Solution
At Least Two of Three
Show the problem again
Three players agree on a strategy, after which each receives a hat that is red or blue, chosen independently by a fair coin. Each player sees the other two hats but not their own, and no communication is allowed once the hats are on. Simultaneously and silently, each player must guess the colour of their own hat (passing is not allowed). The team wins if at least two of the three guesses are correct. What is the largest possible probability of winning? Give a reduced fraction.
Worked solution
The answer is 3/4. Upper bound. For any fixed view of the other two hats, a player's guess is right for exactly one of the two possible colours of their own hat. Over the 8 equally likely colourings, each player is therefore right in exactly 4, so the total number of correct guesses, summed over colourings and players, is 12. A winning colouring accounts for at least 2 correct guesses, so at most 12 / 2 = 6 of the 8 colourings can win, i.e. a probability of at most 3/4. Achievability. Call the players Alice, Bob and Carol. Alice always guesses red. Bob guesses whatever colour he sees on Carol. Carol guesses blue if Alice's and Bob's hats match and red otherwise. Checking the 8 colourings, exactly two guesses are correct in each of six of them (and none in the other two, namely Alice blue, Bob red, Carol blue, and Alice blue, Bob blue, Carol red). So the bound of 6 winning colourings out of 8 is attained. (Confirmed by exhaustive search over all 16^3 strategies.)
Source: Variant of the classic three-player hat-guessing game; original problem written for AxiomIQ.