Quote:
Originally Posted by Eric
Again: no guarantee of finiteness, but in practice it will end.
Each person flips the coin. If all get the same result, do it again. Otherwise, the odd result is the winner.
OP was asking about solutions that don’t require “do overs” though. In practice you will always get a winner in a finite number of iterations. For any given N, the value 2^N has a remainder of either 1 or 2, so there are either 1 or 2 sequences of N flops that will not be assigned to one of the three players. The probability of such a sequence actually occurring is 2^-N if the remainder is 1 and 2^(-N+1) for a remainder of 2. In either case that probability tends to zero as N goes to infinity, so in practice we can make it as small as we like by increasing N. That doesn’t give a solution to OPs problem though. Since 2^N is never divisible by 3, no solution exists.