Yes, but he lives in NY so will have less travel.

Theoretically I am a career manager but I feel more like a hybrid between that and an entrepreneur. I have done mostly management roles but more on the technical side, some individual contributor stints, and joined my last two companies quite early (8 employees and 27).

I am indeed incredibly lucky to be able to make this choice.

Congrats mate!

Thanks. The more I thought about the choice the easier it became.

Very good! Now how about the hard version?

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?

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)

I think these places are not first choice but depending where you are from may be a way to stay in the country. Where are you graduating?

Idaho

Besides office hours in California NY and London we also have user groups in many cities http://www.10gen.com/user-groups and have (one day, very inexpensive) developer conferences frequently (next two in Dallas and Seattle).

Its the best I've found.

Another way of thinking about the visualization is to make a model of the surface that can fold up to form it. The shortest path (which will be a diagonal if not a straight line) on any of those is the answer.

No, it does not. And while someone suggested that it could drop to the floor without injury, you could consider it to be afraid to do so :)