# Hat and Earring Guessing Puzzle, with penalty for passing

You are in the final round of a game show with 3,999 other conteseants.

Here is the game: The host has blindfolded you at put a red, green or blue hat and red, green or blue earrings (not necessarily matching) on each of you. Then he says truthfully, "I see a blue hat and blue earrings."

Now you will each stand in a line and each of you can see everyone else's hats and earrings except yours. Each of you must then either say two colours, say one colour or pass.

A two-colour guess is correct if one is your hat colour and one is your earring colour (in any order). A one-colour guess is correct if it is either your hat colour or your earring colour. The earrings of a player match each other.

If someone guesses wrong, the prize pool is divided by ten, rounded down.

If someone with a blue hat or blue earrings passes, £80 is subtracted from the prize pool.

If someone with neither a blue hat nor blue earrings passes, £25 is subtracted from the prize pool.

If someone guesses only one colour, and guessess correctly, the prize pool is multiplied by 0.95, rounded down.

The game ends when the prize pool is empty or everyone guessess correctly, then the prize pool is distributed to the players.

The players know the amount of money in the prize pool at all times.

What is the strategy to maximize your expected winnings if they go in the same order each round, and every round, each player gets one turn with unknown order?

• Do the earrings match each other? Nov 19, 2022 at 20:51
• If someone guesses only one colour... This triggers no matter if the guess is right or wrong? And if it is wrong does also the divide by 10 trigger? Nov 19, 2022 at 21:30
• All contestants know the current prize money (at all times)? Nov 19, 2022 at 21:32
• There are a lot of interacting components here - do you have reason to believe there's a "nice" solution to this?
– Deusovi
Nov 19, 2022 at 22:17
• "each player gets one turn with unknown order" is not clear to me, is that a separate question, or does this refer to the first round, or..? -- There is a 'nice' solution at least "if they go in the same order each round" Nov 19, 2022 at 23:37

There is a solution with no wrong guesses and only one correct single-color guess, assuming a strategy can be agreed upon beforehand.
Such a solution is optimal.

Note:
- Passing is the cheapest option by far, even an entire round of passing costs less than 5% of the price money.
- Only passing does not allow any signaling (assuming e.g. timing, pitch or fomulation of the "I pass" message cannot be used for signaling)
- All persons know if they wear blue after one pass (due to the different money deduction), and possibly immediately if they are the only one with a blue hat or the only one with blue earrings.
- No one will know both its colors through passing only.
- Odd/even signaling can be used: If a person/group of persons signals the other persons both
- if there is a odd number of green wearing persons in that 'other' group
- if there is a odd number of red wearing persons in that 'other' group
Then those others can reason out their colors after one pass (which let them know they have blue or not)

Phase one:
First round: Everyone passes
everyone now knows if blue is worn
Phase two:
Everyone passes except possibly the first 3 blue wearing persons, from them:
- single blue, means an odd number of green and an odd number of red*
- pass, single blue means an odd number of green and an even number of red*
- pass, pass, single blue means an even number of green and an odd number of red*
- pass, pass, pass means an even number of green and an even number of red*
*Counting the number of persons wearing that color outside the signaling group of 3.
Now all persons except the three signalers know their 2 colors.
Phase three:
The first two persons that are not one of the three earlier signalers will signal them their odd/evenness
- the first passes if the 3 have an odd number of red -- otherwise guesses 2 colors correctly.
- the second passes if the 3 have an odd number of green -- otherwise guesses 2 colors correctly.
Now also the first 3 signalers know their 2 colors.
Phase four:
Everyone guesses both their colors correctly

notes:
If there are only 2 blue wearing persons, the first on them is used twice in phase 2 (and phase 2 then ends in round 3) If there is only 1 blue wearing person, that person knows that already in round 1, and can start dignaling in round 1 already (and phase 2 then ends in round 3) Both these special cases have so many cheaper non-blue passes that the contestents end with more money than if everyone would wear blue