(Infinite) Hat and Box Riddles u/Skaib1 Abstract This is just a list of personal favorites. Sorry for not crediting the original creators in most cases. I claim no originality 1 I tried to explicitly mention any prerequisites beyond "school-knowledge". Know any other infinite ones? I’d love to hear them! The most paradoxical ones are 2 α , 3b ′′′′ , 3c ′ , 4b and 5. Oh and for a plot twist, 6 is not even hat-related; you’re welcome. 1) Hat riddle warm-up (finite, no prerequisites). Hint: solution to version of (b) [Wikc] and (c) [Wikb]. Each riddle has their infinite counterpart. a) Two players, each is given a black or white hat to wear. Each player sees the other’s hat but not their own. They must guess their own hat color simultaneously. Exactly one player must guess correctly. 2 a ′ ) Three players, each is given a hat. There are three possible hat colors. Each player sees the others’ hats. Exactly one player must guess correctly. a ′′ ) Now with 100 players and 100 hat colors. - infinite version: riddle (2a). b) 100 players, hats are black or white. They must either all guess correctly or all guess incorrectly. b ′ ) Now with 7 hat colors. - infinite version: riddle (2b). c) There are 100 players standing in a line. Hats are black or white. Each player sees only those in front of them. There is a player at the back who sees everyone. He guesses first, then the next, and so on. Everyone hears all prior guesses. At most one guess may be wrong. c ′ ) Now with 7 hat colors. - infinite version: riddle (2c). 1 Maybe (3b ′′′ ). It looks a bit ugly (but the solution is neat! The ’naïve’ one doesn’t work). 2 Rules: In all hat/beanie riddles the players may agree on a strategy among themselves before they receive the hats. Your goal is to come up with a winning strategy. After the meeting, no communication of any kind is possible and all players must guess simultaneously, unless stated otherwise. They must always guess the color of their own hat(s). There are no further restrictions on the hats, e.g. they could all be wearing the same color, even if more colors are allowed. If relevant: each player has a distinct name and knows everyone else’s name. They know the possible colors and the rules. ’Guaranteeing’ something or having a ’winning strategy’ means that they win in every single possible scenario. 1 2) Hat riddle (infinite 3 , Axiom of Choice is allowed 4 ). α ) Countably infinitely many players. Each player sees all the others. Hats are black or white. All guess simultaneously. Only finitely many may guess incorrectly. (Hint: solution to ( α ) is here [Wika]. With that in hand, the rest of (2) is solvable-ish. α ′ ) Each player may choose whether to guess or remain silent; all but finitely many must guess. At most one guess may be wrong. α ′′ ) As in ( α ), but they stand in a line: each sees only those in front, and there is a first player who sees everyone else. All guess simultaneously , and only finitely many may guess incorrectly. i) As in ( α ′′ ), but each player in the line is replaced by a countably infinite line of players as in ( α ′′ ), each facing in the same direction and seeing all players ahead of them, including those in subsequent lines. ii) (this might require some familiarity with ordinals...) As in ( α ′′ ), but the line of players is replaced by an arbitrary (even uncountable) well- ordered set of players, each seeing the hats of all players larger in the ordering. a) Riddle (1a) but with countably infinitely many players and colors b) Riddle (1b) but with countably infinitely many players. b ′ ) . . . and countably infinitely many colors. c) Riddle (1c) but with countably infinitely many players. c ′ ) . . . and countably infinitely many colors. 3) Beanie riddle. a) (no Axiom of Choice, only “school-level” knowledge) Two players, each wearing a tower of countably infinitely many beanies, each beanie white or black. Both see the other’s beanies. They must simultaneously each point to one of their own beanies. They win if both point to a black one. Goal: 1 / 3 chance of winning (each beanie’s color is determined by an independent fair coin flip). a ′ ) Now with three players. All must point to a black beanie. Goal: 1 / 4 chance of winning. a ′′ ) n players. 1 / ( n + 1) chance of winning. - Fun fact: It’s an open problem what the optimal strategy is. It’s conjectured that the probability goes to 0 with n → ∞ The conjecture for n = 2 is 7 / 20 . See [Alo+23]. b) Two players, each wearing one beanie. Beanies are black or white. They must simultaneously guess the color of their own beanie. One must guess correctly. 3 We assumes the players each have the ability to memorize an arbitrary (infinite, if necessary) amount of information and perform ’infinitely complex computations’. 4 They have access to a choice function. 2 b ′ ) Now both are wearing n beanies. Beanies come in three colors. Both must simultaneously guess a color for each of their beanies (all 2 n guesses at the same time). At least one of the 2 n guesses must be correct. Show that there exists a n for which this works. b ′′ ) Same as (b ′ ), but with k colors. b ′′′ ) (no Axiom of Choice, but still a bit more than school-knowledge) As in (b ′′ ), but player A has countably infinitely many beanies and player B has continuum-many beanies. Beanies come in countably many colors. One of the 2 × ∞ many guesses must be correct. The rest is the same. b ′′′′ ) (Axiom of Choice + Continuum Hypothesis + some familiarity with ordi- nals) Both wear a tower of countably infinitely many beanies. The color is replaced by a real number (we have continuum-many colors). The rest is the same. (Hint: riddle 4b) c) (no Axiom of Choice, only “school-level” knowledge) n players, each wearing a tower of countably infinitely many beanies, each beanie white or black. Each player sees the others’ beanies. They must simultaneously each point to three of their own beanies and guess the color of each. They win if at least one of the n players guesses all three correctly. Show that for n = 8 there is a winning strategy. c ′ ) (Axiom of Choice + Continuum Hypothesis + some familiarity with ordi- nals) Show that for n = 4 there is a winning strategy (Hint: riddle 4b)). This riddle is due to Elliot Glazer. 4) Wizard riddle. a) Two wizards each receive a hat labeled with a natural number. Each sees the other’s number but not their own. Simultaneously, each submits a list of finitely many natural numbers. They win if at least one submitted list contains the wizard’s own number. Find a winning strategy. b) (Axiom of Choice + Continuum Hypothesis + some familiarity with ordinals) Now there are three wizards and each hat bears a real number. 5) Box riddle (Axiom of Choice is allowed). a) A team of 100 players waits outside a room containing a countably infinite sequence of boxes, each holding a real number. One by one, each player enters, may open arbitrarily many (even infinitely many) boxes, but must then guess the value in one box that he didn’t open. The player closes all boxes and exits. No communication is allowed once the game begins. The team wins if at least 99 players guess correctly. b) Now each player must guess the content of infinitely many unopened boxes; all but at most one player must have all their guesses correct. c) Now in addition to (b) there are countably infinitely many players. 3 - Fun fact: The riddle is also solvable without the Axiom of Choice if the number of boxes is increased (requires cardinality the power set of R ). See [Gla22]. 6) Light Bulbs. ( copied word by word from [ulo]. Note: In this riddle ’guaranteeing success’ means ’guaranteeing success with probability 1’. Prerequisites: some basic probability theory and limits... ) a) There are 100 prisoners isolated from one another. Their only form of communica- tion is a room with a single light bulb. Everyday one prisoner is chosen uniformly at random to go to the light bulb room, and they can either toggle the light or leave it alone. They are then lead back to their cell. Once one of the prisoners know that all 100 have visited the room, then they are all freed. They can meet briefly beforehand to discuss a strategy. How can they guarantee eventual success (with probability 1)? b) We add an additional constraint: The prisoners cannot keep track of days. They don’t even have a rough idea of what day it is. That part of their brain is fried. So counting days from the meeting, or counting days between visits to the room, is not allowed. They are perfect logicians otherwise. c) In addition to (b), just to be a jackass, the warden will, with some small constant probability, just turn off the light instead of letting a prisoner into the room. The warden will never turn on the light. References [Alo+23] Noga Alon et al. The success probability in Levine’s hat problem, and independent sets in graphs . 2023. arXiv: 2208.06858 [math.CO] url : https://arxiv.org/ abs/2208.06858 [Gla22] Elliot Glazer. A choiceless box game paradox . 2022. arXiv: 2211.10474 [math.LO] url : https://arxiv.org/abs/2211.10474 [ulo] u/lordnorthiii. “100 prisoners, a light bulb, and a jackass warden” url : https: //www.reddit.com/r/mathriddles/comments/przpqu/100_prisoners_a_ light_bulb_and_a_jackass_warden/ (visited on 08/02/2025). [Wika] Wikipedia contributors. Countably infinite Hat variant url : https : / / en . wikipedia.org/wiki/Induction_puzzles#Countably_infinite-Hat_variant_ without_hearing (visited on 08/02/2025). [Wikb] Wikipedia contributors. Ten Hat variant hearing url : https://en.wikipedia. org/wiki/Induction_puzzles#Ten-hat_variant (visited on 08/02/2025). [Wikc] Wikipedia contributors. Ten Hat variant no hearing url : https://en.wikipedia. org/wiki/Induction_puzzles#Ten- hat_variant_without_hearing (visited on 08/02/2025). 4