ANANymous

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1 hour ago, bonanova said:Following the lead of the 3person example, look for strategies where everyone will guess wrong in a very few special cases, i.e. 1111 and 0000. If you can get all the wrong individual guesses into those two cases you can achieve 14/16 = 0.875 success probability. And you can't do better than that. 1111 and 0000 have the same likelihood of success, and they together comprise 12.5% of the cases. That's the smallest available container for housing the wrong individual responses.
Herein lies my initial question: How do we achieve 14/16, what is the selection method used by the 4 hat wearing participants? Is that solvable?

I appreciate the responses araver; however this link does not provide the solution I am looking for:
http://brainden.com/forum/topic/13034
And the answer I believe is greater than the answer to 3 people. I believe it is not 50% based on the person who asked this question initially (who also knew the answer, but whom I can no longer locate).
SpoilerI am almost certain the answer is 7/8ths, but I am unsure how the selection process works (my confidence for the answer comes from the question giver acknowledging this as the right answer). The trick is I cannot explain how the selection process works for n = 4.

Can someone solve the following brainteaser with an explanation of how the people in the problem choose hat color:
There are 4 people wearing black or white hats. They do not know the color of their own hat. They cannot communicate. They must come up with a strategy for simultaneously shouting out a hat color or keeping their mouth shut. Your goal is to come up with the solution that yields the highest probability of getting at least one hat color correct and no hat colors incorrect. For example, if the hats are black black black white in that order, and you guess black, doesn't guess, doesn't guess, doesn't guess, then this would count as a successful attempt; however if you guessed black, black, black, black, this would be unsuccessful.
Semi spoiler belowthe solution for 3 people:
Using 0 as representation for white and 1 for black, you have the following table seating available. 000, 001, 010, 011, 100, 101, 110, 111
If every time you are at the table and you see two hats of the same color, you guess the opposite color, you will guess correctly in 6 out of 8 attempts, and if you do not see two hats of the same color, then you pass/ do not guess. This yields a 75% win rate for this game among the 3 participants.
Thank you in advance for answers.
4 hats puzzle
in New Logic/Math Puzzles
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This would be considered cheating I believe, it is not the solution being sought, but was a nice try