

A fun math puzzle whose solution will surprise you - mschireson
http://maxschireson.wordpress.com/2012/01/24/how-the-heck-a-puzzle-with-a-very-surprising-answer/

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SatvikBeri
My best guess for #1:

First, pretend that the Mathematicians know there are an even number of white
hats, and an even number of black hats. In that case it's easy for each one of
them to guess what color hat they're wearing. So have the Mathematicians
assume there are an even number of each type of hat. There's a 50% chance that
the number of hats actually is even (non-trivial, but I'll leave out details
for now), and therefore a 50% overall chance that the Mathematicians get every
guess correct (and a 50% chance they get every guess wrong!)

~~~
mschireson
Very good! Now how about the hard version?

~~~
SatvikBeri
It's trivially at least 50%, and I can't see how it can get above that.

This problem can be represented mathematically as a function _f_ : X->Y,
where:

X = {-1,1}^16 represents the set of hats the Mathematicians are actually
wearing. Say -1 is a white hat and 1 is a black hat.

Y = {-1,0,1}^16 represents the guesses made by the Mathematicians, where -1 is
a white hat, 1 is a black hat, and 0 is an "I don't know."

We can also write _f_ (x_0,x_1,...,x_15) -> (y_0,y_1,...,y_15) . Then for the
fixed set _f_ (a_0,...,a_15) = (b_0,...,b_15), the function _f_ is a correct
guess if and only if

1\. For all k, b_k = 0 or a_k .

2\. For at least one k, b_k != 0.

Assume _f_ (a_0,...,a_15) = (b_0,...,b_15) is a correct guess. There must be
some b_k != 0 where b_k = a_k, so we can assume WLOG that b_0 = 1 = a_0. Now
note that if f(1,a_1,...,a_15) = (b_0,b_1,...,b_15) and _f_ (-1,a_1,...,a_15)
= (c_0,c_1,...,c_15), then b_0 = c_0 . This is because mathematician 0 must
make his guess based only on the colors of the other mathematicians' hats.
Thus _f_ (-1,a_1,...,a_15) is an incorrect guess.

Thus for every set (x_0,...,x_15) where _f_ is a correct guess, we can come up
with at least one set where _f_ is an incorrect guess. Thus _f_ cannot be
correct more than 50% of the time.

...so, what am I missing?

~~~
mschireson
yes, each mathematician guesses incorrectly as often as they guess correctly.
but that is not the same as saying that there is a 50% chance the group "wins"
on any given round. remember with the possibility of passing the number of
guesses on a round need not be uniform...

Hope that helps get you on the right track,

\-- Max (the original poster)

