One of these threads again

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I am aware that this has a solution got an infinitely long line of people as well, but I still don't understand it.

This is the worst chapter of Kaiji yet

The ten person solution is the same as the infinite one. The number is arbitrary.

I assume there are 5 red and 5 green hats, otherwise it cannot be done

I am aware, maybe I phrased my post wrong.

Silly user.

If there's the same number of hats for each color, then the person in the back should choose the hat of the person in front of him.

That way the others will be able to keep track of the hats.

Posting some of mine that have never had a good answer before.

The contestands shoud ask it from the other contestans/audience.

No there's definitely an simple answer. It's just not easy to find because of the terrible background info.

First one presses red if there are more hats or green if there are more green hats.
The remaing contestants know their hat by comparing how many hats they can see and how many times red or green were pressed.

Tell the person in front of you their hat colour.

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youtube.com/watch?v=N5vJSNXPEwA

Also not anime.

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>not posting IUT runes

What if there are 3 red hats and 7 green hats. Pressing green only tells you that there are more green. The next person pressing could be thinking that they are the green hat in the group.

Neither anime nor manga.

Sage and Report.

That only gets you five.

>again

>implying I can even read that gobbledygook

no it doesn't out of 10 9 people have someone behind them to tell them their color

Agree to track one colour of hat
first guy counts up that number, He signals even or odd by sacrificing his vote (everyone knows that one button means an even number of whatever kind of hat, and one means an odd number)

The entire group can then count down the number of hats and assume their own

This works, nice. I hate that the problem doesn't define how we can communicate though. Because as said you can just tell the person in front of you what to press.

not always
if the first person presses green and you can see a majority of green hats already in front of you then you could be either have a red hat or a green hat

Then nobody gets it right because everybody is busy wasting their their chance by telling someone else what they have.

god i want to fuck rin

This method is quite simple :

The first guy answer green if the numbers of green hats, besides the one he is wearing, is an odd number. Therefore, the others guy can know, counting the others hats (except the first one), which colour they wear.

This works as well if they are in a line and answer one after the other, because this way they learn the colours before their own listening to the previous correct answers.

How do you tell if there is an odd or even number of infinite hats?

Since there's no rule against turning heads, I'd just ask the person in front of me what color my hat is.

Can you solve it, Sup Forums?

Are the mods dead?

>google the answer
>turns out bitch cant even spell
Fuck this game.

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What is this bullshit, is that even math?

Absolutely brilliant

Not this guy, but there is a silly "answer" using the choice axiom :

Assume you agree with everyone and all chose the same infinite sequences of hats as representants of the equivalence classes with the relation of sequences equal after some finite time.

All the player can see infinitely far away, and therefore they know which equivalence class their real sequence of hats belongs to.

Each of them says the color he would wear if the sequence was the chosen representant.

Therefore, only finitely many of them will state the wrong color.

>using the choice axiom

I basically don't treat anything that depends on this being true as having any practical value whatsoever. Any meaningful problem that might be relevant to reality will never depend on having to select something from a set without any definable entities and will always be able to be reduced to something that does.