Riddle

[wild nova]

There are 3 people captured in a room, they have hats number 0, 1, 2 (in any permutation)(for example,all of the hats can be numbered 0). Each person can see the number on the others' hats but not their own. Atleast 1 person needs to get the number on their own hat correctly without communicating with the others and answering simultaneously to win. Is there a strategy that guarantees a win?

This is possible with 2 people and hats numbered 0 and 1, I was pondering about 3 people and hats numbered 0, 1 and 2 but didn't find any solutions yet.

[misty wing]

1. Wait patiently for a helper to come along.
2. Once someone helps you, say thank you and close the thread with:

+close

3. Feel free to nominate the person for helper of the week in #helper-nominations

[modest light]

in any combination

[wild nova]

Yes

[toxic briar]

So any 3 hats can be 0/1/2, with no restrictions?

[wild nova]

Yes

[modest light]

i remember the solution for 2 depends on one person assuming theyre the same colour and 1 person assuming theyre different colours

[toxic briar]

And none of the hats number affect other numbers

[modest light]

leading to 1 having to be right

[wild nova]

Could be 000 001 021

[wild nova]

Yes

[wild nova]

The number on the hats are independent

[modest light]

maybe we can make a table of all 21 possibilities

*27

then make a list of all 27 triples of possible (i see a, b so i say my hat is c) that must have at least 1 element in the solution

[wild nova]

I did that

[modest light]

so that would produce a list of 81 things

[digital fiber]

Modular arithmetic ftw

[modest light]

base 3 counting

[digital fiber]

This is trivial done by modular arithmetic.

Just compute the checksum

[modest light]

oh yeah

add up the other 2 hats

[wild nova]

I tried taking the sum of other 2 hats mod 2

[modest light]

shouldnt it be mod 3

[wild nova]

It won't work for all combinations

Mod 3*

Typo

[digital fiber]

Yep you need further tricks

[modest light]

one person assumes all 3 hats add up to 0, one assumes they add up to 1, one that they add up to 2

[wild nova]

You can see the hats on the other 2 people

[modest light]

lets say the hats are 0 1 1 right

A will assume they add up to 0, B that they add up to 1, C that they add up to 2

so then A says "im 1"
B says "im 0"
and C says "im 1"

[wild nova]

Oh

Let me try that for all combinations

[modest light]

well

sure

[digital fiber]

It's interesting how sum of all hats are shared among all people

[modest light]

is it true that for n people with n hats from 0 to n - 1 there cannot be expected to get more than 1 correct

[toxic briar]

But the riddle only asks for one

[modest light]

ik

[toxic briar]

And I think that this solution would work for any n

[modest light]

yes

[toxic briar]

Like say 10 ppl and 10 hats

Real nice

[modest light]

im extending the riddle though

[digital fiber]

Hmm, it sucks that only one person could get the hat right with this solution

You cannot have many getting it right by luck

[wild nova]

Hez it works

Yes(

Yes*

[modest light]

conjecture by me, no, there is no strategy that always gets at least 1 and sometimes doesnt get 1

[wild nova]

I noticed for any n for this type of a problem the way to maximise the chance of getting 1 right minimises the chance of getting more than 1 right

+close