Another hat puzzle (six colors, six players)
June 8th, 2012 at 8:20:33 PM
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I do like these sorts of puzzles. Here is a similar puzzle that I like (not original or anything).
A team of six players is told the following rules: They will each be given a hat in one of six specified colors (black, white, red, yellow, green, and blue). There is no guarantee that all six colors will be used: it might be all one color, it might be five of one color and one other color, etc. They each will be able to see all the other hats, but not their own. Absolutely no communication will be allowed. Each player secretly and independently writes down a guess for his or her own hat color. All six guesses must be made before any are checked. If any player guesses their own hat color correctly, each player wins $1,000. If none do, they each lose $19,000.
Having been told the rules, they are given the opportunity to confer and decide on a strategy and on whether or not to play.
What is the best strategy? Should they play?
A natural question is how the hat colors will be selected. If you think this affects the best strategy, you can choose either an easy or a hard scenario, where the players are told the rule in each case:
Easy: Each hat color is selected independently and uniformly at random.
Hard: An opponent listens to their entire discussion about their strategy, and, knowing what their strategy will be, decides how to chose the hat colors (which may still be random), so as to minimize their probability of winning. In this case, the solution does not involve tricking their opponent--this is a logic puzzle--just assume that their opponent is a demon who can read their minds (but not up to the point of predicting their random choices after they're done conferring).
"Easy" just means that it's easy to analyze the probability of winning for a given strategy, not the figuring out the best strategy is easy. I will say that figuring out the best strategy does not require a computer or anything like that.
Note that the method of selecting the hat colors might not matter in the first place. For example, with the simple strategy where each player guesses a color independently and uniformly at random, the probability of winning is 1-(5/6)^6=66.5%, no matter how the hat colors were selected.
(Also, you can assume that they are adequately bankrolled to maximize their EV. The gambling is just flavor for the importance of a high probability of winning, since it's not hard for someone to luck out.)
Please use a spoiler for solutions that beat the offered odds (95%) or proofs/arguments that no such solution exists.
A team of six players is told the following rules: They will each be given a hat in one of six specified colors (black, white, red, yellow, green, and blue). There is no guarantee that all six colors will be used: it might be all one color, it might be five of one color and one other color, etc. They each will be able to see all the other hats, but not their own. Absolutely no communication will be allowed. Each player secretly and independently writes down a guess for his or her own hat color. All six guesses must be made before any are checked. If any player guesses their own hat color correctly, each player wins $1,000. If none do, they each lose $19,000.
Having been told the rules, they are given the opportunity to confer and decide on a strategy and on whether or not to play.
What is the best strategy? Should they play?
A natural question is how the hat colors will be selected. If you think this affects the best strategy, you can choose either an easy or a hard scenario, where the players are told the rule in each case:
Easy: Each hat color is selected independently and uniformly at random.
Hard: An opponent listens to their entire discussion about their strategy, and, knowing what their strategy will be, decides how to chose the hat colors (which may still be random), so as to minimize their probability of winning. In this case, the solution does not involve tricking their opponent--this is a logic puzzle--just assume that their opponent is a demon who can read their minds (but not up to the point of predicting their random choices after they're done conferring).
"Easy" just means that it's easy to analyze the probability of winning for a given strategy, not the figuring out the best strategy is easy. I will say that figuring out the best strategy does not require a computer or anything like that.
Note that the method of selecting the hat colors might not matter in the first place. For example, with the simple strategy where each player guesses a color independently and uniformly at random, the probability of winning is 1-(5/6)^6=66.5%, no matter how the hat colors were selected.
(Also, you can assume that they are adequately bankrolled to maximize their EV. The gambling is just flavor for the importance of a high probability of winning, since it's not hard for someone to luck out.)
Please use a spoiler for solutions that beat the offered odds (95%) or proofs/arguments that no such solution exists.
June 9th, 2012 at 12:04:47 PM
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I was thinking of posting a problem like this, but with 10 hats.
Note that this works regardless of the number of people, if the number of people is the same as the number of colors.
Assign each person a number from 0 to 5, and each color a number from 0 to 5.
Each person guesses the color which, if correct, makes the sum of the numbers of all six hats equal to their own number modulo 6.
The sum of the numbers of all six hats, modulo six, has to equal one of the six persons' numbers, so somebody has to be right.
Example: number the colors 0 for black, 1 for white, 2 for red, 3 for yellow, 4 for green, and 5 for blue.
Let persons 0-5 have hats white, white, red, blue, white, and black; the numbers are 1 + 1 + 2 + 5 + 1 + 0 = 10.
Person 0 sees 1 + 2 + 5 + 1 + 0 = 9; to get it to equal 0 mod 6, person 0's hat would have to be 3, so he guesses "yellow".
Person 1 sees 1 + 2 + 5 + 1 + 0 = 9; to get it to equal 1 mod 6, person 1's hat would have to be 4, so he guesses "green".
Person 2 sees 1 + 1 + 5 + 1 + 0 = 8; to get it to equal 2 mod 6, person 2's hat would have to be 0, so he guesses "black".
Person 3 sees 1 + 1 + 2 + 1 + 0 = 5; to get it to equal 3 mod 6, person 3's hat would have to be 4, so he guesses "green".
Person 4 sees 1 + 1 + 2 + 5 + 0 = 9; to get it to equal 4 mod 6, person 4's hat would have to be 1, so he guesses "white".
Person 5 sees 1 + 1 + 2 + 5 + 1 = 10; to get it to equal 5 mod 6, person 5's hat would have to be 1, so he guesses "white".
Person 4 is correct.
Assign each person a number from 0 to 5, and each color a number from 0 to 5.
Each person guesses the color which, if correct, makes the sum of the numbers of all six hats equal to their own number modulo 6.
The sum of the numbers of all six hats, modulo six, has to equal one of the six persons' numbers, so somebody has to be right.
Example: number the colors 0 for black, 1 for white, 2 for red, 3 for yellow, 4 for green, and 5 for blue.
Let persons 0-5 have hats white, white, red, blue, white, and black; the numbers are 1 + 1 + 2 + 5 + 1 + 0 = 10.
Person 0 sees 1 + 2 + 5 + 1 + 0 = 9; to get it to equal 0 mod 6, person 0's hat would have to be 3, so he guesses "yellow".
Person 1 sees 1 + 2 + 5 + 1 + 0 = 9; to get it to equal 1 mod 6, person 1's hat would have to be 4, so he guesses "green".
Person 2 sees 1 + 1 + 5 + 1 + 0 = 8; to get it to equal 2 mod 6, person 2's hat would have to be 0, so he guesses "black".
Person 3 sees 1 + 1 + 2 + 1 + 0 = 5; to get it to equal 3 mod 6, person 3's hat would have to be 4, so he guesses "green".
Person 4 sees 1 + 1 + 2 + 5 + 0 = 9; to get it to equal 4 mod 6, person 4's hat would have to be 1, so he guesses "white".
Person 5 sees 1 + 1 + 2 + 5 + 1 = 10; to get it to equal 5 mod 6, person 5's hat would have to be 1, so he guesses "white".
Person 4 is correct.
June 9th, 2012 at 12:12:26 PM
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Thanks; I'm working on it.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
June 9th, 2012 at 6:08:56 PM
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Good. Since I think not many people are working on it, and I left the problem very open, here are some answers that don't give the solution but do narrow down how well the best solution does, in case anyone is intimidated by all the possibilities.
The following sections may or may not be identical, so as not to reveal the previous answer if you don't click it.
Easy hat color selection:
Hard hat color selection:
Another hint that doesn't ruin it, which is independent of the information above:
No.
The following sections may or may not be identical, so as not to reveal the previous answer if you don't click it.
Easy hat color selection:
Yes.
100% (without player error)
Hard hat color selection:
Yes.
100% (without player error)
Another hint that doesn't ruin it, which is independent of the information above:
Start out with two players and two hat colors, and/or three players and three hat colors. Solving that will help.
June 9th, 2012 at 6:16:33 PM
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Here's a puzzle (with apologies to Sheldon Cooper):
In what universe is this a hat anyone should wear in public?
In what universe is this a hat anyone should wear in public?
Donald Trump is a fucking criminal
June 9th, 2012 at 6:35:55 PM
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she wore the hat to troll the wedding,
because the family didn't invite her mother.
i wish she had gone even further over the top.
because the family didn't invite her mother.
i wish she had gone even further over the top.
In a bet, there is a fool and a thief.
- Proverb.
June 9th, 2012 at 7:01:01 PM
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That's not a hat atop Princess Beatrice's head, it's a holy vision of The Flying Spaghetti Monster.
His Noodliness be praised!
His Noodliness be praised!
June 9th, 2012 at 7:02:53 PM
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Quote: WizardThanks; I'm working on it.
In the meantime, here's a slightly modified version of a classic:
There are 100 people, each of which is wearing either a red hat or a blue hat. (Note that they are not necessarily all the same color.) They are in a line, such that each person can see the hats of the people in front of them - Person 1 can see every hat but his own; person 2 can see the hats of persons 3-100, and so on, up to person 100, who cannot see any hats.
In order, starting with person 1, each person will guess what color his own hat is. All guesses can be heard by all 100 people.
The object is to get at least a certain number of hats correct. The group is allowed to specify a "target"; if they reach the target, they share a prize, which depends on the target (the higher the target, the larger the prize), but if they do not, they get nothing. What target should they set?
June 9th, 2012 at 7:05:47 PM
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whatever number
Person 1 says
Person 1 says
In a bet, there is a fool and a thief.
- Proverb.
June 9th, 2012 at 7:12:21 PM
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Quote: UP84That's not a hat atop Princess Beatrice's head, it's a holy vision of The Flying Spaghetti Monster.
That actually makes more sense than the woman wearing it thought it looked good ;)
But then you should see some of the things I have in my closet...
Donald Trump is a fucking criminal
June 9th, 2012 at 7:31:07 PM
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Quote: NareedHere's a puzzle (with apologies to Sheldon Cooper):
In what universe is this a hat anyone should wear in public?
It's not a hat, it's a fascinator.
The whole assemble looked bloody awful though. Really, I'm not a fashionista, and I thought it was horrible. So horrible, if I was dating her, when she said "what do you think?", I'd be saying "love, don't wear that to the wedding, it doesn't work", even if I was going to be sleeping on the couch.
"Then you can admire the real gambler, who has neither eaten, slept, thought nor lived, he has so smarted under the scourge of his martingale, so suffered on the rack of his desire for a coup at trente-et-quarante" - Honore de Balzac, 1829
June 9th, 2012 at 8:21:21 PM
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Quote: WongBowhatever number Person 1 says
Person 1 says "red" or "blue" - not a number.
June 9th, 2012 at 8:35:11 PM
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"The object is to get at least a certain number of hats correct. "
NUMBER
NUMBER
In a bet, there is a fool and a thief.
- Proverb.
June 10th, 2012 at 8:55:05 AM
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Quote: WongBo"The object is to get at least a certain number of hats correct. "
NUMBER
Yes - the object is for a certain number of people to guess correctly the color of their own hats. Each person guesses whether or not their hat is red or blue.
There are different strategies based on what they want the "target number" to be - what are they?
June 10th, 2012 at 1:16:04 PM
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my apology, i must have completely misread/misunderstood the original question.
In a bet, there is a fool and a thief.
- Proverb.
June 10th, 2012 at 1:47:20 PM
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Quote: ThatDonGuy
There are different strategies based on what they want the "target number" to be - what are they?
Huh? Why would they want the target number to be less then 99?
"When two people always agree one of them is unnecessary"
June 10th, 2012 at 3:07:03 PM
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Quote: weaselmanHuh? Why would they want the target number to be less then 99?
They wouldn't - but...
If the prize for having a target of 100 is more than twice what the prize for the target of 99 is, they might decide to risk going for 100
Here's the answer:
They can guarantee 99 correct, and have a 50% chance of getting all 100, with the same strategy.
Person 1: "If I see an even number of red hats, I say 'red', and if I see an odd number of red hats, I say 'blue'"
Person 2: "I can see every hat person 1 sees, except for my own. If I see an even number of red hats and P1 says 'red', then my hat is blue; if I see an even number of red hats and P1 says 'blue', then my hat is red. Similarly, if I see an odd number of red hats, then if P1 says 'red', then my hat is red, and if P1 says 'blue', then my hat is blue."
Person 3: "I will assume P2 is correct, so I know every hat that P1 can see except for my own, and use the same strategy as P2."
Persons 4-100: "I will assume everyone behind me except P1 is correct, so I know every hat that P1 can see except for my own, and use the same strategy as P2."
Persons 2 through 100 are always correct. Since the initial placement of hats is arbitrary, I am assuming that Person 1's guess is equally as arbitary and assigning a probability of 1/2 that it is correct.
Person 1: "If I see an even number of red hats, I say 'red', and if I see an odd number of red hats, I say 'blue'"
Person 2: "I can see every hat person 1 sees, except for my own. If I see an even number of red hats and P1 says 'red', then my hat is blue; if I see an even number of red hats and P1 says 'blue', then my hat is red. Similarly, if I see an odd number of red hats, then if P1 says 'red', then my hat is red, and if P1 says 'blue', then my hat is blue."
Person 3: "I will assume P2 is correct, so I know every hat that P1 can see except for my own, and use the same strategy as P2."
Persons 4-100: "I will assume everyone behind me except P1 is correct, so I know every hat that P1 can see except for my own, and use the same strategy as P2."
Persons 2 through 100 are always correct. Since the initial placement of hats is arbitrary, I am assuming that Person 1's guess is equally as arbitary and assigning a probability of 1/2 that it is correct.
