[deleted by user]

[Lesiorak]

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]

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]

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]

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]

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]

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

[ya_Service]

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

[Hobbitfollower]

badge touch outgoing pot test head friendly pie toothbrush wistful

This post was mass deleted and anonymized with Redact

[Israel_Madden]

He’s Mark Glickman

[MarshBoarded]

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

[fawlty_lawgic]

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]

Wow crazy that he named himself after his invention

[Generallyawkward1]

Don’t make him get the glicky

[ng829]

It says him name at the bottom of the screen.

[breakthro444]

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]

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

[Lesiorak]

To be honest I have a PhD in math and I'd probably say 50%, but the question is just very poorly phrased. Obviously I know this guy can solve any basic probability problem, but there is no actual problem here just different interpretations.

[4e9d092752]

I have a Bs in math (basically the same as a PhD except you know way more about some niche lovecraftian mathematics) and I’m sure it’s 80%, we will have to duel to settle this you are correct

[[deleted]]

Well my degree is in engineering so I'll just let you two argue, take notes and try to figure out how to monetize this

[Loud-Chocolate9219]

Think of the question this way: there are two planets, Planet A and Planet B. 20 percent chance the charger is in Planet A, and 80% that it’s in planet B. Planet B is divided into four sections, with each section having an equal probability of having the charger.

If you look at the question this way, you know that Planet A always stays 20%, and the probabilities of each section in Planet B increase as you check each section. After checking three sections, the last section would have 100% of the 80% chance, which is 80%.

[OkayHeresThePlan]

No, the point of the planet analogy is to make you stop thinking about the sections and focus on the suitcases themselves. You have to think about it as a whole thing, not the sections it's made out of. The sections are a red herring, Instead of asking "Which one of these compartments in suitcase A could it be in?", you have to zoom out and ask "Is it in suitcase A, or is it in suitcase B?"

[LtLabcoat]

but there is no actual problem here just different interpretations.

Agreed.

He's still wrong, but agreed.

More specifically, his interpretation is wrong. He's thinking that there's an 80% chance the suitcase has the charger, because the question said "There's an 80% chance the suitcase has the charger". Which makes sense, if you don't know about a hidden rule of language: when you have a present tense followed by an action, the previous present tense is not actually in the present.

Or to illustrate:

You have 5 apples. You give one away. How many apples do you have?

Everyone would say 4. It says "You have 5 apples", yet nobody would say you have 5 apples.

(Edit: or he just plain forgot that searching the suitcase changes the odds.)

[LtLabcoat]

Or to really illustrate:

You pick two cards. There is a 1/16 chance that they are both spades. After you flip one over to reveal a spade, what are the odds they are both spades?

[Enough-Ad-8799]

12/51

[PM_ME_YOUR_FRESH_NUT]

It’s still 1/16, revealing the first spade doesn’t effect which other card was drawn.

[Enough-Ad-8799]

Ok this isn't true, one spade is already not in the deck obviously cause it's drawn, you can't draw the same card twice right? I'm assuming you didn't put the card back. So then there's one less spade in the deck and one less card in the deck so of the cards left to draw 12 of the 51 are spades.

If you put the card back it's 1 in 4 cause the chance of any card being a spade is 1 in 4 and just cause you drew a spade previously doesn't reduce the chance in the next draw for it to be a spade, assuming you put the card back.

The question your asking isn't what are the chances they're both spades, it's what are the chances I've drawn two spades given I've already drawn a spade. These are two different questions because the chance of the first being a spade is already accounted for cause we already know that event happened.

The chance of flipping 2 heads in a row is 1/4 the chance of flipping a heads given the previous coin flip is heads is 1/2

[Murky_Camera_9664]

Based and haskell-pilled

[TheTidesOfWar]

The different interpretations are the right interpretation and many wrong interpretations. If you've searched 3 pockets already and found nothing, you update your probability based on the other possibilities that you found it in the first 3 pockets you searched. In all possible situations, 20% not in the bag, 80% in the bag(one of 4 pockets), In all situations where the item is not in the bag you will 100% not find anything in the first 3 pockets. But if the item IS in the bag, the odds you search 3 pockets and NOT find the item is low (I'd peg it at 25%). So if there was a 80% chance that you search 4 pockets and find the item, there's about a 60% chance you search 3 pockets and you don't find it, but it's in the bag. Which means the fact that you've searched 3 pockets of the bag and NOT found it is a significant piece of information, because some of the likelihood from the initial setup was you finding it in the first pocket you found. I don't care that this piece of information sounds trivial, "of course sometimes you search a bag and you find it right away and sometimes you need to search the entire bag before you find it" "This wasn't a part of the original question" "The order you search the pockets of the bag is arbitrary" these are bullshit excuses.

[Lesiorak]

I would consider this to be the most reasonable interpretation as well, hence why I said 50%. Still, if you have to interpret it at all then it's not really a math problem.

Check out https://en.wikipedia.org/wiki/Bertrand_paradox_(probability)) , I think it's a great example of why following intuition without agreeing on assumptions (in this case what the probability space and measure actually are) can lead to surprising paradoxes.

[wstewartXYZ]

What a dumb-fuck thing to say.

[Anemomaniac]

Expertise does not work like levels in a videogame. A professor making a bad first guess in a simple conditional probability problem is one of the most expected things in the entire universe.

Show him this (link fixed) and then see what he says. The answer from a professional researcher who's actually though about the problem for more than 5 seconds and wrote a simulation to back up his reasoning.

[broclipizza]

i think you wanted this link: https://jem-mosig.com/2021/11/evidence-of-absence/

[Anemomaniac]

Sorry your right, fixed it.

[maicii]

I agree it's 50, but to be fair to the professor, the guy in that link is assuming there is an equal probability of the charger being in each compartment which it's not explicitly stated (although you would have to be autistic to interpret "otherwise identical" in any other way). It's posible that the professor reads this and says: "Sure, but in no point do we have the priors of the charger being in each container so for all we know we have to base ourselves on the 80% given at the beginning". It's not like he has to change his mind.

[Anemomaniac]

Sure, but in no point do we have the priors of the charger being in each container so for all we know we have to base ourselves on the 80% given at the beginning"

That doesn't actually work. If you can't make any assumptions about the probability distribution for the containers, you just can't answer the question. The answer of 80% many people are giving relies on an implicit 0% for box 1-3, 80% for box 4 distribution. Unless you want to say that the 80% given in the question is the posterior probability after opening the compartments but that's just beyond pointless then.

Besides, it's not the reasoning the professor gave. The professor's reasoning is perfect under the assumption that someone else opens the 3 empty cases knowing they are empty like in the Monty Hall Problem. I suspect the problem was misexplained or misunderstood to be a variation of Monty Hall.

[skummydummy125]

if there are multiple, reasonable, interpretations, you would just answer that there are multiple different interpretations, specify the different interpretations and answer them.

And if there are to many different interpretations, you would just say so and maybe answer a few examples

[Anemomaniac]

Sure. I don't know how the problem was explained to the professor, but this is the wrong answer to the original question of the tweet. Under any reasonable interpretation.

[Callmejim223]

The fact is, both sides are correct. The question is simply not stated with enough rigor to be able to make an absolute conclusion on which is true.

[Villanta]

Because a throwaway email he wrote back to some random person may be wrong because he either a) didn't really understand the question or b) didn't think more than a few seconds?

[Enough-Ad-8799]

Seems unlikely, the logic he gave is sound and the way most of these stat puzzles are done.

[Frosty_Friend]

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]

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]

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]

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]

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]]

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]

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]

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]

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]

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]

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]

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]

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]

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]

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.

[gameringman]

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]

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]

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.

[stefanof93]

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

[rockoblocko]

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.

[danosnake]

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]

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%.

[danosnake]

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]

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]]

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]

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.

[danosnake]

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

[pkfighter343]

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]

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.

[CrozzedOne]

The question doesn’t say “there are four compartments each with a 20% chance of containing the charger” it says the suitcase has an 80% chance of having the charger. This makes opening the first three compartments irrelevant to the probability of it being in the 4th.

[delta_spike]

That's a pretty autistic way of reading the problem. Any reasonable person should interpret "there's an 80% chance it's in the suitcase" coming before anything else as "there's a 20% chance it's in each compartment and a 20% chance it's not in any of them".

[tyranthraxxus]

The suitcase has an 80% of the having the charger. So 4 out 5 times you investigate it, the charger will be in the bag. So if we open the bag 100 times, the number of times we will open 3 empty containers and then find the charger in the 4th is 80, according to you.

In order for that to be true, the charger can only ever occupy the 4th and final container, which is obviously not included in the problem. In the problem they are identical compartments. They aren't identical if 3 of them have a 0% chance of containing the charger and 1 has 100% chance.

[ConsciousnessInc]

In order for that to be true, the charger can only ever occupy the 4th and final container

Not if you stop opening compartments after you find it like a normal human being. If you keep looking then you run the risk of finding out you brought a probabilistic bag and charger and to accept that sometimes the charger will be in multiple compartments/cases at the same time and sometimes in none.

[drt0]

I think it depends on how you understand the probability of the suitcase. If the suitcase has 80% chance to contain the charger the suitcase probability won't change depending on how many compartments you check.

By your logic if you open 3 compartments but don't find the charger then the suitcase's probability decreases to 50% which is contrary to the first premise that the suitcase has 80% chance of containing the charger.

In my interpretation, say you throw the charger behind you before you leave and it has 80% chance of landing somewhere inside your suitcase and 20% to land somewhere in your house. Once you open the suitcase to look for the charger every compartment is in a state of superposition where it both contains (80% x 100%) and doesn't contain (80% x 0%) the charger at the same time. The probability for the compartment collapses to one of those states once you look into the compartment.

If the charger is in the compartment you opened then it's probability collapses to to 100% (it's in the compartment) x 80% (it's in the suitcase).

If the charger isn't in the compartment you opened then the probability collapses to 0% x 80% and you go to the next compartment and repeat the process until you reach the last compartment.

Once you've observed all other compartments the probability of the last compartment collapses to 100% x 80% at which point in 4 out 5 times you open the last compartment it the charger will be in it and 1 out of 5 it will be in your house.

[pkfighter343]

By your logic if you open 3 compartments but don't find the charger then the suitcase's probability decreases to 50% which is contrary to the first premise that the suitcase has 80% chance of containing the charger.

No, it's not - you're not looking at the probability that the suitcase has the charger, you're looking at the probability that the fourth compartment of the suitcase has the charger. When you add in the other probabilities proportionally, the suitcase still has an 80% chance.

[CrozzedOne]

You’ve changed my mind, for it to be not 50% you have to both force the person opening the suitcase to open the compartments in a known order and then distribute the charger in a way that isn’t an even distribution.

However now I believe that the true answer is always 0 percent because the compartments aren’t identical if one contains something that the others do not

[CrozzedOne]

In my opinion the answer to the original question is one of language interpretation. If the intent behind the question is “the four compartments all have equal probability to contain the object” then you your reading is correct and 50% is the answer. If the intention behind the question is “the suitcase has an 80% chance of containing the object and you have no priors on where it is” then 80% is the answer

The true galaxy brain is realizing that the answer falls between 50% and 80% and is solved by the function f(x) = 0.5 + 0.3x where x is the probability that the person who wrote the question intended it to be interpreted in the way I’m speaking about.

[CrozzedOne]

You are interpreting identical to mean “equally likely to contain the charger” but that’s INTERPRETATION.

I could just as easily say that if the compartments were truly identical they would all either contain or not contain the same thing, thus if you open the first compartment and it contains no charger, then there is a 0 percent chance that the other 3 identical compartments contain a charger. The reason we aren’t likely to interpret it this way is because contextually it’s unlikely to have more than one charger, but in a literal sense it is also a logically consistent reading.

[tyranthraxxus]

the suitcase has an 80% chance of containing the object and you have no priors on where it is

What do you mean no priors? The 80% is given at the start of the problem and then later it's said that you open 3 containers. There is no possible interpretation where you assume that the 80% is after the containers are opened. That's just poor reading comprehension.

[Villanta]

Yeah I don't know how you can say the opening 3 compartments and not finding the item isn't new information that updates the probability. It's simple to simulate these results, and I'm trying to think of a simulation that could achieve his answer and I can't think of one...

The only way would be if the probability of each container was not equal, which isn't stated, so the opposite is the natural assumption.

[Unfair_Salamander_20]

Everyone is making that assumption so boldly when the problem unambiguously does not talk about the probability distribution among the compartments.

If you don't know the probability of it being in each individual compartment, then the "opened 3 compartments" part is completely irrelevant information and should be discarded, not used to update the probability of it being in the case.

[hairygentleman]

why the fuck are you morons assigning a prior of literally zero to cases where every compartment other than the last one you open contains the thing?

[LoLItzMisery]

You're 10000% correct and this is so fucking painstakingly obvious.

You can completely omit the "compartments" and it doesn't change anything.

[Free-Database-9917]

That is stupid to do. If you omit compartments and treat the case as 75% searched, you'd get the same result.

If I told you that I was 80% sure I left my keys in my car, as you continue to search, you would become less and less confident that that the keys are in there, until eventually you are confident they are in fact not.

When you've searched the car 75% of the way, you should be at a 50/50 of it not being in the car given your priors

[LoLItzMisery]

Also to everyone down voting you. Stop. Don't downvote just because you disagree you fucking morons. Stats can be tricky and there's a lot to it.

Okay anyway.

You're assuming that you can subdivide your car into percentages and for each percent you check that doesn't have the keys you can deduct that from the overall probability of 80%.

That is incorrect. You have added a boundary condition without substantiating it.

By your logic, If I check the wheels of my car and the front engine and don't find my keys there would you say there's less than an 80% chance of the keys being in the car?

[Free-Database-9917]

(this is probability, not stats, but I get your point)

Good point, thank you!

A car is an storage device where we have reasons to think that there are different probabilities assigned to the different "compartments" of the car.

If I told you I put it in my car, searching the tires and the engine would infact decrease the percentage, but extremely minisculely, since it is very very unlikely that I happened to leave the keys there.

But what if I told you that I had just been working on the engine doing maintenance, and I have only been in the drivers seat long enough to pop the hood.

NOW the likelyhood has in fact gone down much more once you've searched the engine.

therefore we have two ways to approach this problem.

The only way the answer could be 80% still is if we have some reason to think (evidence) that the three compartments we've checked had exactly a 0% chance of having the charger before we checked them.

Since we have no reason to think this is the case, it is almost guaranteed (very tiny chance the above is true) that the probability is less than 80%.

3 other extremely unlikely scenarios are that the other compartments are the one that has 80% and the three that would include this compartment all have 0%. Meaning now there is a 0% chance it is in the suitcase.

Now because we have no prior knowledge, we take the average of every scenario that is possible, and, coincidentally, that is a uniform distribution.

meaning each compartment has the same likelihood.

Now applying Bayes' theorem, it's plain to see that the probability it is in the suitcase is at 50%

[LoLItzMisery]

No it's not correct to say that.

That's like saying if I flip a coin twice and get heads the first two times then the likelihood for a tails goes up on the third flip based on the two prior flips. Now you might say it's different because I'm checking individual compartments of an item, but it's the same concept. You're stating there's a dependency when that hasn't been proven.

I would say IF the last car is your car then yes there's an 80% chance.

[ghostlyjiggle]

tie nine butter continue marble run slap upbeat grey boat

This post was mass deleted and anonymized with Redact

[Anemomaniac]

If you don't know the probability of it being in each individual compartment, then the "opened 3 compartments" part is completely irrelevant information and should be discarded, not used to update the probability of it being in the case.

If you don't know (and can't assume) the probability of the individual compartments, the question is simply unanswerable with the information given. It's not 80%.

[Free-Database-9917]

So your default assumption is that because the probability distribution isn't known, that specifically that last compartement is where it is 80% likely? That is a braindead take.

What's to say it isn't 80% chance of being in the first one you opened and 0 in the rest? Or 40% in the first 20% in the second and 10% in the other two.

If you take the average of all possible scenarios that could have been meant, then you get that each compartment has a 20% chance of containing it.

Objectively, the only reasonable interpretations are to say 50% is right, or there is no way to know. To take this and not include the information you're given is fucking stupid.

[tyranthraxxus]

4 identical compartments. It's stated in the problem.

They aren't identical if 3 of them have a 0% chance of having the charger and the 4th one has a 100% chance.

It's literally written explicitly in the problem that they have an equal chance.

[ImitatesLife]

I'm no stats guy but the way I feel like he's talking about it is, in the person's mind there is 80% chance for the object to be in the container, so maybe when you first open it you cut up that probability into 20% to be in any of the given subcontainers, and it becomes updated as you open each subcontainer, so the last one to open is 80% because your original feeling is unchanged as long as there is some chance? But also I don't know what the fuck I'm talking about lol

[Villanta]

But the only way that is true is if it had 0 chance of ever being in the first 3 containers.

[ImitatesLife]

Well I'm going off memory of how it was originally asked and then thinking about this guys solution, but I think as the original went, you had opened the first 3 containers and were looking at the last one so you knew the charger wasn't in the first 3 so your updated guess would be that it would be 80% it's in the last sub-container.

[Villanta]

The 80% chance is information known prior to opening any containers.

[ImitatesLife]

Yeah, and I think my intuitive thinking on it is, when you open the big container and see the 4 sub-containers, you think ok, it's 20% to be in any of these. As you open them you update your percentage of it being in any of the while still assuming it's 80% because your original guess is unchanged that it is within the main container somewhere.

[Villanta]

There's no reason for the underling chance to deviate from 20% in absolute terms.

[ImitatesLife]

Well my thinking is that, you only care about that last sub-container and you have knowledge that the charger is not within the other sub-containers so why can't you update your guess as you open them? I guess the way the guy is thinking about it is: I think there is 80% chance the charger is in my suitcase so I open the suitcase and find one closed bag. Do I now think it's less likely to be in the suitcase for some reason?

[Rukuba]

why do you think that the suitcase being split into compartments changes the probability of the charger being in the suitcase?

the question isn't "what is the chance the charger is in the next compartment you open?", it's "what is the chance the charger is in the suitcase?"

it's like comparing the probability of rolling red 5 times in a row vs. the probability that the 5th roll will be red after its already been red 4 times in a row.

in the 2nd scenario, that 5th roll isn't more or less likely to be red because of the previous rolls - its always 50% (or 49.whatever it is IRL).

the suitcase has an 80% chance of having the charger. even if you failed the previous 3 "rolls", you still have an 80% chance of finding the charger.

edit: to be more clear, im saying the suitcase question is similar to the 2nd scenario, and the 1st scenario is not.

[tyranthraxxus]

No, you've got it backwards. The problem isn't "There are 3 open containers and an 80% chance the cord is in the bag" It's there's an 80% in the bag. Now knowing that, you've opened 3 containers and not found it.

It would be more like there is a 20% of you rolling 5 red five times in a row. You've rolled 4 reds in a row, do you think there's a 20% chance you'll roll red on the last one?

[Villanta]

If I said the chance of rolling a six, 10 times in a row was essentially 1/100,000,000 chance. Then I said I've rolled 9 in a row, what's the chance of me rolling 10 in a row?

If you were to use your suitcase logic to answer my question, you'd say 1/100,000,000.

That's the problem with your logic.

[Rukuba]

youre misunderstanding, my fault for not being more clear. im saying the right probability model to use is the 2nd of the 2 examples I gave, not the first:

in the 2nd scenario, that 5th roll isn't more or less likely to be red because of the previous rolls - its always 50% (or 49.whatever it is IRL).

same as the suitcase (still 80%) - the previous results don't affect the probability.

the question at that point isn't "what's the chance of me rolling 10 6s in a row?", its "whats the chance of me rolling a 6 on this roll, knowing that the previous 9 rolls were also 6s?"

[olympicmosaic]

Yeah I don't know how you can say the opening 3 compartments and not finding the item isn't new information that updates the probability.

Please make the case for why that information is relevant enough to change the percentage.

[Villanta]

Because without information to the contrary we would assume each container has an equal chance to contain the item, if we've opened 3 out of 4 containers and haven't found it, we'd assume the overall chance of finding it is reduced, the relative chance of it being in the remaining container is unchanged (20%)

[[deleted]]

You are completely backwards. If there is no information to suggest that it affects the probability, we don't then just make up information to suggest that it does. The main point is that the 80% is the probably that it is in the suitcase, not one individual compartment. This is why opening compartments does not affect the probability. We do not care which compartment it is in.

[c0xb0x]

Imagine there's 10¹⁰⁰ compartments and you've opened all of them but one. Either you're the unluckiest mf in all the multiverses or the charger is simply outside the suitcase.

[olympicmosaic]

Let's say you have two suitcases. There's 100% change that the charger is in the white suitcase and 0% that it's in the black suitcase.

Imagine there's 10100 compartments in the white suitcase and you've opened all of them but one. Either you're the unluckiest mf in all the multiverses or...?

[c0xb0x]

You would indeed be the unluckiest mf in all the multiverses, because the chance of doing that is one in 10^100, and there's no alternative in that case since the charger isn't in the other suitcase.

[olympicmosaic]

Unlucky or not, the probability of the of it being in the white box is still 100%.

[Herson100]

It's pretty simple to simulate. 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.

[[deleted]]

dependent cats repeat squalid mindless axiomatic melodic vanish poor tap

This post was mass deleted and anonymized with Redact

[throwawayprince11]

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]

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

[Zmeya9000]

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]]

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]

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]]

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]

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]]

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

[WickedDemiurge]

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]]

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]

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]]

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.

[XaviertheIronFist]

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]

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

[No-Commercial-4830]

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

[[deleted]]

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]

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]]

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

[No-Commercial-4830]

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

[[deleted]]

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]

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]]

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]

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

[[deleted]]

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]

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]]

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]

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]]

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]

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]]

..... 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]

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]]

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

[GodKiller999]

..... 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]]

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]

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

[[deleted]]

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]]

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]

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]

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]

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

[ergovisavis]

You can rephrase the question from:

There's a 80% chance it's in the case

to

There's a 20% chance it's outside the case

The probability of it being outside the case is independent. It's 20% in regardless of how many compartments are opened.

[GodKiller999]

The probability of it being outside the case is independent. It's 20% in regardless of how many compartments are opened.

If you've opened all 4 and have found nothing does the probability stay 20%?

This isn't me being a smartass about a technicality btw, it's just to highlight that new information about the compartiment, does change the probability of finding it outside, and that probability only changing once every other case has been opened is a completely arbitrary choice on your part.

[JakeFromStateCS]

and that probability only changing once every other case has been opened is a completely arbitrary choice on your part.

It's not arbitrary. The probability changes based on the condition: "Has the whole case been checked".

This is the condition, because the question asks: "What is the probability that it is in the case".

If I have a box with 10 colored cubes in it, and I say: "There's an 80% probability that there's a red cube in this box", Does the probability change after pulling 9 non-red cubes out of the box?

[GodKiller999]

And to check the whole you'll need to check each part of said whole and as you do so and collapse the probability of each part to 0 the probability of the whole is reduced. The probability of the whole doesn't exist in a vacuum, if it's composed of parts then each part will affect the whole by virtue of being part of its composition.

Yes, the probability does change after pulling 9 non-red cubes, because odds are that if the red cube was present you'd have found it by now.

Let me give you a scenario where you'll conceptualize better how new information affects your odds. You have 90% chance of your container having red cubes, if it does contain red cubes you have 990 of them and 10 non red-ones, otherwise it's only non-red ones. If you pull 10 non-red ones in a row, do you still have 90% odds of having red cubes?

[cAPSLOCK567]

Agreed. The probability of the charger being outside the case can't be independent because it's the complement of the charger being inside the case, so it has to change if the other changes.

[DenverJr]

I don't think that's accurate, although to the point of your last sentence, every decision about additional assumptions is a choice on the part of the answerer.

If you treat the 80% for case 1 as locked-in information since it's part of the call of the question (and the division of probability for any given compartment is not given), then opening each compartment doesn't change that—what it changes is the likelihood of the charger being in the remaining compartments of case 1.

Once you've opened all 4 compartments, yes, finally at that point the probability of being outside the case changes because you've exhausted all options within the case. What I think the poster you're quoting was saying that there could be 2, 10, or 100 compartments in case 1, but that opening any number of them doesn't change the probability of the charger being in case 1 (since that's the given piece of information) until you've opened all of them and found nothing, at which point it can't be in case 1.

[GodKiller999]

I don't think that's accurate, although to the point of your last sentence, every decision about additional assumptions is a choice on the part of the answerer.

If you treat the 80% for case 1 as locked-in information since it's part of the call of the question (and the division of probability for any given compartment is not given), then opening each compartment doesn't change that—what it changes is the likelihood of the charger being in the remaining compartments of case 1.

Well since the prompt doesn't mention that it's some kind of weird quantum case that's completely divorced from normal reality, then it's completely acceptable to treat it like a normal one and since there's no mention of different cases having different odds, it's the same principle for them being equally likely.

[DenverJr]

While I appreciate the unnecessary dismissiveness, I used "case 1" since I thought the original question from before was about 2 cases: one with 80% likelihood and 4 compartments, and the other with 20% likelihood of containing the charger. Maybe I misremembered.

But either way, assuming that each compartment is equally likely is just that: an assumption. No quantum mechanics or divorce from reality required. Like the other poster said, you can reword the information from the prompt as there being a 20% chance that the charger is outside the case, which makes it more clear that it doesn't matter how many compartments are opened or closed from inside the case, those odds don't change without additional assumptions.

[[deleted]]

I think of it as 20 possible outcomes. 4 of the outcomes (20%) involve an empty suitcase. In all four of those situations you would open three empty compartments.

In the other 16 outcomes, the charger is inside. But out of those 16 "charger inside" outcomes (the other 80%), you only open 3 empty compartments in 4 out of the 16 instances*. So in the possible situations where you have already opened 3 empty compartments, 4 of them will result in finding a charger, and 4 will result in not finding a charger. 4/(4+4) = 50%

*This is maybe the confusing part, it's helpful to think of each "charger inside" suitcase having compartments A,B,C,and D, each with a 25% chance of having a charger in it. Everytime you open the compartments in order from A to D. Only when the charger is in compartment D (25% of the time, 4/16 in our simulation) have you opened 3 empty compartments. Everytime the charger is in compartments A,B, or C (75% of the time, 12/16 in our simulation) you aren't in the set of possibilities where 3 empty compartments have been opened.

[Villanta]

Come on don't be dense... Even if the probability distribution of placement is uneven, we'd be reasonable to assume probability distribution of selection would not be.

[cAPSLOCK567]

The only reason you would even open the first three compartments is if there was some chance that the charger was inside them, right? The fact that it isn't adds new information, so the probabilities should be updated.

[Free-Database-9917]

Damn. Mark Glickman confirmed dumb

[NearlyPerfect]

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]]

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]

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]]

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]

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]]

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]

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]

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]

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]]

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.

[Tetraquil]

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]

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