A logician has a favorite game to play at parties. He shows a set of solidly colored stickers to all his logician friends. Each logician, without looking, puts a random sticker on his/her own back. Each logician can only see the stickers on other people's backs, and no one can look at the unused stickers. The logicians take turns announcing whether they can deduce their own color. The game ends when someone announces he/she can deduce his/her own color.
One time while playing this game, no one had yet ended the game even though everyone had a turn. Should they continue to take second turns, or should they just give up and start a new game? Prove that it is impossible for a game that hasn't ended after everyone's first turn to ever end, or provide a counterexample.
3 logicians, labelled a,b,c
6 stickers: 3 red, 3 blue
imagine if they were distributed as:
a: blue
b: red
c: blue
right now, the logicians knows two things:
a) these stickers were randomly distributed
b) have a 50% chance to guess what their colour is.
Simple Scenario One: The Silent Treatment
Assumptions:
a) logicians may pass their turn
b) logicians will not announce they can deduce their own colour unless they are 100% certain
logician a sees a red/blue pattern, and does not guess, since a has a
2/4 chance of having red and 2/4 of having blue. logician a
announces that he/she cannot deduce his/her own colour.
logician b sees blue on both sides, and has a 1/4 chance of blue and
3/4 chance of red. does this mean logician b is justified in
guessing that the sticker is red? if by "announcing they can
deduce their own color" means, they know for sure, then 3/4 may not be
good enough. logician b believes that logician a could be in the
same spot; if logician a sees blue on both sides, would logician
a answer? logician b announces that he/she cannot deduce his/her
own colour based on logician a's silence.
logician c sees a red/blue pattern, and has a 2/4 chance of red or
blue. logician c thinks, is it possible that both a and b to see
a red/blue pattern? no. now obviously, c has one colour in
common with a or b, which means one of them saw the other two of the
same colour. but this doesn't really help c make a decision, so
logician c announces that he/she cannot deduce his/her own colour based
on logician a and b's silence.
in this case, the game will not end after the first round, because no one will ever be sure how the randomness works
Simple Scenario Two: The Guessing Game
Assumptions:
a) logicians may pass their turn
b) logicians may attempt their deduction, but will be told whether they are wrong or not.
c) there is no penalties in guessing
logician a sees the same setup, but since there are no penalties in guessing, guesses red, and is wrong.
logician b sees both blue, and guesses blue, and is wrong.
logician c sees red and blue, and seeing that both a and b were wrong, guesses red, and is wrong.
round one is over.
round two ... since logician a got red wrong the first time, announces
that the sticker can be deduced through trial and error in the first
round. wins.
My Thoughts:
a) i keep thinking back to the question where it says, "a
logician has a favourite game to play at parties." if this was
some sort of a trick, such as scenario two, his logician friends would
have found out a long time ago, and would know the trick
b) logicians are trying to deduce what they are randomly given.
with a diverse set of stickers, it would be just as impossible as
pasting randomly generated numbers on their backs and trying to deduce
each one. most of the games would either be won in round one (3
logicians, 2 red, 3 blue), or go on indefinitely (3 logicians, 3 red, 3
blue, 3 pink, 3 black, 3 orange, etc) if they are not allowed to guess,
be wrong, and try again.
Would this round be possible?
