Several different "hat games" have recently received a fair amount of attention. Typically, in a hat game, one or more players are required to correctly guess their hat colour when given some information about other players' hat colours. Some versions of these games have been motivated by research in complexity theory and have ties to well-known research problems in coding theory, and some variations have led to interesting new research. In this paper, we review Ebert's Hat Game, which garnered a considerable amount of publicity in the late 90's and early 00's, and the Hats-on-a-line Game. Then we introduce a new hat game which is a "hybrid" of these two games and provide an optimal strategy for playing the new game. The optimal strategy is quite simple, but the proof involves an interesting combinatorial argument.
A few years prior to the introduction of Ebert's Hat Game, in 1994, a similar game was described by Aspnes, Beigel, Furst and Rudich @cite_7 . In their version of the game, players are not allowed to pass, and the objective is for a majority of the players to guess correctly. For the three-player game, it is easy to describe a strategy that will succeed with probability @math , just as in Ebert's game: Alice votes the opposite of Bob's hat colour; Bob votes the opposite of Charlie's hat colour; and Charlie votes the opposite of Alice's hat colour.
@cite_7
1990538552
In a two-player game, two cooperating but non communicating players, Alice and Bob, receive inputs taken from a probability distribution. Each of them produces an output and they win the game if they satisfy some predicate on their inputs outputs. The entangled value ω *(G) of a game G is the maximum probability that Alice and Bob can win the game if they are allowed to share an entangled state prior to receiving their inputs. We show a parallel repetition theorem for the entangled value @math of any two-player one-round game @math where the questions @math to Alice and Bob are drawn from a product distribution on @math . We show that for the @math -fold product @math of the game @math (which represents the game @math played in parallel @math times independently), @math , where @math and @math represent the sets from which the answers of Alice and Bob are drawn. We study the following one-person game against a random graph process: the Player's goal is to @math -colour a random sequence of edges @math of a complete graph on @math vertices, avoiding a monochromatic triangle for as long as possible. The game is over when a monochromatic triangle is created. The online version of the game requires that the Player should colour each edge as it comes, before looking at the next edge.While it is not hard to prove that the expected length of this game is about @math , the proof of the upper bound suggests the following relaxation: instead of colouring online, the random graph is generated in only two rounds, and the Player colours the edges of the first round before the edges of the second round are thrown in. Given the size of the first round, how many edges can there be in the second round for the Player to be likely to win? In the extreme case, when the first round consists of a random graph with @math edges, where @math is a positive constant, we show that the Player can win with high probability only if constantly many edges are generated in the second round. In the generalized Russian cards problem, Alice, Bob and Cath draw a, b and c cards, respectively, from a deck of size a+b+c. Alice and Bob must then communicate their entire hand to each other, without Cath learning the owner of a single card she does not hold. Unlike many traditional problems in cryptography, however, they are not allowed to encode or hide the messages they exchange from Cath. The problem is then to find methods through which they can achieve this. We propose a general four-step solution based on finite vector spaces, and call it the ''colouring protocol'', as it involves colourings of lines. Our main results show that the colouring protocol may be used to solve the generalized Russian cards problem in cases where a is a power of a prime, c=O(a^2) and b=O(c^2). This improves substantially on the set of parameters for which solutions are known to exist; in particular, it had not been shown previously that the problem could be solved in cases where the eavesdropper has more cards than one of the communicating players.