This problem involves a classic puzzle related to probability, strategy, and information theory. To maximize the probability that each player guesses their hat color correctly, they can agree on a strategy beforehand, even though they cannot communicate once the hats are placed.

Key Facts:

• There are three players.
• Each player wears either a red or blue hat, determined by a fair coin toss (so each hat has a 50% chance of being red or blue).
• Each player can see the hats of the other two players but not their own.
• They must guess their hat color simultaneously, with no communication after seeing the hats.
• They are allowed to meet beforehand to devise a strategy.

Strategic Insight:

A clever strategy the players can use is based on the parity of the number of red hats. Here's the strategy:

1. Each player observes the other two players' hats.
2. If a player sees two hats of the same color, they should guess their own hat is the opposite color.
3. If a player sees two hats of different colors, they should guess their own hat is red.

Why this strategy works:

The players' goal is to maximize the number of correct guesses, not necessarily ensuring that all players guess correctly. If the total number of red hats is even (i.e., 0 or 2 red hats), exactly one player will guess correctly under this strategy. If the number of red hats is odd (i.e., 1 or 3 red hats), all three players will guess incorrectly. However, this strategy ensures that at least one player always guesses correctly.

Probability Calculation:

• There are $2^3 = 8$ possible combinations of hat colors for the three players, since each hat can be either red or blue.
• The parity of red hats in each case determines the outcome:
  • If the number of red hats is even (0 or 2), exactly one player will guess correctly.
  • If the number of red hats is odd (1 or 3), all players will guess incorrectly.

We can list out the possible combinations:

1. (Red, Red, Red) → 3 reds (odd), all guess wrong.
2. (Red, Red, Blue) → 2 reds (even), one player guesses correctly.
3. (Red, Blue, Red) → 2 reds (even), one player guesses correctly.
4. (Red, Blue, Blue) → 1 red (odd), all guess wrong.
5. (Blue, Red, Red) → 2 reds (even), one player guesses correctly.
6. (Blue, Red, Blue) → 1 red (odd), all guess wrong.
7. (Blue, Blue, Red) → 1 red (odd), all guess wrong.
8. (Blue, Blue, Blue) → 0 reds (even), one player guesses correctly.

Out of these 8 possibilities:

• In 5 cases, one player guesses correctly.
• In 3 cases, all players guess wrong.

Thus, the probability that each individual player guesses correctly is the probability that they are in one of the cases where the total number of red hats is even, which occurs in 5 out of 8 cases.

Therefore, the probability that each player guesses correctly (not necessarily all at the same time) is:

$\boxed{\frac{5}{8}}$

This is the highest probability achievable using a coordinated strategy like the one described above.

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Would you like further details or have any other questions?

Here are five related questions to explore:

1. What would happen to the probability if there were four players instead of three?
2. How would the strategy change if communication were allowed between guesses?
3. What is the minimum number of players required to guarantee at least one correct guess in every scenario?
4. Can this strategy be generalized for more players or different hat colors?
5. How does the fairness of the coin toss affect the probability outcome?

Tip: In many problems involving independent events, focusing on symmetries and patterns like parity (odd vs. even) can simplify strategy design.