Best Strategy ---If you see two blue or two red hats, then write down the opposite color, otherwise write down "pass".

BBB

RRR

BBR

BRB

RBB

RRB

RBR

BRR

Out of the 8 possible scenarios, this strategy will work in 6/8 situations. Only in scenario 1 and 2 where all hats are 1 color will all of them answer wrongly. Which is optimal strategy since all the wrong answers are bunched together.

75% chance of success...

First, let's come up with a strategy that seems effective. Then, let's prove that it's *the

Optimal Strategy:

If you see two BLUE hats guess RED ,
If you see two RED hats guess BLUE .
Otherwise, pass.

Proof that this strategy wins 6/8 times:

Here are all 8 equally likely possibilities and their results with this strategy:

• red red red (EVERYONE guesses WRONG)
• red red BLUE (person 3 guesses CORRECTLY)
• red BLUE red (person 2 guesses CORRECTLY)
• BLUE red red (person 1 guesses CORRECTLY)
• blue blue RED (person 3 guesses CORRECTLY)
• blue RED blue (person 2 guesses CORRECTLY)
• RED blue blue (person 1 guesses CORRECTLY)
• blue blue blue (EVERYONE guesses WRONG)

This strategy gives a 6/8 chance of winning.

Proof that this strategy is optimal:

Because the hats are drawn independently, seeing the other two people's hats doesn't give you any information about your own. Thus if you decide to guess your own hat color, you MUST have exactly a 50% chance of being correct. Note that the solution that I'm proposing doesn't violate this. For example, let's say that you are person 1 and you see blue blue . According to the strategy proposed here, you should guess red . But there are two equally likely cases where you would see blue blue , namely red blue blue and blue blue blue . Thus you actually only have a half chance of being correct. But the reason that this strategy can win 6/8 times and not just 1/2 of the time is because it bunches together mistakes and spreads out successes .

Generally, with any strategy, if you if you list out all 8 equally likely possibilities, then for each person who guesses correctly in one of the 8 possibilities, he will guess wrong in another possibility (the one where he got the other color of hat). With our strategy, there are 6 correct guesses and 6 incorrect guesses. But the 6 correct guesses are spread out among 6 of the different 8 possibilities (one correct guess per possibility). On the other hand, the 6 incorrect guesses are concentrated in the remaining 2 or the 8 possibilities ( RRR and BBB ). This strategy is optimal because it spreads out the correct guesses and concentrates the incorrect guesses as much as possible. Specifically, there cannot be any strategy that wins in 7/8 possibilities because that would require having 7 correct guesses. But this of course implies that there must also be 7 incorrect guesses. And there is no way to concentrate 7 incorrect guesses on the remaining 1 of the 8 possibilities. Only 3 incorrect guesses can be concentrated into a possibility. Thus no strategy can achieve 7/8 and therefore 6/8 is optimal.

Extensions

How good is the optimal strategy for 4 people? 5? 6? In the limit as the number of people goes to infinity?

Try using the spreading+concentrating notion presented above to generate an upper limit on the performance for each of these problems, then see if you can actually construct a strategy to achieve this performance.

See if you can find the connection between the solutions to these puzzles and Hamming Codes.

Acknowledgements

Thanks Satyen Nabar for your excellent solution. I've written a somewhat more complete proof of optimality here.

The percentage chance of winning should be >50% if a fixed solution was discussed before the game: the probability for each to win is 50%,

A --> 50% B --> 50% C --> 50%

The best is only 1 guessed correctly and the other 2 persons chosen 'pass', the other case if more than 1 guessed correctly will not be consider, as mentioned if answer is not true, then is false, because the probability of winning in this case< first case.

so, the best solution/strategy they discussed must be {win, pass, pass}, rearrange we got 3 sets.

0.5^3=0.125, 0.125*3=0.375 (3 sets included)

if they lose, minimum probability also =0.375

optimum chance of winning: 0.375(from best strategy)+0.375(if they lose//worst strategy)=0.75(winning probability)