1 The 2-hat, 2-player game Consider the following infinite hat game. There are two players, Alice and Bob. Each wears an infinite sequence of hats, each hat either black or white. Think of each stack as a function A, B : N → { 0 , 1 } Each player sees the other’s entire stack, but not their own. Simultaneously, each player: – chooses two positions on their own head (two natural numbers), and – announces a color for each of those positions. The game is won if at least one player’s two announced colors match exactly the actual colors of the two hats chosen on that player’s head. A strategy consists of a deterministic rule for Alice and a deterministic rule for Bob: given the observed infinite hat stack on the other head, each player chooses two indices and two colors. The question is: Is there a strategy for Alice and Bob that guarantees a win for every possible pair of infinite hat stacks? The answer is no . Moreover, the non-existence of such a strategy can be proved in ZF (Zermelo–Fraenkel set theory, without the Axiom of Choice); once it is a theorem of ZF, the same statement remains true in any stronger theory such as ZFC or ZFC+CH, by monotonicity of entailment; the argument can be arranged so that no issues of measurability arise, even if strategies are allowed to be arbitrarily pathological functions. The remainder explains this in detail and then shows how moving to four players changes the situation completely: with four players, there is a simple finitary winning strategy. 2 Formalising strategies (deterministic only) Let Alice’s stack be A ∈ { 0 , 1 } N and Bob’s be B ∈ { 0 , 1 } N A strategy for Alice is a function S A : { 0 , 1 } N → N 2 × { 0 , 1 } 2 which, given Bob’s infinite sequence B , outputs S A ( B ) = ( i 1 , i 2 , α 1 , α 2 ) , meaning: “the hats at positions i 1 , i 2 on Alice’s head are α 1 , α 2 .” A strategy for Bob is likewise S B : { 0 , 1 } N → N 2 × { 0 , 1 } 2 , S B ( A ) = ( j 1 , j 2 , β 1 , β 2 ) Randomised strategies do not change the analysis: if a randomised strategy for Alice and Bob succeeded with probabil- ity 1 on every hat assignment, then fixing the internal random bits to some particular outcome would yield a deterministic strategy that also wins on every assignment. Therefore it suffices to show that no deterministic strategy can always win. So suppose arbitrary deterministic strategies S A and S B are given. The goal is to derive a contradiction. 1 3 Reducing to a finite probability space The key idea is: If a deterministic strategy for Alice and Bob won on every hat assignment, then for every probability distri- bution on hat assignments the probability of winning would be 1. To refute the existence of such a strategy, it is enough to exhibit a single finite probability space of assignments on which the win probability is < 1. The crucial point is that this probability space will be finite and built from a small number of independent fair coin flips. In a finite sample space, every subset is measurable; thus no measurability or outer-measure subtleties appear. 3.1 Fixing the structure of the sample space Fix the two strategies S A , S B Define a finite sample space of hat assignments as follows. 1. First, choose a fair bit C ∈ { 0 , 1 } . This bit determines all of Alice’s hats: A ( n ) = C for all n. Thus Alice always has a monochromatic stack: all white ( C = 0) or all black ( C = 1). 2. Next, consider Bob’s strategy on the two monochromatic stacks: S B (all 0s) = ( j (0) 1 , j (0) 2 , β (0) 1 , β (0) 2 ) , S B (all 1s) = ( j (1) 1 , j (1) 2 , β (1) 1 , β (1) 2 ) The indices Bob might guess in these two cases form a finite set J B := { j (0) 1 , j (0) 2 , j (1) 1 , j (1) 2 } , so | J B | ≤ 4. 3. For Bob’s hats, only the positions in J B will be random. For each k ∈ J B , choose a fair bit X k ∈ { 0 , 1 } independently. For every position n / ∈ J B , set B ( n ) = 0 (say, white). Thus a hat assignment in this probability space is determined by: one bit C for Alice’s color; one bit for each position in J B on Bob’s stack (at most four more bits). In total there are at most 2 1+ | J B | ≤ 2 5 possible assignments, each occurring with equal probability. Everything is finite. Note that Alice’s stack and Bob’s stack are independent in this construction: the bit C is independent of the collection ( X k ) k ∈ J B 4 Bounding the success probabilities Let: E A be the event “Alice’s two guesses are both correct”, E B be the event “Bob’s two guesses are both correct”. The game is won exactly on the event E A ∪ E B . The next step is to show: P ( E A ) ≤ 1 2 , P ( E B ) = 1 4 Then the union bound gives P ( E A ∪ E B ) ≤ P ( E A ) + P ( E B ) ≤ 1 2 + 1 4 = 3 4 < 1 , so the strategy cannot win for all assignments. 2 4.1 Alice cannot exceed probability 1 / 2 Alice’s hats are all equal to the single bit C . For any realised B , the strategy S A ( B ) outputs some indices and colors S A ( B ) = ( i 1 , i 2 , α 1 , α 2 ) For this assignment, Alice’s actual hats at positions i 1 , i 2 are both equal to C : A ( i 1 ) = A ( i 2 ) = C. Thus Alice’s two guesses are both correct if and only if ( α 1 , α 2 ) = ( C, C ) The crucial observation: the pair ( α 1 , α 2 ) is a function of B only. Since B is determined solely by the independent bits ( X k ) k ∈ J B , and C is an independent fair bit, the vector ( α 1 , α 2 ) is independent of C Consequently, for each fixed B , either α 1 ̸ = α 2 , in which case Alice’s probability of success (over the random bit C ) is 0; or α 1 = α 2 = α , in which case Alice’s probability of success is P ( C = α ) = 1 2 In all cases, P ( E A | B ) ≤ 1 2 Taking the expectation over all possible B gives P ( E A ) ≤ 1 2 Thus even with full access to Bob’s stack and arbitrary dependence on it, Alice cannot beat the one-bit information limit: there is only one unknown bit C describing an entire monochrome stack, so the probability that two independent guesses of that same bit are both correct cannot exceed 1 / 2. 4.2 Bob cannot exceed probability 1 / 4 Consider Bob’s performance. For each monochrome stack on Alice’s side: if C = 0, then Alice’s stack is all zeros and S B (all 0s) = ( j (0) 1 , j (0) 2 , β (0) 1 , β (0) 2 ) , if C = 1, then Alice’s stack is all ones and S B (all 1s) = ( j (1) 1 , j (1) 2 , β (1) 1 , β (1) 2 ) For a fixed value of C , the indices j ( C ) 1 , j ( C ) 2 and guesses β ( C ) 1 , β ( C ) 2 are now just definite numbers determined by the strategy. On the other hand, the hats at those positions on Bob’s head are: B ( j ( C ) 1 ) = X j ( C ) 1 , B ( j ( C ) 2 ) = X j ( C ) 2 , and the bits X k for k ∈ J B were chosen independently and uniformly, independently of C Thus the pair ( B ( j ( C ) 1 ) , B ( j ( C ) 2 ) ) is a pair of independent fair bits, independent of the guesses ( β ( C ) 1 , β ( C ) 2 ), which are fixed once C is fixed. Therefore, conditional on C , P ( ( B ( j ( C ) 1 ) , B ( j ( C ) 2 )) = ( β ( C ) 1 , β ( C ) 2 ) ∣ ∣ ∣ C ) = 1 4 Averaging over C gives P ( E B ) = 1 4 So Bob cannot beat the basic limit for guessing two independent unseen fair bits. 3 4.3 Combining the bounds The event that the game is won is E A ∪ E B . For any two events, P ( E A ∪ E B ) ≤ P ( E A ) + P ( E B ) , without any independence assumption between the events themselves. Using the computed bounds, P ( E A ∪ E B ) ≤ 1 2 + 1 4 = 3 4 < 1 Therefore, in this finite probability space, the strategy given by S A , S B fails with probability at least 1 4 . That implies that there is at least one specific hat assignment in the finite sample space on which both players are wrong in at least one of their guesses. But the sample space contains only assignments that are valid configurations of the original game (two infinite stacks of hats). If a strategy failed even once inside this finite set, it cannot be a winning strategy that succeeds on every possible assignment. Hence: No deterministic strategy can guarantee a win in the 2-hat, 2-player game. 5 Why ZF suffices, and why the result persists in stronger theories The argument above uses only: basic finite combinatorics; a finite probability space with at most 2 5 elements; simple inequalities for probabilities in finite spaces. All of this is formalizable in ZF. No use is made of the Axiom of Choice, nor of constructions (such as arbitrary choice functions) that would require it. Moreover, there is no appeal to infinite product measures or non-measurable sets: The only randomness arises from finitely many independent fair coin flips. Every subset of the finite sample space is measurable. The success events E A and E B are simply finite unions of atoms in this finite sample space; their probabilities are exactly the sums of the relevant atomic probabilities. Thus the non-existence of a winning 2-player, 2-hat strategy is a theorem of ZF. Once a statement is provable in ZF, it remains valid in any stronger axiomatic system that contains ZF as a subsystem, by monotonicity of entailment: If ZF ⊢ “no winning strategy exists”, then there is no model of ZF in which a winning strategy exists. Any model of a stronger theory such as ZFC or ZFC+CH is, in particular, a model of ZF. Therefore the stronger theory cannot have a model in which a winning strategy exists either; otherwise that model would contradict the ZF theorem. Hence the conclusion: There is no winning strategy in the 2-hat, 2-player game in ZF, and therefore no such strategy exists in ZFC, ZFC+CH, or any other extension of ZF. The subtle measure-theoretic issues that arise in other infinite hat problems (for example, when there are three or more players, or when outer measure is used to control non-measurable strategies) simply do not appear here, because the argument never leaves the realm of finite probability spaces. 4 6 Increasing the number of players: a winning strategy for 4 players When the number of players increases, the situation changes dramatically. For the same “2 hats guessed” rule, there is a winning strategy with four players, again completely finitary and formalizable in ZF. 6.1 Statement of the 4-player game The game is now: There are four players, labelled 0 , 1 , 2 , 3. Each wears an infinite binary hat stack. Each player sees all other stacks, but not their own. Simultaneously, each player chooses two positions on their own head and guesses the colors there. The team wins if at least one player is correct on both guessed hats. The aim is to exhibit a deterministic strategy that works for every hat assignment. 6.2 Coding two hats as a number mod 4 Fix, once and for all, the bottom two hats as the positions to be guessed: For player k , positions 1 and 2 (say, the first two hats from the bottom) will always be the guessed positions. Encode the colors of these two hats as a number in { 0 , 1 , 2 , 3 } : white = 0, black = 1; if the bottom two hats of player k are ( h k, 1 , h k, 2 ) ∈ { 0 , 1 } 2 , define c k := 2 · h k, 1 + h k, 2 ∈ { 0 , 1 , 2 , 3 } Thus each player’s bottom two hats determine an integer code c k between 0 and 3. Let S := c 0 + c 1 + c 2 + c 3 (mod 4) This total sum modulo 4 is an element of { 0 , 1 , 2 , 3 } 6.3 The strategy The strategy is symmetric and uses only the bottom two hats of each player. For each player n ∈ { 0 , 1 , 2 , 3 } : 1. Player n observes all other players’ bottom two hats, and so knows the codes c k for all k ̸ = n 2. Player n computes the partial sum s n := ∑ k ̸ = n c k (mod 4) 3. Player n now guesses that own code is g n := n − s n (mod 4) That is, player n guesses the two bottom hats on their own head are the binary expansion of the number g n 4. Concretely, player n guesses: position 1 has color ⌊ g n / 2 ⌋ , position 2 has color g n mod 2. This is a purely local rule: each player uses only the labels 0 , 1 , 2 , 3, the visible bottom hats of the other three players, and arithmetic mod 4. 5 6.4 Verification Let S be the true total sum modulo 4 of all four codes: S = c 0 + c 1 + c 2 + c 3 (mod 4). Consider the player whose label equals this sum: player S For this player, s S = ∑ k ̸ = S c k = S − c S (mod 4) Therefore g S = S − s S = S − ( S − c S ) = c S (mod 4) So player S guesses own code correctly: the guessed integer g S equals the actual code c S , and both bottom hats are correctly identified. Thus: For every possible configuration of four infinite hat stacks, exactly one of the four players (namely the one whose label equals the total sum of the codes mod 4) is guaranteed to guess both of their chosen hats correctly. This is a winning strategy for the 4-player, 2-hat game. The construction uses only: the bottom two hats of each player; elementary arithmetic mod 4; no Axiom of Choice or other strong set-theoretic principles. Consequently, the existence of this winning strategy is a theorem of ZF, and therefore also holds in ZFC, ZFC+CH, and similar extensions. 7 Summary In the 2-player, 2-hat infinite game, no deterministic strategy can guarantee that at least one player always guesses both hats correctly. – This is shown by constructing a finite probability space on at most 2 5 hat assignments and demonstrating that, for any fixed choice of strategies for Alice and Bob, the probability of success is at most 3 / 4. – The argument uses only finite combinatorics and finite probability; there is no interaction with non-measurable sets or outer measure. – Thus the impossibility is provable in ZF, and hence also in ZFC and ZFC+CH by monotonicity of entailment. In contrast, in the 4-player, 2-hat game, there is a simple winning strategy: – Encode each player’s bottom two hats as an integer in { 0 , 1 , 2 , 3 } ; – let players coordinate their guesses so that the player whose label equals the total sum mod 4 guesses own code correctly. – This provides a finitary ZF proof of solvability for four players, again automatically valid in stronger systems. The 2-player case therefore illustrates a clean boundary: with two players, no amount of cleverness or choice-based construction can defeat the basic probabilistic obstruction; with four players, a simple modular-arithmetic construction suffices to guarantee success in all cases. 6