Easy hat puzzle
December 5th, 2015 at 3:49:57 PM
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Okay, I think I am finally get this. Thank you Charlie for all your patience with me. I think I will make a future "Ask the Wizard" question out of this one.
Time for a new "hat" puzzle.
Time for a new "hat" puzzle.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
December 5th, 2015 at 4:20:43 PM
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Quote: WizardTime for a new "hat" puzzle.
Okay; here's an "easy" one.
Ten people, and each one is given a hat that can be one of 10 colors (black, brown, red, orange, yellow, green, blue, purple, gray, white); note that not every color needs to be on a hat, and one or more colors can be on one or more hats. Again, each person cannot see his own hat, but can see all nine of the other hats, and knows whose hat is what color. Each person then tries to guess the color of their own hat; this is done simultaneously - no information is available to anyone except the hat colors of the other nine people. The group needs only one guess to be correct to "win". What is the best strategy?
December 5th, 2015 at 4:58:18 PM
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So far this one doesn't seem so easy to me.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
December 5th, 2015 at 5:46:40 PM
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Let the colours be numbered 0 thru 9 (rather than their colours), and let the people stand in a row being numbered 0 thru 9.
Consider the actual total of the hats ABCDEFHGHIJ be a total of T. Then consider that total mod 10.
Instruct every player to guess a hat colour number, that given they can see the others, would total (Mod10) to their position number.
Suppose the total was 2, then this means that the 2nd person in the row will guess correctly, since they'll see the other nine hats, try and create a hat that gives a total of 2 (Mod 10) and guess correctly.
You can see that 0th player is 2 out, 1st player is 1 out, 2nd player is correct etc.
Instruct every player to guess a hat colour number, that given they can see the others, would total (Mod10) to their position number.
Suppose the total was 2, then this means that the 2nd person in the row will guess correctly, since they'll see the other nine hats, try and create a hat that gives a total of 2 (Mod 10) and guess correctly.
Hats : 2 7 9 1 4 3 8 2 0 6 - total is 42 = 2 Mod 10
Seen : 40 35 33 41 38 39 34 40 42 36
Target 40 41 42 43 44 45 36 47 48 39
Guess: 0 6 9 2 6 6 2 7 6 3
You can see that 0th player is 2 out, 1st player is 1 out, 2nd player is correct etc.
December 5th, 2015 at 5:48:03 PM
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Pick one person to be the guesser. All the other players will turn the brims of their hats to point to another person with the same hat color. If no one else has the same hat color as the guesser, they will turn the brim of their hat toward the back, and the guesser will have to select from a color that he does not see.
Simplicity is the ultimate sophistication - Leonardo da Vinci
December 5th, 2015 at 5:50:08 PM
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So if a person had a blue hat with a white feather and a green band, would "blue", "white", and "green" all be considered correct answers for that person? With a white hat with red pinstripes in the fabric, would "red" and "white" be both valid? And a person would have to state just one color, not all colors present?
I think since there's multiple colors on multiple hats, I would have each person look around and count how many of each color they can see, and either have them guess a color they can't see if there is one, or if all colors are in sight, have them guess the color they see on the most hats.
I think since there's multiple colors on multiple hats, I would have each person look around and count how many of each color they can see, and either have them guess a color they can't see if there is one, or if all colors are in sight, have them guess the color they see on the most hats.
If the House lost every hand, they wouldn't deal the game.
December 5th, 2015 at 8:41:27 PM
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Quote: beachbumbabsSo if a person had a blue hat with a white feather and a green band, would "blue", "white", and "green" all be considered correct answers for that person? With a white hat with red pinstripes in the fabric, would "red" and "white" be both valid? And a person would have to state just one color, not all colors present?
You misunderstand the question - each hat is one color. Yes, I did say, "one or more colors can be on one or more hats," but that was just to clarify that not all ten colors have to be present, and some could appear more than once (in fact, all ten hats could be the same color); otherwise, the problem is a little too easy.
Charlie's answer is correct
December 6th, 2015 at 12:45:47 AM
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Quote: charliepatrickLet the colours be numbered 0 thru 9 (rather than their colours), and let the people stand in a row being numbered 0 thru 9.
Consider the actual total of the hats ABCDEFHGHIJ be a total of T. Then consider that total mod 10.
Instruct every player to guess a hat colour number, that given they can see the others, would total (Mod10) to their position number.
Suppose the total was 2, then this means that the 2nd person in the row will guess correctly, since they'll see the other nine hats, try and create a hat that gives a total of 2 (Mod 10) and guess correctly.
Hats : 2 7 9 1 4 3 8 2 0 6 - total is 42 = 2 Mod 10
Seen : 40 35 33 41 38 39 34 40 42 36
Target 40 41 42 43 44 45 36 47 48 39
Guess: 0 6 9 2 6 6 2 7 6 3
You can see that 0th player is 2 out, 1st player is 1 out, 2nd player is correct etc.
First of all....how the f......wait just....WHAT!? I don't disagree with you, I'm just stunned you or anyone could come up with the answer. Although I am a little confused as to how it'd work. Let me give it a try -- see if I can re-explain your answer and see if I get it right?
Each person wearing a different numbered shirt, 0-9 (this represents their position).
Each person wears a # on their head that they cannot see (ie: color of hat).
So basically everyone adds up all the other #'s on everyone else's head.
The 0'th person comes up with a number that'll make the total #'s divisible by 10 (number % 10).
The 1'th person comes up with a number that'll make the total #'s divisible by 10 + 1.
Pos: 0 1 2 3 4 5 6 7 8 9
Num: 1 2 3 4 5 6 7 8 9 0 [ex]
Seen 44 43 42 41 40 39 38 37 36 35
Guess 6 8 0 2 4 6 8 9 2 4
0: 44+6 = 0
1: 43+8 = 1
2: 42+0 = 2
3: 41+2 = 3
4: 40+4 = 4
oh a winner!
edit: currently being edited i messed up
Here's a SUPER SIMPLE hat-question:
No prove Charlie's answer works!
EDIT: Wow, I can't believe I was smart enough to figure out how to edit this. PS: wasn't easy. PSS: If you double-spoiler you'll break the forum, it was kinda cool.
double super mega edit: it appears doing a CODE within SPOILER breaks the forum if you don't /CODE.
December 6th, 2015 at 12:52:25 AM
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hmm, it appears I've broken the forum and can't edit my own post. unable to click the spoiler because of the ads. :(
Edit: I have rescued the forumz. All is fair and well now. Nothing to worry about.
Edit: I have rescued the forumz. All is fair and well now. Nothing to worry about.
December 6th, 2015 at 2:19:58 AM
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Quote: AyecarumbaPick one person to be the guesser. All the other players will turn the brims of their hats to point to another person with the same hat color. If no one else has the same hat color as the guesser, they will turn the brim of their hat toward the back, and the guesser will have to select from a color that he does not see.
You can join Babs in the sin bin for trying to cheat. There is no communication allowed to be conveyed other than what is allowed in the declaration of hat color. Please think inside the box.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
December 6th, 2015 at 2:21:47 AM
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What if you simplify the question to two people and two hat colors? There is a strategy to guarantee a victory.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
December 6th, 2015 at 4:10:07 AM
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I choose the opposite of your hat and you choose the same as my hat.
Our hats are either 00, 01, 10, or 11.
Answers would be 10, 11, 00, 01, respectively.
In each case at least one person is correct.
Our hats are either 00, 01, 10, or 11.
Answers would be 10, 11, 00, 01, respectively.
In each case at least one person is correct.
December 6th, 2015 at 7:46:20 AM
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Quote: RSI choose the opposite of your hat and you choose the same as my hat.
Our hats are either 00, 01, 10, or 11.
Answers would be 10, 11, 00, 01, respectively.
In each case at least one person is correct.
Correct!
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
December 6th, 2015 at 9:00:09 AM
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Here is the proof that my earlier answer works.
In the problem there are ten people in a row (in my case numbered 0 to 9 for simplicity) and they all wear various digits which can be 0 thru 9. Each one can only see the other people's hat.
Suppose there was another person in the room who could see all ten hats and so told everyone the total, then everyone would be able to work out their own and get it correct. Similarly that other person could give them the total (Mod 10), they would also all get it correct. Thus if you are given the correct total (Mod 10) you can work your hat out correctly.
Now consider a situation where that other person in the room (regardless of the actual total) always informs person 0 the total is 0 (Mod 10), person 1 the total is 1 (Mod 10) ...etc. then the only person being told the truth would be the person whose figure matches with the truth. In fact this would be the Nth person since the total was N (Mod 10), and they would calculate their hat number correctly.
Now flip it around, and you can see that if the ten adopt the principle of assuming the total is <their position> (Mod 10), then one of them must be correct (the Nth person) and hence guess their hat number correctly.
Another way of thinking about it is there are ten possible states for the total. As long as each player selects a different one and all ten states are covered, then one of them must be correct. Thus if James picks 0, Albert picks 1, Brian picks 2 etc. they must win.
Suppose there was another person in the room who could see all ten hats and so told everyone the total, then everyone would be able to work out their own and get it correct. Similarly that other person could give them the total (Mod 10), they would also all get it correct. Thus if you are given the correct total (Mod 10) you can work your hat out correctly.
Now consider a situation where that other person in the room (regardless of the actual total) always informs person 0 the total is 0 (Mod 10), person 1 the total is 1 (Mod 10) ...etc. then the only person being told the truth would be the person whose figure matches with the truth. In fact this would be the Nth person since the total was N (Mod 10), and they would calculate their hat number correctly.
Now flip it around, and you can see that if the ten adopt the principle of assuming the total is <their position> (Mod 10), then one of them must be correct (the Nth person) and hence guess their hat number correctly.
Another way of thinking about it is there are ten possible states for the total. As long as each player selects a different one and all ten states are covered, then one of them must be correct. Thus if James picks 0, Albert picks 1, Brian picks 2 etc. they must win.
December 6th, 2015 at 9:26:09 AM
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RE: proof for 10 hats 10 colours
Very clever! Also seems oddly reminiscent of the solution for the Seven Hat puzzle.
Very clever! Also seems oddly reminiscent of the solution for the Seven Hat puzzle.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
