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In this economic downturn, Santa is having a bit of financial trouble preparing for his yearly trip. There are a couple of outstanding lawsuits against Santa Inc. for broken roofs and busted chimneys; the North Pole Toy-Assembling Elves Union (NPTEU) is demanding a raise; prices for toy raw materials are skyrocketing; and Mrs. Claus just blew the family budget on a bunch of iPads for the reindeers. Knowing this, the Easter Bunny is offering Santa a chance to make money through a game. The game is as follows:

1) The Easter Bunny is the game host. Santa and his 9 reindeers (Dasher, Dancer, Prancer, Vixen, Comet, Cupid, Donner, Blitzen, Rudolph) will be the participants. Santa and his reindeers has to pay a total of 1 gold coin to play the game.

2) The Easter Bunny will blindfold all participants, and then place either a RED or GREEN Santa hat on each participant's head.

3) The Bunny will then randomly arrange all participants in a circle in such a way that each participant can only see the 7 immediate neighbors in the clockwise direction.

4) The Bunny will then present each player with a list of 10 names, one for each participant. Each player is then requested to guess the hat color of ALL participants and write them down on this list. (Note: Each player can already see 7 other participants, so each player effectively only need to guess his own hat and the hats of 2 immediate neighbors in the counterclockwise direction). Each player must write down either RED or GREEN next to each name on the list, and all players must write at the same time.

5) If ALL of the 10 lists are correct, then Santa and company will win 32 gold coins. If 1 or more list is incorrect, then Santa and company wins nothing. A correct list means that all ten names has the correct hat color written next to them.

Please assume that Santa and company do not cheat (e.g. trying to exchange information through words, utterances, signs, facial expressions, delays in writing their answer, etc.). The reindeers don't want to get on the naughty list, and Santa of course would never be naughty.

Santa and the reindeers can discuss a strategy before the game, and the Easter Bunny said that he is willing to play this game as many times as Santa wants. Please help fund Santa's yearly trip by determining a strategy with positive expected winnings. Good children around the world (and the elves in the NPTEU) eagerly await your input.

Edited by bushindo
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I'm sorry but...

it can't be solved. It's like the first prisoner in "Hats on the death row" but ten times (no one knows the others' answers and everyone has to guess three colours). Or maybe the names are the clue? I don't know, but logically there's no possible solution. Or is it about the fact, that reindeer's eyes point left and right?

At least that's what I think.

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freekyme, while it is perfectly OK to share your opinions (gently) in the thread if you think the puzzle is broken/flawed somehow, it is not OK to:

1) spam the thread with unnecessary and off-topic comments.

2) vote down on a puzzle or a post unless either you are absolutely sure that the puzzle is not solvable/wrong or you feel offended by the puzzle.

Voting a puzzle should follow the

Thank you.

In case you were wondering why those red votes mark your posts, it is because at least two members found them off-topic and/or unnecessary.

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Is the number of red and green hats from game to game constant? (5 red and 5 green always, or 2 and 7 or whatever the numbers are)

I believe, in the spirit of other puzzles from bushindo, that there is no restriction on the number of red/green hats - the Easter bunny draws 10 hats randomly - independent coin flip for each hat.

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I forgot to say that at the beginning of step 4, the Easter Bunny then removes everybody's blindfold. That shouldn't detract from the puzzle, though, as that fact is implied in later details.

The Easter Bunny does draw random hat color for everybody based on independent coin flips. Please assume that the reindeers are sentient and can write. No cheating using ipads, reindeer eyes, Rudolph's big red nose, etc.

I'm sorry but...

it can't be solved. It's like the first prisoner in "Hats on the death row" but ten times (no one knows the others' answers and everyone has to guess three colours). Or maybe the names are the clue? I don't know, but logically there's no possible solution. Or is it about the fact, that reindeer's eyes point left and right?

At least that's what I think.

It is possible to find a strategy with positive expected winning. The strategy does not require much charting or computation. In fact, you just may go 'ah-ha' once you find the solution.

Edited by bushindo
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If each person uses the strategy in the following table, the probability of winning

is 14/128 which gives an expected win of 167/64 gold coins/game. In the following

table, let 0 be Red and 1 be Green. Under the 7 is the color of the hat 7 away

from you on your left, 6 is the color of the hat 6 away from you on your left, etc.

Under the '|' is what color you should predict for yourself. The 1 and 2 on the

right is what you should predict for the hats one and two away from you to your right.

There are 14 winning 10-long hat assignments which win using this strategy. They are

0000000000 0011100111 0101010101 0101101011 0110101101 0111001110 1001110011

1010101010 1010110101 1011010110 1100111001 1101011010 1110011100 1111111111


left |right
7654321 012
0000000 000
0000001 100
0000010 101
0000011 000
0000100 111
0000101 110
0000110 011
0000111 011
0001000 000
0001001 111
0001010 000
0001011 011
0001100 000
0001101 001
0001110 011
0001111 010
0010000 100
0010001 001
0010010 111
0010011 010
0010100 010
0010101 110
0010110 000
0010111 101
0011000 101
0011001 101
0011010 101
0011011 010
0011100 111
0011101 010
0011110 011
0011111 101
0100000 110
0100001 001
0100010 111
0100011 001
0100100 011
0100101 001
0100110 111
0100111 010
0101000 010
0101001 101
0101010 101
0101011 000
0101100 111
0101101 011
0101110 111
0101111 111
0110000 010
0110001 000
0110010 100
0110011 001
0110100 110
0110101 101
0110110 100
0110111 101
0111000 000
0111001 110
0111010 011
0111011 011
0111100 101
0111101 001
0111110 011
0111111 000
1000000 100
1000001 110
1000010 010
1000011 000
1000100 011
1000101 111
1000110 001
1000111 111
1001000 101
1001001 110
1001010 001
1001011 010
1001100 001
1001101 110
1001110 011
1001111 001
1010000 100
1010001 010
1010010 101
1010011 011
1010100 100
1010101 010
1010110 101
1010111 111
1011000 011
1011001 111
1011010 110
1011011 101
1011100 110
1011101 000
1011110 001
1011111 001
1100000 101
1100001 001
1100010 001
1100011 110
1100100 110
1100101 000
1100110 000
1100111 001
1101000 100
1101001 010
1101010 011
1101011 010
1101100 110
1101101 000
1101110 110
1101111 110
1110000 011
1110001 101
1110010 100
1110011 100
1110100 010
1110101 001
1110110 101
1110111 110
1111000 100
1111001 001
1111010 111
1111011 111
1111100 000
1111101 000
1111110 101
1111111 111

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Each participant writes down (at the bottom of the page) the name & colour of the hat worn by the participant in front of them. They then fold the page up to cover what they've written and pass the sheet to the participant in front of them. Repeat 10 times and you have 10 complete and correct lists without any disclosure of the others knowledge.

In this economic downturn, Santa is having a bit of financial trouble preparing for his yearly trip. There are a couple of outstanding lawsuits against Santa Inc. for broken roofs and busted chimneys; the North Pole Toy-Assembling Elves Union (NPTEU) is demanding a raise; prices for toy raw materials are skyrocketing; and Mrs. Claus just blew the family budget on a bunch of iPads for the reindeers. Knowing this, the Easter Bunny is offering Santa a chance to make money through a game. The game is as follows:

1) The Easter Bunny is the game host. Santa and his 9 reindeers (Dasher, Dancer, Prancer, Vixen, Comet, Cupid, Donner, Blitzen, Rudolph) will be the participants. Santa and his reindeers has to pay a total of 1 gold coin to play the game.

2) The Easter Bunny will blindfold all participants, and then place either a RED or GREEN Santa hat on each participant's head.

3) The Bunny will then randomly arrange all participants in a circle in such a way that each participant can only see the 7 immediate neighbors in the clockwise direction.

4) The Bunny will then present each player with a list of 10 names, one for each participant. Each player is then requested to guess the hat color of ALL participants and write them down on this list. (Note: Each player can already see 7 other participants, so each player effectively only need to guess his own hat and the hats of 2 immediate neighbors in the counterclockwise direction). Each player must write down either RED or GREEN next to each name on the list, and all players must write at the same time.

5) If ALL of the 10 lists are correct, then Santa and company will win 32 gold coins. If 1 or more list is incorrect, then Santa and company wins nothing. A correct list means that all ten names has the correct hat color written next to them.

Please assume that Santa and company do not cheat (e.g. trying to exchange information through words, utterances, signs, facial expressions, delays in writing their answer, etc.). The reindeers don't want to get on the naughty list, and Santa of course would never be naughty.

Santa and the reindeers can discuss a strategy before the game, and the Easter Bunny said that he is willing to play this game as many times as Santa wants. Please help fund Santa's yearly trip by determining a strategy with positive expected winnings. Good children around the world (and the elves in the NPTEU) eagerly await your input.

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If each person uses the strategy in the following table, the probability of winning

is 14/128 which gives an expected win of 167/64 gold coins/game. In the following

table, let 0 be Red and 1 be Green. Under the 7 is the color of the hat 7 away

from you on your left, 6 is the color of the hat 6 away from you on your left, etc.

Under the '|' is what color you should predict for yourself. The 1 and 2 on the

right is what you should predict for the hats one and two away from you to your right.

There are 14 winning 10-long hat assignments which win using this strategy. They are

0000000000 0011100111 0101010101 0101101011 0110101101 0111001110 1001110011

1010101010 1010110101 1011010110 1100111001 1101011010 1110011100 1111111111


  left  |right

7654321 012

0000000 000

0000001 100

0000010 101

0000011 000

0000100 111

0000101 110

0000110 011

0000111 011

0001000 000

0001001 111

0001010 000

0001011 011

0001100 000

0001101 001

0001110 011

0001111 010

0010000 100

0010001 001

0010010 111

0010011 010

0010100 010

0010101 110

0010110 000

0010111 101

0011000 101

0011001 101

0011010 101

0011011 010

0011100 111

0011101 010

0011110 011

0011111 101

0100000 110

0100001 001

0100010 111

0100011 001

0100100 011

0100101 001

0100110 111

0100111 010

0101000 010

0101001 101

0101010 101

0101011 000

0101100 111

0101101 011

0101110 111

0101111 111

0110000 010

0110001 000

0110010 100

0110011 001

0110100 110

0110101 101

0110110 100

0110111 101

0111000 000

0111001 110

0111010 011

0111011 011

0111100 101

0111101 001

0111110 011

0111111 000

1000000 100

1000001 110

1000010 010

1000011 000

1000100 011

1000101 111

1000110 001

1000111 111

1001000 101

1001001 110

1001010 001

1001011 010

1001100 001

1001101 110

1001110 011

1001111 001

1010000 100

1010001 010

1010010 101

1010011 011

1010100 100

1010101 010

1010110 101

1010111 111

1011000 011

1011001 111

1011010 110

1011011 101

1011100 110

1011101 000

1011110 001

1011111 001

1100000 101

1100001 001

1100010 001

1100011 110

1100100 110

1100101 000

1100110 000

1100111 001

1101000 100

1101001 010

1101010 011

1101011 010

1101100 110

1101101 000

1101110 110

1101111 110

1110000 011

1110001 101

1110010 100

1110011 100

1110100 010

1110101 001

1110110 101

1110111 110

1111000 100

1111001 001

1111010 111

1111011 111

1111100 000

1111101 000

1111110 101

1111111 111

Good work, superprismatic. Looks like the Easter Bunny is going to file for bankruptcy very soon. This puzzle didn't last that long. I'll have to think of a harder one next time.

Edited by bushindo
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It is possible to find a strategy with positive expected winning. The strategy does not require much charting or computation. In fact, you just may go 'ah-ha' once you find the solution.

Do you have an 'Ah-Ah' solution? I'd like to see it, if you don't mind explaining it. I enjoyed the nice 'Ah-Ah' you had for your 12 friends problem.

I just solved this Christmas problem with a boring old discrete hill-climb.

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Do you have an 'Ah-Ah' solution? I'd like to see it, if you don't mind explaining it. I enjoyed the nice 'Ah-Ah' you had for your 12 friends problem.

I just solved this Christmas problem with a boring old discrete hill-climb.

I was hoping to save the ah-ha trick for the next puzzle, but it's Christmas, no point being a Scrooge :lol: . Here it is

Let the hat states be denoted by 0 and 1. Let the participants be indexed by i = 1, .., 10, and let hi represent the hat state of participant i. All the participants should assume that the following three equations are true

( h1 + h4 + h7 + h10 ) mod 2 = 0

( h1 + h2 + h5 + h8 ) mod 2 = 0

( h2 + h3 + h6 + h9 ) mod 2 = 0

Assuming that the hat states are determined by independent coin toss, the chance that the above three equations are true is (1/2)3 = 1/8. During the game, each participant will know 7 variables out of the set (h1, .., h10 ). If they substitute those 7 known variables into the above, they are left with 3 equations and 3 unknown, which is solvable. The winning rate here is 1/8.

Edited by bushindo
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Bushindo, your solution seems to only apply to one list. If you can't see 3 hats there are 8 possible combos. but the problem requires all 10 lists to be accurate. I'm not sure how to calculate the statistic for it but if it's 1/8 to the 10th power, then the winning percentage would be far below the 32:1 payout being offered.

I am also struggling with an understanding of superprismatics solution. I only see all red and all green hats as being consistent winners with his chart. Could you expand on how each player would write down the same solution? Example if the hat layout was 0000011000 your list shows the player seing 0000011 guessing 000 but the player seeing 0000000 would also guess 000. The 2 lists would not match and thus the game is lost.

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Bushindo, your solution seems to only apply to one list. If you can't see 3 hats there are 8 possible combos. but the problem requires all 10 lists to be accurate. I'm not sure how to calculate the statistic for it but if it's 1/8 to the 10th power, then the winning percentage would be far below the 32:1 payout being offered.

I am also struggling with an understanding of superprismatics solution. I only see all red and all green hats as being consistent winners with his chart. Could you expand on how each player would write down the same solution? Example if the hat layout was 0000011000 your list shows the player seing 0000011 guessing 000 but the player seeing 0000000 would also guess 000. The 2 lists would not match and thus the game is lost.

You are correct that the hat layout of 0000011000 is a loser. I specified the

only winning layouts when I said

"There are 14 winning 10-long hat assignments which win using this strategy. They are

0000000000 0011100111 0101010101 0101101011 0110101101 0111001110 1001110011

1010101010 1010110101 1011010110 1100111001 1101011010 1110011100 1111111111".

This is 14 winners out of 128 possibilities. All the others are losers.

I hope this helps you understand my strategy. By the way, I ran my program a bit longer

and I got a much better score of 24 out of 128. But this doesn't help in the

understanding of what I did.

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Bushindo, your solution seems to only apply to one list. If you can't see 3 hats there are 8 possible combos. but the problem requires all 10 lists to be accurate. I'm not sure how to calculate the statistic for it but if it's 1/8 to the 10th power, then the winning percentage would be far below the 32:1 payout being offered.

I struggled with that at first too but then I re-read and thought about it. The 1/8 chance doesn't refer to the chance of a list being correct according to those equations. The 1/8 chance refers to the chance that the equations hold at all. If the equations hold then every individual list has a 100% chance to be correct if solved using algebra. I know this is necessarily if they all use the same numbers to refer to the same contestant, I'm not sure if it's true otherwise (i.e if Santa's "h1" isn't the same as Vixen's "h1"). So if they somehow define which position is h1 (and thus the rest in clockwise order) before they are arranged randomly, then they will all be using the same equations with different unknowns. They will all still be able to solve the equations though and get the same answers.

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freekyme, while it is perfectly OK to share your opinions (gently) in the thread if you think the puzzle is broken/flawed somehow, it is not OK to:

1) spam the thread with unnecessary and off-topic comments.

2) vote down on a puzzle or a post unless either you are absolutely sure that the puzzle is not solvable/wrong or you feel offended by the puzzle.

Voting a puzzle should follow the

Thank you.

In case you were wondering why those red votes mark your posts, it is because at least two members found them off-topic and/or unnecessary.

sorry if u tuk it in that way, but i never voted for or against any comment. Any way, sorry if anyone disliked the comment, i'll try to be more careful next time. :)

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