Guess simultaneously, color of your hat, no passing allowed
$begingroup$
1000 logicians standing in a circle are blindfolded and a hat of either black or white color is placed on each person's head. The blindfolds are then removed and each person can now see everybody's hat color but his own. After exactly one minute, they all have to simultaneously guess their own hat color.
No communication in any form is allowed after the hats have been placed. However, before the hats are placed, the logicians are allowed to gather and devise a strategy for maximising the number of correct guesses.
Questions :
1) What should be their strategy to get the maximum guaranteed
correct guesses ?
2) Had the total number of different hat colors been three instead
of two, then what strategy should they have devised to get the
maximum guaranteed correct guesses ?
P.S: I do not know the answer of this puzzle but I have been
wondering about the answer for very long . It is based on a
similar well known puzzle with infinite hats where the aim is
to make only a finite number of incorrect guesses. The infinite
version and it's solution can be read here :
https://tinyurl.com/infinitehats
mathematics logical-deduction hat-guessing
New contributor
Hemant Agarwal is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
1000 logicians standing in a circle are blindfolded and a hat of either black or white color is placed on each person's head. The blindfolds are then removed and each person can now see everybody's hat color but his own. After exactly one minute, they all have to simultaneously guess their own hat color.
No communication in any form is allowed after the hats have been placed. However, before the hats are placed, the logicians are allowed to gather and devise a strategy for maximising the number of correct guesses.
Questions :
1) What should be their strategy to get the maximum guaranteed
correct guesses ?
2) Had the total number of different hat colors been three instead
of two, then what strategy should they have devised to get the
maximum guaranteed correct guesses ?
P.S: I do not know the answer of this puzzle but I have been
wondering about the answer for very long . It is based on a
similar well known puzzle with infinite hats where the aim is
to make only a finite number of incorrect guesses. The infinite
version and it's solution can be read here :
https://tinyurl.com/infinitehats
mathematics logical-deduction hat-guessing
New contributor
Hemant Agarwal is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Note: They are superhuman logicians and for them, having just 1 minute is more than enough to see all the 999 hat colors around them and then come up with their answer using the most optimal strategy that exists .
$endgroup$
– Hemant Agarwal
8 hours ago
add a comment |
$begingroup$
1000 logicians standing in a circle are blindfolded and a hat of either black or white color is placed on each person's head. The blindfolds are then removed and each person can now see everybody's hat color but his own. After exactly one minute, they all have to simultaneously guess their own hat color.
No communication in any form is allowed after the hats have been placed. However, before the hats are placed, the logicians are allowed to gather and devise a strategy for maximising the number of correct guesses.
Questions :
1) What should be their strategy to get the maximum guaranteed
correct guesses ?
2) Had the total number of different hat colors been three instead
of two, then what strategy should they have devised to get the
maximum guaranteed correct guesses ?
P.S: I do not know the answer of this puzzle but I have been
wondering about the answer for very long . It is based on a
similar well known puzzle with infinite hats where the aim is
to make only a finite number of incorrect guesses. The infinite
version and it's solution can be read here :
https://tinyurl.com/infinitehats
mathematics logical-deduction hat-guessing
New contributor
Hemant Agarwal is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1000 logicians standing in a circle are blindfolded and a hat of either black or white color is placed on each person's head. The blindfolds are then removed and each person can now see everybody's hat color but his own. After exactly one minute, they all have to simultaneously guess their own hat color.
No communication in any form is allowed after the hats have been placed. However, before the hats are placed, the logicians are allowed to gather and devise a strategy for maximising the number of correct guesses.
Questions :
1) What should be their strategy to get the maximum guaranteed
correct guesses ?
2) Had the total number of different hat colors been three instead
of two, then what strategy should they have devised to get the
maximum guaranteed correct guesses ?
P.S: I do not know the answer of this puzzle but I have been
wondering about the answer for very long . It is based on a
similar well known puzzle with infinite hats where the aim is
to make only a finite number of incorrect guesses. The infinite
version and it's solution can be read here :
https://tinyurl.com/infinitehats
mathematics logical-deduction hat-guessing
mathematics logical-deduction hat-guessing
New contributor
Hemant Agarwal is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Hemant Agarwal is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 1 min ago
Hemant Agarwal
New contributor
Hemant Agarwal is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 12 hours ago
Hemant AgarwalHemant Agarwal
192
192
New contributor
Hemant Agarwal is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Hemant Agarwal is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Hemant Agarwal is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
Note: They are superhuman logicians and for them, having just 1 minute is more than enough to see all the 999 hat colors around them and then come up with their answer using the most optimal strategy that exists .
$endgroup$
– Hemant Agarwal
8 hours ago
add a comment |
$begingroup$
Note: They are superhuman logicians and for them, having just 1 minute is more than enough to see all the 999 hat colors around them and then come up with their answer using the most optimal strategy that exists .
$endgroup$
– Hemant Agarwal
8 hours ago
$begingroup$
Note: They are superhuman logicians and for them, having just 1 minute is more than enough to see all the 999 hat colors around them and then come up with their answer using the most optimal strategy that exists .
$endgroup$
– Hemant Agarwal
8 hours ago
$begingroup$
Note: They are superhuman logicians and for them, having just 1 minute is more than enough to see all the 999 hat colors around them and then come up with their answer using the most optimal strategy that exists .
$endgroup$
– Hemant Agarwal
8 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
First, I'd note three things:
- The usual solution to the hat puzzle cannot be used, since the
guesses must be made simultaneously, meaning nobody could pass on
any information to the other players their guess. - With only a minute to count 1000 hats, you wouldn't have enough time
to work out a definite answer by elimination, even assuming you knew
the totals in advance. - You would have enough time to count a fairly large sample which, if selected
sufficiently at random, ought to give you a fairly accurate estimate of the overall
populations. Hopefully, being logicians, they'd know how to do this.
Bearing that in mind, my strategy would be:
They could all give the answer as the hat they can see the most of. That way, in aggregate, they'll at least beat random guesswork. For instance if there are 600 black hats and 400 white ones, they'll be correct 60% of the time rather than just 50%, with the accuracy rising to nearly 100% if there's a high proportion of one over the other.
The worst case is when it's 500 of each, where everyone will see slightly more of the hat that isn't their own. However, its unlikely that everyone could count to such accuracy so the results should still be about as good as guesswork in that limit.
The strategy remains the same if there are more kinds of hat added; just keep going with the one that has the highest count.
$endgroup$
$begingroup$
I have added a comment to my question after reading your answer. They are all superhumans and 1 minute of time does not hinder them in anyway to use the most optimal strategy that exists.
$endgroup$
– Hemant Agarwal
8 hours ago
add a comment |
$begingroup$
Assuming:
The colour of each hat is picked randomly from the available colours with equal chance of it being either colour (e.g. 50% for 2, and 33.3% for 3 different colours).
Then:
Once they've counted the total of hats of each colour on all other heads, then there is a higher chance of them having the colour that has the least amount.
E.g if there are 499 white and 500 black, then they are more likely to have a white hat.
In the case of the 3 colours this breaks down if the total is exactly 333 in each direction, as one colour will need 334 to round it up, but if the outcome is not actually evenly spread (e.g 331, 333, 335) then the strategy still holds for being one of the lowest counted colours.
However:
I believe this falls into the common misconception that if you have flipped 10 coins in a row and they've all been heads, then the next one has a higher chance of being tails. Which is obviously wrong as it will continue to be a 1 in 2 chance.
In this case, the chance of your hat being a specific colour is still that 1 in 2 or 1 in 3, despite what everything else is.
Therefore:
I believe there is no such strategy that can maximise this, given my assumption.
$endgroup$
$begingroup$
Note that the question explicitly asks for a strategy that gives the maximum "guaranteed" correct guesses. According to your answer , the maximum guaranteed correct guesses is 0 for both questions 1 and 2. However, 0 is not the correct answer .
$endgroup$
– Hemant Agarwal
2 hours ago
$begingroup$
@HemantAgarwal then without knowing the number of hats of each colour, there is no guarantee that anyone can get it right.
$endgroup$
– AHKieran
1 hour ago
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
First, I'd note three things:
- The usual solution to the hat puzzle cannot be used, since the
guesses must be made simultaneously, meaning nobody could pass on
any information to the other players their guess. - With only a minute to count 1000 hats, you wouldn't have enough time
to work out a definite answer by elimination, even assuming you knew
the totals in advance. - You would have enough time to count a fairly large sample which, if selected
sufficiently at random, ought to give you a fairly accurate estimate of the overall
populations. Hopefully, being logicians, they'd know how to do this.
Bearing that in mind, my strategy would be:
They could all give the answer as the hat they can see the most of. That way, in aggregate, they'll at least beat random guesswork. For instance if there are 600 black hats and 400 white ones, they'll be correct 60% of the time rather than just 50%, with the accuracy rising to nearly 100% if there's a high proportion of one over the other.
The worst case is when it's 500 of each, where everyone will see slightly more of the hat that isn't their own. However, its unlikely that everyone could count to such accuracy so the results should still be about as good as guesswork in that limit.
The strategy remains the same if there are more kinds of hat added; just keep going with the one that has the highest count.
$endgroup$
$begingroup$
I have added a comment to my question after reading your answer. They are all superhumans and 1 minute of time does not hinder them in anyway to use the most optimal strategy that exists.
$endgroup$
– Hemant Agarwal
8 hours ago
add a comment |
$begingroup$
First, I'd note three things:
- The usual solution to the hat puzzle cannot be used, since the
guesses must be made simultaneously, meaning nobody could pass on
any information to the other players their guess. - With only a minute to count 1000 hats, you wouldn't have enough time
to work out a definite answer by elimination, even assuming you knew
the totals in advance. - You would have enough time to count a fairly large sample which, if selected
sufficiently at random, ought to give you a fairly accurate estimate of the overall
populations. Hopefully, being logicians, they'd know how to do this.
Bearing that in mind, my strategy would be:
They could all give the answer as the hat they can see the most of. That way, in aggregate, they'll at least beat random guesswork. For instance if there are 600 black hats and 400 white ones, they'll be correct 60% of the time rather than just 50%, with the accuracy rising to nearly 100% if there's a high proportion of one over the other.
The worst case is when it's 500 of each, where everyone will see slightly more of the hat that isn't their own. However, its unlikely that everyone could count to such accuracy so the results should still be about as good as guesswork in that limit.
The strategy remains the same if there are more kinds of hat added; just keep going with the one that has the highest count.
$endgroup$
$begingroup$
I have added a comment to my question after reading your answer. They are all superhumans and 1 minute of time does not hinder them in anyway to use the most optimal strategy that exists.
$endgroup$
– Hemant Agarwal
8 hours ago
add a comment |
$begingroup$
First, I'd note three things:
- The usual solution to the hat puzzle cannot be used, since the
guesses must be made simultaneously, meaning nobody could pass on
any information to the other players their guess. - With only a minute to count 1000 hats, you wouldn't have enough time
to work out a definite answer by elimination, even assuming you knew
the totals in advance. - You would have enough time to count a fairly large sample which, if selected
sufficiently at random, ought to give you a fairly accurate estimate of the overall
populations. Hopefully, being logicians, they'd know how to do this.
Bearing that in mind, my strategy would be:
They could all give the answer as the hat they can see the most of. That way, in aggregate, they'll at least beat random guesswork. For instance if there are 600 black hats and 400 white ones, they'll be correct 60% of the time rather than just 50%, with the accuracy rising to nearly 100% if there's a high proportion of one over the other.
The worst case is when it's 500 of each, where everyone will see slightly more of the hat that isn't their own. However, its unlikely that everyone could count to such accuracy so the results should still be about as good as guesswork in that limit.
The strategy remains the same if there are more kinds of hat added; just keep going with the one that has the highest count.
$endgroup$
First, I'd note three things:
- The usual solution to the hat puzzle cannot be used, since the
guesses must be made simultaneously, meaning nobody could pass on
any information to the other players their guess. - With only a minute to count 1000 hats, you wouldn't have enough time
to work out a definite answer by elimination, even assuming you knew
the totals in advance. - You would have enough time to count a fairly large sample which, if selected
sufficiently at random, ought to give you a fairly accurate estimate of the overall
populations. Hopefully, being logicians, they'd know how to do this.
Bearing that in mind, my strategy would be:
They could all give the answer as the hat they can see the most of. That way, in aggregate, they'll at least beat random guesswork. For instance if there are 600 black hats and 400 white ones, they'll be correct 60% of the time rather than just 50%, with the accuracy rising to nearly 100% if there's a high proportion of one over the other.
The worst case is when it's 500 of each, where everyone will see slightly more of the hat that isn't their own. However, its unlikely that everyone could count to such accuracy so the results should still be about as good as guesswork in that limit.
The strategy remains the same if there are more kinds of hat added; just keep going with the one that has the highest count.
answered 8 hours ago
Matthew BarberMatthew Barber
2662
2662
$begingroup$
I have added a comment to my question after reading your answer. They are all superhumans and 1 minute of time does not hinder them in anyway to use the most optimal strategy that exists.
$endgroup$
– Hemant Agarwal
8 hours ago
add a comment |
$begingroup$
I have added a comment to my question after reading your answer. They are all superhumans and 1 minute of time does not hinder them in anyway to use the most optimal strategy that exists.
$endgroup$
– Hemant Agarwal
8 hours ago
$begingroup$
I have added a comment to my question after reading your answer. They are all superhumans and 1 minute of time does not hinder them in anyway to use the most optimal strategy that exists.
$endgroup$
– Hemant Agarwal
8 hours ago
$begingroup$
I have added a comment to my question after reading your answer. They are all superhumans and 1 minute of time does not hinder them in anyway to use the most optimal strategy that exists.
$endgroup$
– Hemant Agarwal
8 hours ago
add a comment |
$begingroup$
Assuming:
The colour of each hat is picked randomly from the available colours with equal chance of it being either colour (e.g. 50% for 2, and 33.3% for 3 different colours).
Then:
Once they've counted the total of hats of each colour on all other heads, then there is a higher chance of them having the colour that has the least amount.
E.g if there are 499 white and 500 black, then they are more likely to have a white hat.
In the case of the 3 colours this breaks down if the total is exactly 333 in each direction, as one colour will need 334 to round it up, but if the outcome is not actually evenly spread (e.g 331, 333, 335) then the strategy still holds for being one of the lowest counted colours.
However:
I believe this falls into the common misconception that if you have flipped 10 coins in a row and they've all been heads, then the next one has a higher chance of being tails. Which is obviously wrong as it will continue to be a 1 in 2 chance.
In this case, the chance of your hat being a specific colour is still that 1 in 2 or 1 in 3, despite what everything else is.
Therefore:
I believe there is no such strategy that can maximise this, given my assumption.
$endgroup$
$begingroup$
Note that the question explicitly asks for a strategy that gives the maximum "guaranteed" correct guesses. According to your answer , the maximum guaranteed correct guesses is 0 for both questions 1 and 2. However, 0 is not the correct answer .
$endgroup$
– Hemant Agarwal
2 hours ago
$begingroup$
@HemantAgarwal then without knowing the number of hats of each colour, there is no guarantee that anyone can get it right.
$endgroup$
– AHKieran
1 hour ago
add a comment |
$begingroup$
Assuming:
The colour of each hat is picked randomly from the available colours with equal chance of it being either colour (e.g. 50% for 2, and 33.3% for 3 different colours).
Then:
Once they've counted the total of hats of each colour on all other heads, then there is a higher chance of them having the colour that has the least amount.
E.g if there are 499 white and 500 black, then they are more likely to have a white hat.
In the case of the 3 colours this breaks down if the total is exactly 333 in each direction, as one colour will need 334 to round it up, but if the outcome is not actually evenly spread (e.g 331, 333, 335) then the strategy still holds for being one of the lowest counted colours.
However:
I believe this falls into the common misconception that if you have flipped 10 coins in a row and they've all been heads, then the next one has a higher chance of being tails. Which is obviously wrong as it will continue to be a 1 in 2 chance.
In this case, the chance of your hat being a specific colour is still that 1 in 2 or 1 in 3, despite what everything else is.
Therefore:
I believe there is no such strategy that can maximise this, given my assumption.
$endgroup$
$begingroup$
Note that the question explicitly asks for a strategy that gives the maximum "guaranteed" correct guesses. According to your answer , the maximum guaranteed correct guesses is 0 for both questions 1 and 2. However, 0 is not the correct answer .
$endgroup$
– Hemant Agarwal
2 hours ago
$begingroup$
@HemantAgarwal then without knowing the number of hats of each colour, there is no guarantee that anyone can get it right.
$endgroup$
– AHKieran
1 hour ago
add a comment |
$begingroup$
Assuming:
The colour of each hat is picked randomly from the available colours with equal chance of it being either colour (e.g. 50% for 2, and 33.3% for 3 different colours).
Then:
Once they've counted the total of hats of each colour on all other heads, then there is a higher chance of them having the colour that has the least amount.
E.g if there are 499 white and 500 black, then they are more likely to have a white hat.
In the case of the 3 colours this breaks down if the total is exactly 333 in each direction, as one colour will need 334 to round it up, but if the outcome is not actually evenly spread (e.g 331, 333, 335) then the strategy still holds for being one of the lowest counted colours.
However:
I believe this falls into the common misconception that if you have flipped 10 coins in a row and they've all been heads, then the next one has a higher chance of being tails. Which is obviously wrong as it will continue to be a 1 in 2 chance.
In this case, the chance of your hat being a specific colour is still that 1 in 2 or 1 in 3, despite what everything else is.
Therefore:
I believe there is no such strategy that can maximise this, given my assumption.
$endgroup$
Assuming:
The colour of each hat is picked randomly from the available colours with equal chance of it being either colour (e.g. 50% for 2, and 33.3% for 3 different colours).
Then:
Once they've counted the total of hats of each colour on all other heads, then there is a higher chance of them having the colour that has the least amount.
E.g if there are 499 white and 500 black, then they are more likely to have a white hat.
In the case of the 3 colours this breaks down if the total is exactly 333 in each direction, as one colour will need 334 to round it up, but if the outcome is not actually evenly spread (e.g 331, 333, 335) then the strategy still holds for being one of the lowest counted colours.
However:
I believe this falls into the common misconception that if you have flipped 10 coins in a row and they've all been heads, then the next one has a higher chance of being tails. Which is obviously wrong as it will continue to be a 1 in 2 chance.
In this case, the chance of your hat being a specific colour is still that 1 in 2 or 1 in 3, despite what everything else is.
Therefore:
I believe there is no such strategy that can maximise this, given my assumption.
answered 3 hours ago
AHKieranAHKieran
5,0811040
5,0811040
$begingroup$
Note that the question explicitly asks for a strategy that gives the maximum "guaranteed" correct guesses. According to your answer , the maximum guaranteed correct guesses is 0 for both questions 1 and 2. However, 0 is not the correct answer .
$endgroup$
– Hemant Agarwal
2 hours ago
$begingroup$
@HemantAgarwal then without knowing the number of hats of each colour, there is no guarantee that anyone can get it right.
$endgroup$
– AHKieran
1 hour ago
add a comment |
$begingroup$
Note that the question explicitly asks for a strategy that gives the maximum "guaranteed" correct guesses. According to your answer , the maximum guaranteed correct guesses is 0 for both questions 1 and 2. However, 0 is not the correct answer .
$endgroup$
– Hemant Agarwal
2 hours ago
$begingroup$
@HemantAgarwal then without knowing the number of hats of each colour, there is no guarantee that anyone can get it right.
$endgroup$
– AHKieran
1 hour ago
$begingroup$
Note that the question explicitly asks for a strategy that gives the maximum "guaranteed" correct guesses. According to your answer , the maximum guaranteed correct guesses is 0 for both questions 1 and 2. However, 0 is not the correct answer .
$endgroup$
– Hemant Agarwal
2 hours ago
$begingroup$
Note that the question explicitly asks for a strategy that gives the maximum "guaranteed" correct guesses. According to your answer , the maximum guaranteed correct guesses is 0 for both questions 1 and 2. However, 0 is not the correct answer .
$endgroup$
– Hemant Agarwal
2 hours ago
$begingroup$
@HemantAgarwal then without knowing the number of hats of each colour, there is no guarantee that anyone can get it right.
$endgroup$
– AHKieran
1 hour ago
$begingroup$
@HemantAgarwal then without knowing the number of hats of each colour, there is no guarantee that anyone can get it right.
$endgroup$
– AHKieran
1 hour ago
add a comment |
Hemant Agarwal is a new contributor. Be nice, and check out our Code of Conduct.
Hemant Agarwal is a new contributor. Be nice, and check out our Code of Conduct.
Hemant Agarwal is a new contributor. Be nice, and check out our Code of Conduct.
Hemant Agarwal is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
Note: They are superhuman logicians and for them, having just 1 minute is more than enough to see all the 999 hat colors around them and then come up with their answer using the most optimal strategy that exists .
$endgroup$
– Hemant Agarwal
8 hours ago