>>9489479

I have had the same result.

how do any of the hints help at all? it's a toss-up

D4, right?

Since Albert knows first that Bernard doesn't know, and Albert doesn't know, there must be ambiguity for the both of them. It can't be tracks 5 and 6 because Albert would know immediately. It also can't be track B, because if, according to the first premise, Albert knew that Bernard couldn't know, and it was along track B, Albert would know which it is.

By letting Bernard know of this ambiguity, and Albert admitting his own, the only one pad that is made unambiguous enough for Bernard to be able to know, and by extension, Albert to deduce since it only took one step of elimination, would be D4- after removing the B track, the ambiguity between B4 and D4 was removed, meaning that the only way to definitively know which one it is would require it be on the only non-ambiguous pad.

you lost me in the second paragraph
I can see why track B,4,5,6 are eliminated

Doesn't track A have the same situation as B?

Can't be D4. If it was D4 there would be no way of Albert knowing that Bernard doesn't know the right pad. If Albert had D, he would know that if Bernard has 4 he'd know the answer.

Tracks 5 and 6 are eliminated right away because otherwise Bernard would know the answer. If Albert had track B he would then know that the pad is B4, since he know it can't be 5 or 6 (because if that was the case Bernard would know the pad).

I agree with you. C3.

yeah I can see that
but what of finding the answer?

>If Albert had track B he would then know that the pad is B4
citation needed.

Explain your logic using this. You faggots have flaws in your logic.

remove A and B you fag, if albert has A then he cannot state "i know you don't know" since Bernard can still single A6. same case with B5.

You are wrong. All that does is remove b5 and b6 as a posibility.
It doesn't make it impossible for it to be anything in A or B just because it is know to not be A6 and B5.

The time it takes to close each new gap is an infinite series that converges rather quickly, proving that infinities exist in the real world.

Hints are not related at all, kek the pseuds replying to a bait/joke

he cannot say that he knows that bernard doesn't know if he has A or B for the simple fact that bernard can still have 5 for B and 6 for A. since he says he is CERTAIN then neither A nor B can be a possibility.

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The prompt doesn't say it is impossible for letter to know numbers which are not the number, only that he doesn't know the number.

Pretty sure its B4
200IQ btw

you have to remove at least B you dumbo. if albert was certain that bernard did not know then he cannot have A or B, since both A and B give opportunity for Bernard to know from just his number.

Imagine you are Albert and you are told the right track is A. How would you be able to state for sure that Bernard doesn''t know the answer if that was the case? You woulnd't. I had trouble seeing it at first too, but it makes logical sense.

>It can't be tracks 5 and 6 because Albert would know immediately.
lol are you serious or just larping pseuds

please use the name of the hints, i cannot understand what you are saying. What i am saying is that since Albert knows and is certain that Bernard doesn't know which pad it is then Albert knows it cannot be anything that gives opportunity for pad A6 and B5 since those are by themselves for Bernard

albert cannot know shit you brainlet, it is bernard who would be able to know instantly if he has either 5 or 6

There is only a single option for 5 and 6, and they know the possible positions.
If there is a single option available for 5 and a single option available for 6, then number would would immediately know the correct position.

There is only one pad on each track 5 and 6. How would he not know straight away? That's literallyy the first thing you eliminate.

What I'm saying is that you assume that "I know you don't know" is the result of a logical deduction, that is not stated.

What else could it be?

the best assumption is to assume they make perfect logical decisions and statements.
Albert says "i know that you don't know" it means that after thinking Albert is 100% certain Bernard knows jack shit

"I wasn't told the answer, and you weren't told the answer"

that is not how these games work, man
most of the time the characters are fucking geniuses

C3

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D4, and here's why
1) Letter doesn't know the correct pad, but he knows 100% that number doesn't know either. If his letter was A or B, then this would be false because 5 and 6 are deductible from number alone, so he wouldn't be sure. If his letter was C or D then he is 100% sure that number doesn't know.
2) At first number didn't know but now he does. Getting rid of A B 5 and 6 gives 2 1's, 2 2's, 2 3's and only one 4.
3) unnecessary info.
So it's D4 you're welcome brainlets

d2, c3, and d4 are indistinguishable with the information available.

With your logic you get a 2 a 3 and a 4 for bernard. He knows which it is since he knows his number.

>Albert: I don't know
Excludes B
>Bernard: I know
has to be D4 (the only pad that is now alone on a number track (having excluded B4)

Eliminated 5 & 6 easily, then I kinda half-assed it and made an intuition jump to C3. It doesn't make sense for it to be D-aything, there's not enough for information for the 2nd guy to get which one it is.

Turns out I was right: en.wikipedia.org/wiki/Cheryl's_Birthday#Solution

July = C
16 = 3

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But what about the other two alone?

/thread

nobrainz detected

what do you mean?

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...

youtube.com/watch?v=MiEYCXPI-qY
you can test it out

the low-pixel frowning man never fails to make me lose

While the puzzle is fun albeit a bit too easy (just repeated applications of elimination) can I just point out the fact that this kinda elegantly demonstrates human distractability when confronted with novel challenges?

Remember people, this is still a trolley problem! You have to decide whether to actively crash 20 people together or let Cheryl kill 50

I feel the puzzle demonstrates a case where philosophical dilemmas are entirely forgotten when there are puzzles to solve, which is an interesting psychological phenomenon. Not exactly entirely new, but an old one given new perspective.

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the answer is indisputably no, anyone who disagrees is wrong

C3.

If Albert knows that Bernard doesn't know, then Albert's letter contains no pads that are the only one on its number row. This rules out letter rows A and B.

Bernard is able to deduce his pad from this info. That means that he was stuck between a pad on letters A/B and a pad on letters C/D, and with Albert's new information he realized that it had to be the one on C/D.

Now, if Albert knew the letter was D then D1 would be ruled out because his last statement wouldn't have given Bernard any new information to rule any pads out. But D2 and D4 would still work and he wouldn't know the correct pad.

But because he also now knows the correct pad his letter had to be C and, to fulfill the requirement that the number row had a pad on A/B and a pad on C/D, it had to be C3.

I'd say who the fuck ever put you in this situation would have killed him.

That, or all of the people who switch out the parts complying with this plan and not just using them to stop the train. If they have the knowledge to take it apart they probably know how to stop its acceleration.

>the answer is indisputably no, anyone who disagrees is wrong
The correct answer is it's an unanswerable question and anyone who answers is wrong.

.

I pick randomly, the expected number of people that will die is 3 which is better than the guaranteed 5 (due to the predictor predicting correctly)

your move predictor

I choose randomly because I have a 50/50 chance of finally killing all six of these annoying motherfuckers and then I can finally be done with this.

I think you misunderstood the problem. If you choose randomly, it either kills five or six people.

nevermind the fact that I could just pull it to track #2 to kill only one person

suddenly I feel confused

Exactly but they are all annoying, insufferable faggots, wasting my time on whether or not I should kill them. I'm just done with the whole thing.

Because of my state of apathetic nihilism, I chose randomly, in the hopes that on a beneficial coin toss I can just misanthropically kill all of them.

>I feel confused
Why? Just read the problem. IN THE CASE YOU CHOOSE RANDOMLY, the Predictor puts five people inside of the box no matter what.

but which box

Yes because of the Plank length.

given that the predictor will always predict correctly
if i switch to the second track, the opaque box will be left empty and a train goes on track no 2 so 1 man in a glass box dies
if i leave the switch alone there will be 5 people put into a black box on a track 1 and trolley/trian whatever goes on track 1. so 5 people are dead
if i choose randomly I have 50/50 chance of 5 people or 6 people dead-> either it goes on track 1 with 5 people or it goes on track 2 and 1 where there will be 6 people in line
so if i want to kill the most i choose third option
if I want to save a man in a glass box I go number two
If i want least people dead I choose one

Yes, that's correct, we know that

correct

I do not see where it says more people will be put in the transparent box
the text was always referring to the opaque box as the "active" one that it is talking about?

the last box in the random choice option is not stated explicitly but I assumed it's opaque. it does not say more people will be in transparent box and i do not know how did you infer that? it always goes either only 1 or 2 and then 1. on 2 there is one man in a transparent box which is removed only if you switch to 1.

it kills 6 in the third option because it goes through second track, killing glassm8 and then goes on track 1 which is with 5 people anyways since i have chosen randomly there right?

>it kills 6 in the third option because it goes through second track, killing glassm8 and then goes on track 1 which is with 5 people anyways since i have chosen randomly there right?
are you referring to the very last sentence in that the predictor has set up another test?
I interpreted this as separate from option 3, like it is a catch-option for anons that try to be clever spying on the predictor or making a conversation with the predictor... meaning the predictor is done here

Well that's stupid of you to do.

I really do not see how you can interpret the new setup that the predictor did to have anything to do with this setup or your choice

I'm on your side. Maybe you're confused. Or maybe you're someone else on my side.

I'm saying it's just one situation. There's no change in whether you choose randomly or not. But if you choose randomly or choose to not flip the switch, the Predictor will put five people in the box.

The only option that results in the Predictor not putting five people in the box is switching the track.

where does the six people death people are talking about come from then?
50% 5 people
50% 1 person

If the predictor put five people in the opaque box, your options are 5 deaths with track 1 or 6 deaths with track 2, and if the predictor did not, your options are 0 deaths with track 1 and 1 death with track 2. In either case, track 1 strictly dominates track 2. It is given that the predictor has already set the contents of the black box, and so its decision does not depend on yours. Thus you rationally minimize deaths by choosing track 1.

>given that the predictor will always predict correctly, your choices are between 1 death on track 2 and 5 deaths on track 1
>the predictor will always predict correctly
Duh, but that's not an assumption you can make

No it's
50% 5 people
50% 6 people.

Just read correctly, basically.

There are no scenarios where 0 deaths occur because the Predictor is never wrong.

>50% 6 people.
why?
opaque box is empty if you set it to 2
opaque box gets people added if you set it/leave it to 1
opaque box gets people added if you random

where does 6 come from

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If only one systematic part switching has occurred, then yes.
All the parts would need to be simultaneously switched, or an infinite number of part switchings would need to occur in order for the trolley to be 100% not the same trolley.

If they do the 1000 miles, they can call helicopters to rescue the person before the trolly comes

no, Ill just get another job

Bullshit

>the Predictor is never wrong.
Read the post you replied to. You can't assume that.

what does this shit mean even?
create a gap thats big enough so a fucking trolley cant catch up to him?
thats dum the trolley just kills the guy he isnt fast enough to do something like that

platonicrealms.com/encyclopedia/zenos-paradox-of-the-tortoise-and-achilles

ugly website and very hard to read words, do you expect me to read through that?
fucking no is my answer

i remain the victor here, what op's image shows its literally impossible

Turns out Albert is a lying cunt and his claim "I know you don't know either" was wrong.
Albert would say it was all a joke but in truthness he is just retarded; Bernard's number was 6. Unfortunately Cheryl really had a trolley problem, Albert's trolley would not start that day and no one lost their lives.

It's an ancient paradox.
Your common sense response is correct, that of course it doesn't matter how many many or how small the divisions you make are, the trolley will quickly catch and crush him.
But explaining that mathematically is much more difficult.

The new gap is created continuously, not discretely. The point it that the clearance of the new gap increases exponentially, resulting in it overtaking the man.

no you are just retarded
"explaining that mathematically" its just you making a bunch of useless noice
you are literally making shit up

except it doesn't because if you played this out for real the man would be crushed