—– SPOILER - possible solution below —–

We’ve solved the riddle if we can find two sets of infinite bit-vectors, called “Zero” and “One”, such that:
1. Zero and One are disjoint and their union contains all possible infinite bit-vectors.
2. For each vector in Zero, flipping one bit produces a vector in One.
3. For each vector in One, flipping one bit produces a vector in Zero.

If we can find two such sets, then the infinite men will agree on these sets before the task begins.
When it begins, the first guy sees the vector of all the hats ahead of him in line. He yells “white” if the vector is in Zero, or “black” if the vector is in One.
Every other guy knows all the bits in the vector except his own (the previous by hearing, the next by seeing). Also, he knows whether the vector is in Zero or in One. Due to properties 2 and 3 of the sets Zero and One, this is enough to determine the missing bit.

Now, to prove the existence of these 2 sets:
At first this seemed grim to me because both sets are of the same cardinality as R: infinite bit vectors are isomorphic to real numbers between 0 and 1, and both sets contain infinite vectors that are isomorphic to irrational numbers.
However, maybe the axiom of choice saves us.
For each infinite bit vector v, define Bubble(v) as the set of all vectors reachable from v by flipping a finite number of bits.
Obviously there are infinite disjoint bubbles. Using AC, we choose one vector to represent each bubble, and define Zero_0 to be the set of all representatives chosen.
Then we continue in the following manner:
One_0 = {all vectors one bit away from vectors in Zero_0}
Zero_1 = {all vectors one bit away from vectors in One_0}
One_1 = {all vectors one bit away from vectors in Zero_1}
…
Zero_k = {all vectors one bit away from vectors in One_k-1}
One_k = {all vectors one bit away from vectors in Zero_k}
…
And finally:
Zero = union Zero_0, Zero_1, …
One = union One_0, One_1, …

Unless I’m wrong, I think it can be shown (though I don’t feel like formulating a proof right now) that these sets fulfill the requirements I set above.

A link to the python source can be found in the body of the article.

Well, I’ll save you the trouble ;-): http://yaniv.leviathanonline.com/blog/code/4d.py

Of course, you need python in order to run it.

Enjoy!

#1:
This is very similar to Yair’s idea, but makes certain useful assumptions: We need to assume that N is at least 4, and that Q=N-1 is a prime number, and choose K=Q. This satisfies that (Q choose C) = Q!/C!(Q-C)!, and for Q>C>0, since Q is prime, Q divides Q! but does not divide C! or (Q-C)!, and hence divides (Q choose C), like we wanted. We can choose the probability P such that f(P)=P^K+(1-P)^K=1/N, because f(P) is a continuous function of P, which satisfies f(1)= 1 > 1/N, f(1/2)= 2*(1/2)^K = 1/2^(N-2), which is not greater than 1/N for all N at least 4. Hence by the mean value theorem, there is a P in (1/2,1) with f(P)=1/N, and we are finished.

Now we only need to understand why the assumption that N-1 is prime does not limit us: Indeed, for all N, by Dirichlet’s Theorem, we know that there are infinitely many primes Q in the arithmetic sequence {m*N-1| m is Natural}, and if we take one such prime that is at least 3, then by the above we have a solution for m*N=Q+1. But then if we treat each person as m different people, we have a solution for N people, and we are happy.

(By the way, such K does not necessarily exists for all N where N-1 is not prime, and in fact no such K exists when N-1 is non-prime, as the following claim is true: The greatest common divisor of all (K choose C), where C ranges from 1 to K-1, equals a prime Q iff K is a power of Q, and if K is not a prime power, it equals 1)

#2:
This has some similarities to #1, but is a little different. We will need to assume that N is large enough, but as shown before, small N can be reduced to the larger m*N case easily.

Let the probability of the coin be P. We will flip the coin K=N-1 times, and write the result of the filp as a string of length K composed of letters H and T. Such a string will be called “bad” if it can be cyclically shifted right less than K times and it will return to itself, i.e. if it is non-trivially periodic. A string will be called “good” iff it is not bad. Then obviously the good strings can be divided into N-1 sets of strings, all with equal total probability, by ensuring that if one good string can be obtained from another by cyclic rotations, then they are in different sets. Now, let f(P) denote the probability of the result string being a bad string. Then we only need to be certain that it’s possible to choose P such that f(P)=1/N.

It is clear that f(P) is a polynomial of order K-1 in P, and thus continuous. If P=1, then the result string is always all heads, which has period 1 and is thus a bad string, so f(1)=1 > 1/N, and we need only see that 1/N > f(1/2), for example. Now, if P=1/2, then f(P) is simply the number of bad strings of length K divided by 2^K. We can count the bad strings by separating them based on their minimal period - the number of bad strings with period d equals 2^d if d is a proper divisor of K or 0 otherwise, and hence the number of bad strings with minimal period d is certainly not greater than 2^d, and 0 when d does not divide K. Hence the total number of bad strings is at most the sum of 2^d over all proper divisor d of K. The greatest proper divisor of K is of course at most K/2, hence the number of bad strings is at most the sum of 2^d over all natural d up to K/2, which is lesser than 2^((K/2)+1), and hence f(1/2) is lesser than 2^(K/2+1)/2^K=1/2^(K/2-1)=1/2^((N-3)/2). Obviously, exponential in N decreases to 0 much faster than 1/N, and so for large enough N we indeed have f(1)>1/N>f(1/2), as required.

#3:
This solution is somewhat long and messy, so I will leave some parts as exercise.

Denote by p the probability of heads, q the probability of tails (p+q=1), and choose them such that p*q=1/N (possible iff N is at least 4). Flip the coin a large number K of times. We will assign the results of the raffle to the results of the coin flips in such a way that if all coin flips were changed, we will choose the same person, with the exception of results with equal numbers of heads and tails (if K is even). We will then have basic “building blocks” with sizes of the form p^l*q^m+p^m*q^l for m+l=K, l>m non-negative integers, or p^l*q^m when l=m=K/2. Using the basic relations p+q=1, pq=1/N, it is possible to partially compute the above expressions and see that they can all be represented as rational numbers with denominator N^[K/2] (where [x]=floor(x)). Furthermore we have all the following properties:

1) The integers (N^[K/2])*(p^l*q^m+p^m*q^l), or half of that for m=l, decrease as m increases, for m up to [K/2];
2) For all integers r from 0 to [K/2]-1, the integer corresponding to m=r is less than N times greater than the integer corresponding to m=r+1;
3) For all r from 0 to [K/2], there are (K choose r) appearances of the integer corresponding to m=r, and hence at least K appearances for all r from 1 to [K/2];
4) The sum of all these integers, counting multiple appearances, is of course N^[K/2] by the binomial theorem;
5) The smallest of these integers (corresponding to m=[K/2]) is exactly 1;

These properties guarantee that as long as K is at least N^2, it is possible to divide the integers above into N sets, such that each set will have sum exactly N^[K/2]-1, which once we return to the original probabilities by dividing by N^[K/2], yields exactly the wanted results.
The properties above imply this by implying something somewhat stronger: They imply that if we try dividing these integers into sets by taking at each step the largest integer not yet used, and trying to place it into a set such that it will not make its sum exceed N^[K/2]-1, we will succeed: Properties 2 and 3 and 4, together with K > N^2, imply that this process will not halt before we reach the smallest integers, and properties 4 and 5 imply that the smallest integers will always fit into the sets, because they are of size 1. This concludes this long proof.

Some analysis and comparison: Solutions #1 and #2 are quite similar, and in fact are completely the same in the case that N-1 is prime. The extension to other cases is quite different though: Solution #2 is efficient in coin tossing, always requiring only N-1 tosses, but the calculation of the polynomial f(P), and thus of P, might be long and annoying. On the other hand, solution #1 always uses a simple polynomial that needs to be solved to find P, but also depends on finding a prime in the arithmetic sequence, which might be difficult and gives no simple estimation of the number of coin tosses as a function of N. Solution 3 only requires solving a quadratic in order to obtain the probability p, and only O(N^2) tosses (possibly significantly less), but actually assigning the results seems more complicated than in #1 and #2.

You are of course right - thanks for your correction!

it goes like this:

we toss the coin K times.
the coin’s odds, P, are such that P^K+(1-P)^K=1/N.
I think we can find such P for every K>1, N>1. but I leave proving that as an exercise

if all tosses are the same, then we choose the number N.

if they’re not, then let’s look at the number of possible combinations that have the same heads/tails count, C.
there are K!/C!/(K-C)! such combinations.
if we chose K right, that divides by N-1 for every C except for 0 and K.
so we can group all of the tosses of same heads/tails count to N-1 equal groups. and each group will give a different number from 1 to N-1.
what K can do the job? I think maybe (N-1)^2 can. but I’ll leave that as an exercise too

maybe my blanks/exercises here can be filled and maybe not

Dan - you are welcome to post your solutions.

Thanks!