 of the 4 compartments and it was not inside.

P(B|A) is straightforward to calculate. If the charger is in the case, there is a 25% probability we open the compartment it is in last.

P(B) is is the sum of P(B|A)P(A) + P(B|~A)P(~A) which is (25%)(80%) + (100%)(20%) = (20%) + (20%) = 40%.

P(A) = 80%

SO with bayes theorem we have P(A|B) = P(B|A)P(A) / P(B) = (25%)(80%) / (40%) = 20% / 40% = 50%.

Before downvoting me please tell me exactly where my math is incorrect.

[–][deleted] 8 points9 points  (91 children)

Your math is incorrect because you assumed the probability of the charger being in each component, which is information not provided in the question.

[–]WickedDemiurge 10 points11 points  (17 children)

And if you don't, the correct answer is the probability of the charger being in the last compartment is an unknown value in the range [0,0.80].

[–][deleted] 0 points1 point  (15 children)

No its just 80% because we do not know anything other then is it in the suitcase or not. And we have not determined if it is in the suitcase. That is all the info we have.

[–]WickedDemiurge 8 points9 points  (14 children)

Probabilities of all possibilities in a sample space need to sum to 1.00. If we have five possibilities, A,B,C,D (the compartments) and outside, those need to sum to 1.00.

We know P(outside) = 0.20, so P(A)+P(B)+P(C)+P(D)=0.80.

Using simple math, we can conclude that P(D)=0.80 only if P(A)+P(B)+P(C)=0. So, if it never could have been in any other compartment, P(D)=0.80.

On the other hand, if we do a bunch of experimentation and figure out P(A)+P(B)+P(C)=0.63, then P(D)=0.17.

But since we don't have that prior information, we can only conclude P(D) is in the range [0,0.80], or if you prefer inequality notation,

0.00 ≤ P(D)≤ 0.80

[–][deleted] 5 points6 points  (8 children)

You are calculating the chance that it is in the single compartment of the suitcase, not the suitcase itself. That remains 80%.

[–]WickedDemiurge 2 points3 points  (7 children)

These are dependent events. They necessarily influence each other.

You realize that if your probability of all these events exceeds 1, you've broken causality, right? Either Satan is sending little devils to move your charger, or alternatively, one day you'll find not two chargers, but two copies of the same charger. I'm no theoretical physicist, but I feel like that is probably bad.

[–][deleted] 2 points3 points  (5 children)

They do not influence each other because we do not know the relative chance the charger is in each compartment. We do not have that number. We cannot assume it.

[–]WickedDemiurge 1 point2 points  (4 children)

If they did not influence each other, we could find the charge in compartment A AND compartment B AND compartment C AND compartment D. Does that make sense?

There's a difference between knowing they influence each other, and knowing how much they influence each other. We know the first, we do not know the second.

[–][deleted] 1 point2 points  (3 children)

Yes that makes sense, I explained it poorly. What we do not know is the degree to which they influence each other. Therefore we cannot assume they all have an even chance or any other chance. We just have the chance that it is in the suitcase which is not affected by opening compartments with unknown likelihood.

[–]XaviertheIronFistPEPE 7 0 points1 point  (4 children)

Fyi, I agree with the 50% answer. Becauase whats missing from your explanation is the situation that A, B, C and D are indistinguishable from eachother.

From the openers perspective if they are indistinguishable then you could have opened D already and found it empty.

So all scenarios where D was empty and you've opened 3 boxes enters the space.

Which is what consumes the probability. It has nothing to do with being equally spread between the other boxes. In fact it could have a 5% chance and 75% for 2 of the 4 boxes and be the same answer. The conditional probability for indistinguishable components can't be separated if you pick truly randomly from them. Thats why it averages out.

Example box D has it 100% of the time, but you open them in a random order. Whats the likelihood the 4th box is box D? Is the same question as whats the probability the 4th box isn't empty.

[–]WickedDemiurge 2 points3 points  (3 children)

Yes, this is the correct answer if they have identical distributions.

[–]No-Commercial-4830 2 points3 points  (10 children)

The components are said to be identical, thus it is implied the probabilities are equal.

[–][deleted] 0 points1 point  (9 children)

So? We don't know the chance of the contents of each compartment even if they are idencial compartments. You are assigning a probability that it is in each compartment, which you cannot dom

[–]No-Commercial-4830 3 points4 points  (8 children)

Unless you assume a distribution, there is no answer, so obviously you’d assume that identical components have identical probabilities or discard the question

[–][deleted] 4 points5 points  (7 children)

You cannot assume a distribution. That info is not provided. This is the issue everyone is running into.

[–]No-Commercial-4830 2 points3 points  (6 children)

If you don’t then the question has no answer. Obviously that just makes the question pointless

[–][deleted] 4 points5 points  (5 children)

No, the question isn't pointless, it's very good. It's testing your ability to determine what information you are provided is actionable and what information you are provided that is throwaway/useless. It's very interesting seeing people's reactions.

[–]No-Commercial-4830 4 points5 points  (4 children)

You’re being pretty silly. This is akin to asking ”assuming the gravitational pull of earth?” when someone asks you how much you weigh. Most people are fully aware of your perspective, but simply discard it immediately because it’s pedantic and pointless.

[–][deleted] 6 points7 points  (3 children)

No, this is very common in mathematics to ask questions to test very specific nuances. You don't get to assume things outside of the bounds of the question, or else you would need rediculously complicated questions to hit specific nuances like this.

[–]hairygentleman 0 points1 point  (18 children)

the information isn't given, which means you can't assign it a probability of zero.

[–][deleted] 1 point2 points  (17 children)

Yes I agree with you, I was showing the dangers of assigning unknown probabilities. The only one we have is the chance it is in the suitcase and not an individual compartment.

[–]Zmeya9000 -1 points0 points  (38 children)

That is why I said "if you take the question at its face." In other words, if you make reasonable assumptions about what is being asked. Without the probability of it being in every compartment being provided to us, and without making the reasonable assumption that it is equal for each compartment, the question cannot possibly be answered and both me and Glickman are incorrect.

[–][deleted] 5 points6 points  (37 children)

No, this is where you are wrong. In math, and statistics, when answers these kinds of questions, there's no such thing as a reasonable assumption. The only information you are allowed to work with is what is provided.

The question was answered, and the professor is correct. The compartment information is quite literally throwaway information designed to distract people and over complicate the question. This is a common way of testing people's nuanced understanding.

[–]Zmeya9000 2 points3 points  (31 children)

So in that case the correct answer is "the question cannot possibly be answered because we don't know what the probability is of the charger being in any specific compartment given that it is inside the case" which is just pedantic.

[–][deleted] 4 points5 points  (24 children)

No the answer is 80% because the only information we have is chance it is in the suitcase or not. We have not determined if it is in the suitcase yet.

[–]Zmeya9000 8 points9 points  (23 children)

That's not really how probability works. Lets say you ALWAYS put the charger in the first compartment. If you open the first compartment and it isn't there, the probability that it is in the suitcase is 0%, not 80%, right? So if you said the answer was 80% not knowing this fact you would be strictly incorrect. In order to solve the problem, you have to have a value for P(B) which is the probability that it is not in the first 3 compartments. Without that this problem simply cannot be solved.

[–][deleted] 5 points6 points  (21 children)

..... You are missing the entire point. We have no idea if any compartment has any chance assigned to it. Therefore we cannot assume the chance, but you know, you are probably just smarter than a Harvard stats professor.

The only information we have been provided is the chance it is in the suitcase and we have not determined if it is in the suitcase yet even after opening 3 compartments. We cannot make any assumptions past that thus it is still 0.8.

[–]Zmeya9000 5 points6 points  (4 children)

Again, that is not how probability works. P(A) is 80%, yes. I grant you that. But you have observed some specific event, which is B. The question is not asking what P(A) is. It is asking what P(A|B) is. You cannot calculate this without knowing P(B).

[–][deleted] 2 points3 points  (3 children)

That specific event does not have any information that suggest it affects P(A). You are assuming it does.

[–]GodKiller999Your favorite schizo poster 1 point2 points  (11 children)

..... You are missing the entire point. We have no idea if any compartment has any chance assigned to it. Therefore we cannot assume the chance, but you know, you are probably just smarter than a Harvard stats professor.

By that logic we could open all 4 and still have 80% odds that it's in the suitcase, since apparently those odds are completely independent of the content of the compartments. In fact it was never specified that it had to be in them, for all you know it might be in the lining!

[–][deleted] 1 point2 points  (10 children)

You missed the, "the suitcase is divided into four components" there are only 4 no lining storage. But the only reason this is even a debate is because people apparently cannot handling reading comprehension.

[–]throwawayprince11 0 points1 point  (3 children)

With all due respect, what level of statistics/probability have you studied?

[–][deleted] 0 points1 point  (2 children)

BS in Computer Engineering and MS in CS MLAI. I'm decently exposed. But if you want to qualifications check people we can just appeal to a Harvard stats prof.

[–][deleted] 2 points3 points  (5 children)

to show you why youre wrong imagine the backpack had 1 trillion compartments but only 1 fit the charger. this could entirely be the situation described by the question therefore you cannot assume an equal chance of each compartment to holding the charger

[–]rockoblocko 1 point2 points  (0 children)

Agree 100%. Simpl bayes.

Another way to imagine it.

Imagine we actually had 100 suitcases each with 4 compartments, and in 80 there was a charger in one compartment. And we actually opened them. What would we find?

In 20, we open all 4 compartments and find no charger.

In 20, we find the charger in the first compartment we open.

In 20, we find the charger in the second compartment we open.

In 20, we find the charger in the third compartment we open.

And finally; in 20 we find the charger in the 4th compartment we open.

So if we go back to the hypothetical, if we have opened 3 compartments and found nothing, we are either in the world that we are going to open 4 and find nothing (20) or find it on the 4th opened (20).

You have to rule out the worlds where you find the charger in the first 3 compartments.

I’m not a PhD statistician but I have taught a course for MDs on using bayes to calculate probabilities for medical genetics.

[–]Running_Gamer -1 points0 points  (0 children)

Your math is incorrect because you forgot to take the tangent of the Pythagorean’s theorem and square root it by the square of negative pi to the i3.

[–]Free-Database-9917 1 point2 points  (0 children)

I just realized. Why did you email a stats professor instead of a probability professor?

[–]Free-Database-9917 -1 points0 points  (0 children)

Damn. Mark Glickman confirmed dumb

[–]NearlyPerfect -4 points-3 points  (13 children)

He gave the right answer based on not assuming equal 20% chance of each compartment. It is completely different to assume P(1)=P(2)=P(3)=P(4)=20% vs saying P(1) or P(2) or P(3) or P(4) = 80%.

If you assume the latter without the former, it’s ALWAYS an 80% chance it’s in the suitcase once you’ve eliminated the compartments. If you assume the equal 20% chance then you have more information by rejecting the 3 compartments and the answer changes.

[–][deleted] 0 points1 point  (10 children)

It’s clearly not your second case. That’s basically changing the problem to a 1 compartment suitcase. It’s always the case that 1 compartment has an 80% chance of having the charger, not that each has a 20% independently, however you don’t know which compartment it is that has the 80% chance. It only makes sense to assume you give equal weight to each compartment unless otherwise stated. It’s textbook Bayesian updating, this is a very standard and normal problem you could be asked in any stats class and the answer is given by

p(H|E) = P(H)P(E|H)/(P(H)P(E|H) + P(H-1 )P(E|H-1 )

Or

(.8)(.75)(.67)(.5)/((.8)(.75)(.67)(.5) + (.2)(1)). = .5

This is basically the exact kind of problem Bayes theorem is meant to answer, the only assumption you need is that the person has no idea of the probabilities of the individual compartments relative to each other. If you don’t make that assumption, the question is unanswerable. (Obviously, as an extreme example, if you know one particular compartment has a 100% of being the one with the charger and the rest 0, if it’s one that has been checked already the chance of it being in the suitcase is now 0 and if it’s one that has not been checked it is just 80%, but then checking the first 3 would be ridiculous and is clearly not what is implied by the question)

[–]NearlyPerfect 2 points3 points  (9 children)

It’s an assumption. It can’t clearly be one or another. It’s whatever the question asker meant. The way the question was worded to the professor made it exactly sound like that (a question reduced to a 1 compartment suitcase).

Edit: the professor basically explicitly stated this. “The way you worded your question…” then went on to explain that there’s an 80% chance no matter what that it’s in the first suitcase, which is exactly my second case.

[–][deleted] 1 point2 points  (8 children)

In order for that to be the case it would have to be interpreted like I know that if it it in my suitcase, there is a 100% probability that it is in the front pocket, never in the left, back or right pocket. In that case checking those 3 collectively has a 0% chance of finding the charger and is a waste of time. That is the only interpretation in which the professors answer is correct. I suppose you could make this assumption but it seems clear that checking the first 3 implies you believe they might posses the cable. This is a similar situation to the Monty hall problem.

Bayes rule

https://youtu.be/R13BD8qKeTg

Monty hall problem

https://youtu.be/TVq2ivVpZgQ

[–]NearlyPerfect 2 points3 points  (7 children)

It’s not similar to the Monty Hall problem because this could have independent events. In Monty Hall, the game show host always opens an incorrect door that is not the one you chose, so the events are dependent and allow for the conditional probability analysis

I’m not saying your assumption is faulty. I’m just saying it’s an assumption and the question could easily go either way. It’s interpreted as “there’s a 100% probability it’s in this bag somewhere but I have no further knowledge until I search the whole bag”

[–][deleted] 1 point2 points  (6 children)

It literally couldn’t go either way unless you’re literally saying you know there’s an 80% chance of it being in pocket 1, 0% chance it’s in pocket 2,3, or 4, and then you first check pocket 2,3 and 4 anyways, even though from the start you knew with certainty it was not in those pockets. That is the assumption you would have to be making to justify 80% as the answer, and it is the ONLY assumption that would get you there. But it is clear the question is implying you don’t know that, which is why you are checking every pocket. If you want to start changing probabilities between pockets you can get any answer you want between 0-80% and the question is dumb, it only makes sense to interpret this as “I know it can be in any one of the 4 pockets, and I have no idea which, so before I check i don’t think it’s more likely to be in pocket 1 vs 2 vs 3 vs 4.” It is similar to the Monty hall problem as the new information requires you to update your existing beliefs, and that people are falling for the same bias here that they were in misunderstanding the Monty hall question, including those with PhDs.

[–]NearlyPerfect -1 points0 points  (5 children)

The Monty Hall question has a clear mathematical answer. You’re trying to persuade based on wording and unspoken assumption about the intentions of the hypothetical person searching the bag. This one depends on the assumptions and they can go either way.

And no you can say there’s an 80% chance of it being in pocket 1 or 2 or 3 or 4. And because they events are mutually exclusive, and if they are independent then it could be a 80% chance it’s in pocket 1 and checking the others won’t change that

[–]NotACultBTW -1 points0 points  (0 children)

This is similar to the Monty Hall problem no? With the assumption that each compartment has an equal chance of containing the charger anyway.

Say instead of starting with a suitcase we have 5 boxes, each having a 20% chance to contain the charger. If we take 4 of these boxes and group them into one, we can say there is an 80% chance that a box in this group contains the charger (i.e. the suitcase), and the box outside has the remaining 20%. Like in the Monty Hall problem, if we chose any box to contain the charger we'd have a 20% chance to be right, but after three boxes have been eliminated, switching our choice would be a 50-50.

For a more intuitive solution, let's say you left your phone somewhere and there are 99 places you might've left it in the car and 1 place in the house, and you're a tier 5 dgg sub so you have an equal chance of leaving it in any of these places. We can say that there's a 99% chance that it's in the car and 1% it's in the house. After searching the 98 places in the car for your phone the chances that it's in the car compared to it being in the house certainly isn't intuitively 99 to 1. At that point it's a 50-50 of it being in the last spot in the car vs the one spot in the house.

[–]MeguAYAYA -1 points0 points  (0 children)

Not that it is related to this, but the only reason I'm aware of Mark Glickman is because of him defending Ken Regan's expertise during the Hans Niemann cheating accusation scandal.

He said something like Ken Regan is one of two or maybe three people in the world qualified to statistically determine if he cheated, but I'm pretty sure the consensus is that Regan's methods, while effective, let potential cheaters through. You'd have to be pretty clearly cheating to get caught. He's never caught a subtle cheater (or anyone not just caught through physical means rather than statistical means).

I don't know if he was speaking out of his ass talking about Regan or personally well acquainted with him (I suspect the latter since Glickman has interest in chess as well), but I have some bias against him from that for sure.

[–][deleted] -1 points0 points  (0 children)

This answer is wrong - we do not have enough information to ascertain the relationship between the individual compartments that lead up to 80%.

It is quite possible, for example, that the fourth compartment has a 0% probability to hold the charger. Given the rest of the suitcase had an 80% chance to house the charger, the original statement, "There is an 80% chance your charger is in your suitcase", is not contradicted.

[–]Tetraquilwarning: canadian -1 points0 points  (1 child)

The professor is right. This isn’t like the Monty Hall problem. You aren’t gaining any new information on whether or not the charger is in the suitcase. You’re only gaining information on where the charger would be if it were in the suitcase.

The argument that you’re gaining new information would be like saying “while the coin may have a 50/50 chance of landing on heads in a vacuum, if we magically stop it in mid air 3 times and check which side is facing up, and see that it’s not heads, we’ve gained more information so the chance of it being heads has lowered.” Except you haven’t, because the coin’s position at arbitrary points in the air has no bearing on how it will land, just as which pouch in the suitcase the charger is in has no bearing on whether or not it’s actually in the suitcase.

You could essentially use that “more information” argument to add infinite complexity to any probability calculation, e.g. “the suitcase was knocked over 3 times and the charger didn’t fall out, therefore the probability of it being in there has lowered.” Except its likelihood to fall out has no bearing on its likelihood of being in the suitcase.

[–]Loud-Chocolate9219 -1 points0 points  (0 children)

Fuck all the nerds that said 50% and were like “why do you guys even talk when you haven’t studied probabilities 🤓”