Can you open the door or die?A game of Green JackSearching for a secret doorColored Pills MADNESSFind the correct door!The Monty Hall Arena12 Birds in the petshopCan you open the bathroom door?Hacking an electronic keypad v233 stones into gold transmuterWhat is the optimal strategy for this triangular board game?
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Can you open the door or die?
A game of Green JackSearching for a secret doorColored Pills MADNESSFind the correct door!The Monty Hall Arena12 Birds in the petshopCan you open the bathroom door?Hacking an electronic keypad v233 stones into gold transmuterWhat is the optimal strategy for this triangular board game?
$begingroup$
You wake up and you found yourself in a very strange room where there is a single door in it. and you found a note on the ground with lots of cards around it:
"There are 66 green and 66 blue door cards on the ground, if you put one blue and one green cards into the slots next to the door, the door will open. If you put the same color, you will hear a fail sound and lose a life! I won't tell you how many lifes you have, but you will die if you fail more than that!"
it would only require one try normally but you remember that you cannot distinguish the color of the stones, especially when they are green and blue because you are colorblind!, and probably whoever put you there knows this very well and that guy also knows what would be the minimum number of tries you need to open the door guaranteed as well so
Can you open it in time?
if so
How many tries do you need to open the door guaranteed?
logical-deduction strategy optimization
asked 8 hours ago
OrayOray
16.5k438165
$endgroup$
add a comment |
$begingroup$
You wake up and you found yourself in a very strange room where there is a single door in it. and you found a note on the ground with lots of cards around it:
"There are 66 green and 66 blue door cards on the ground, if you put one blue and one green cards into the slots next to the door, the door will open. If you put the same color, you will hear a fail sound and lose a life! I won't tell you how many lifes you have, but you will die if you fail more than that!"
it would only require one try normally but you remember that you cannot distinguish the color of the stones, especially when they are green and blue because you are colorblind!, and probably whoever put you there knows this very well and that guy also knows what would be the minimum number of tries you need to open the door guaranteed as well so
Can you open it in time?
if so
How many tries do you need to open the door guaranteed?
logical-deduction strategy optimization
asked 8 hours ago
OrayOray
16.5k438165
$endgroup$
$begingroup$
Can I get the cards back after an attempt?
$endgroup$
– cinico
6 hours ago
1
$begingroup$
@cinico of course
$endgroup$
– Oray
6 hours ago
add a comment |
$begingroup$
You wake up and you found yourself in a very strange room where there is a single door in it. and you found a note on the ground with lots of cards around it:
"There are 66 green and 66 blue door cards on the ground, if you put one blue and one green cards into the slots next to the door, the door will open. If you put the same color, you will hear a fail sound and lose a life! I won't tell you how many lifes you have, but you will die if you fail more than that!"
it would only require one try normally but you remember that you cannot distinguish the color of the stones, especially when they are green and blue because you are colorblind!, and probably whoever put you there knows this very well and that guy also knows what would be the minimum number of tries you need to open the door guaranteed as well so
Can you open it in time?
if so
How many tries do you need to open the door guaranteed?
logical-deduction strategy optimization
asked 8 hours ago
OrayOray
16.5k438165
$endgroup$
You wake up and you found yourself in a very strange room where there is a single door in it. and you found a note on the ground with lots of cards around it:
"There are 66 green and 66 blue door cards on the ground, if you put one blue and one green cards into the slots next to the door, the door will open. If you put the same color, you will hear a fail sound and lose a life! I won't tell you how many lifes you have, but you will die if you fail more than that!"
it would only require one try normally but you remember that you cannot distinguish the color of the stones, especially when they are green and blue because you are colorblind!, and probably whoever put you there knows this very well and that guy also knows what would be the minimum number of tries you need to open the door guaranteed as well so
Can you open it in time?
if so
How many tries do you need to open the door guaranteed?
logical-deduction strategy optimization
logical-deduction strategy optimization
asked 8 hours ago
OrayOray
16.5k438165
asked 8 hours ago
OrayOray
16.5k438165
edited 7 hours ago
Oray
asked 8 hours ago
OrayOray
16.5k438165
asked 8 hours ago
OrayOray
16.5k438165
asked 8 hours ago
OrayOray
16.5k438165
16.5k438165
$begingroup$
Can I get the cards back after an attempt?
$endgroup$
– cinico
6 hours ago
1
$begingroup$
@cinico of course
$endgroup$
– Oray
6 hours ago
add a comment |
$begingroup$
Can I get the cards back after an attempt?
$endgroup$
– cinico
6 hours ago
1
$begingroup$
@cinico of course
$endgroup$
– Oray
6 hours ago
$begingroup$
Can I get the cards back after an attempt?
$endgroup$
– cinico
6 hours ago
$begingroup$
Can I get the cards back after an attempt?
$endgroup$
– cinico
6 hours ago
1
1
$begingroup$
@cinico of course
$endgroup$
– Oray
6 hours ago
$begingroup$
@cinico of course
$endgroup$
– Oray
6 hours ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
I believe I can open it in
66
The reasoning behind it is:
The general strategy I'll be using is just brute-forcing a single card against all others. Pick a card, this'll be the "master" card you'll always use. Then you simply test the master card with another card up to 66 times. If it doesn't work, than you toss out the non-master card, and repeat.
As there are an equal number of green and blue cards (66), we can solve it the same regardless of the first card you chose.
So say your first card is green, then the worst case is that you fail your first 65 tries by getting 65 greens in a row. Then on the 66th try you're guaranteed to have a matching blue card since there are no green cards remaining besides the one in your hand.
The argument applies in exactly the same way if you chose a blue card first, just with blue cards instead.
answered 7 hours ago
JayJay
661
New contributor
Jay is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Ooo, I like it! I'll have to think more on it (Sorry, my discrete math's a bit rough around the edges)
$endgroup$
– Jay
7 hours ago
2
$begingroup$
This seems far too simple to be the answer, but I can't think of a better one.
$endgroup$
– Christian
7 hours ago
$begingroup$
I think you can do it in less if you do it in pairs. I'll try to organize this idea and write up an answer
$endgroup$
– João Bravo
6 hours ago
$begingroup$
seems i did something wrong with my assumptions and failed with question, this answer seems quite right to me now... :/ sorry...
$endgroup$
– Oray
6 hours ago
add a comment |
$begingroup$
This is like a reverse puzzle to the well-known
"How often do you have to break a chocolate apart to end up with 24 pieces?" - The answer is 23 because every act of breaking - no matter how - increases the number of pieces by exactly one
Here in reverse:
Any failed attempt with two cards in effect only tells you that these two cards have the same colour. As they are henceforth equivalent, you can as well dispose one of them - or to stay in the spirit of the chocolate puzzle, glue them together to a larger monochromatic card cluster.
The only move that guarantees that the door opens is when you would upon failure produce a card cluster of 67 or more cards (as a cluster of this size cannot be monochromatic). But producing a cluster of size 67 is easily possible in many different ways with 66 moves, but definitely not with less than 66 moves (cf. chocolate puzzle)
A remark as a thinking outside the box solution:
Most cases of colour blindness are red-green blindness and as far as I know can distinguish blue from green ;)
answered 6 hours ago
Hagen von EitzenHagen von Eitzen
40627
$endgroup$
1
$begingroup$
As a colorblind person myself, I can confirm that yes, I can generally distinguish blue from green. That said, if the green is slightly blue-tinted or vice versa, the two colors would look a lot more similar to me than they would for a normal-sighted person. In general, for a colorblind person, all colors look a bit more like other colors than they should (with the exception of red and green, which look a lot more like each other than they should). So it's possible that the cards are shades of blue and green that would stump only colorblind people.
$endgroup$
– Gilad M
5 hours ago
add a comment |
$begingroup$
If I can get the cards back after an attempt
the minimum number of tries to open the door is 3. Pick 3 cards. Worst case scenario is: you pick green and blue for the first attempt; then you pick one of the first and test it against the third, which fails. Then, you know that the third card will match the card unpicked on the second attempt and you can open the door.
Otherwise,
It is not guaranteed that you can open the door, because there is a chance that you will always pick different colors for the 66 attempts.
answered 6 hours ago
cinicocinico
27416
$endgroup$
3
$begingroup$
This is the right answer, but to a different question (the left and right sock problem). Read again, what condition opens the door.
$endgroup$
– Hagen von Eitzen
6 hours ago
$begingroup$
I'm must be slower than usual today, because I don't understand what I misunderstood from the conditions
$endgroup$
– cinico
5 hours ago
1
$begingroup$
You are thinking that equal colors will open the door. They must be different
$endgroup$
– João Bravo
5 hours ago
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I believe I can open it in
66
The reasoning behind it is:
The general strategy I'll be using is just brute-forcing a single card against all others. Pick a card, this'll be the "master" card you'll always use. Then you simply test the master card with another card up to 66 times. If it doesn't work, than you toss out the non-master card, and repeat.
As there are an equal number of green and blue cards (66), we can solve it the same regardless of the first card you chose.
So say your first card is green, then the worst case is that you fail your first 65 tries by getting 65 greens in a row. Then on the 66th try you're guaranteed to have a matching blue card since there are no green cards remaining besides the one in your hand.
The argument applies in exactly the same way if you chose a blue card first, just with blue cards instead.
answered 7 hours ago
JayJay
661
New contributor
Jay is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Ooo, I like it! I'll have to think more on it (Sorry, my discrete math's a bit rough around the edges)
$endgroup$
– Jay
7 hours ago
2
$begingroup$
This seems far too simple to be the answer, but I can't think of a better one.
$endgroup$
– Christian
7 hours ago
$begingroup$
I think you can do it in less if you do it in pairs. I'll try to organize this idea and write up an answer
$endgroup$
– João Bravo
6 hours ago
$begingroup$
seems i did something wrong with my assumptions and failed with question, this answer seems quite right to me now... :/ sorry...
$endgroup$
– Oray
6 hours ago
add a comment |
$begingroup$
I believe I can open it in
66
The reasoning behind it is:
The general strategy I'll be using is just brute-forcing a single card against all others. Pick a card, this'll be the "master" card you'll always use. Then you simply test the master card with another card up to 66 times. If it doesn't work, than you toss out the non-master card, and repeat.
As there are an equal number of green and blue cards (66), we can solve it the same regardless of the first card you chose.
So say your first card is green, then the worst case is that you fail your first 65 tries by getting 65 greens in a row. Then on the 66th try you're guaranteed to have a matching blue card since there are no green cards remaining besides the one in your hand.
The argument applies in exactly the same way if you chose a blue card first, just with blue cards instead.
answered 7 hours ago
JayJay
661
New contributor
Jay is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Ooo, I like it! I'll have to think more on it (Sorry, my discrete math's a bit rough around the edges)
$endgroup$
– Jay
7 hours ago
2
$begingroup$
This seems far too simple to be the answer, but I can't think of a better one.
$endgroup$
– Christian
7 hours ago
$begingroup$
I think you can do it in less if you do it in pairs. I'll try to organize this idea and write up an answer
$endgroup$
– João Bravo
6 hours ago
$begingroup$
seems i did something wrong with my assumptions and failed with question, this answer seems quite right to me now... :/ sorry...
$endgroup$
– Oray
6 hours ago
add a comment |
$begingroup$
I believe I can open it in
66
The reasoning behind it is:
The general strategy I'll be using is just brute-forcing a single card against all others. Pick a card, this'll be the "master" card you'll always use. Then you simply test the master card with another card up to 66 times. If it doesn't work, than you toss out the non-master card, and repeat.
As there are an equal number of green and blue cards (66), we can solve it the same regardless of the first card you chose.
So say your first card is green, then the worst case is that you fail your first 65 tries by getting 65 greens in a row. Then on the 66th try you're guaranteed to have a matching blue card since there are no green cards remaining besides the one in your hand.
The argument applies in exactly the same way if you chose a blue card first, just with blue cards instead.
answered 7 hours ago
JayJay
661
New contributor
Jay is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I believe I can open it in
66
The reasoning behind it is:
The general strategy I'll be using is just brute-forcing a single card against all others. Pick a card, this'll be the "master" card you'll always use. Then you simply test the master card with another card up to 66 times. If it doesn't work, than you toss out the non-master card, and repeat.
As there are an equal number of green and blue cards (66), we can solve it the same regardless of the first card you chose.
So say your first card is green, then the worst case is that you fail your first 65 tries by getting 65 greens in a row. Then on the 66th try you're guaranteed to have a matching blue card since there are no green cards remaining besides the one in your hand.
The argument applies in exactly the same way if you chose a blue card first, just with blue cards instead.
answered 7 hours ago
JayJay
661
New contributor
Jay is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 7 hours ago
JayJay
661
New contributor
Jay is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 7 hours ago
JayJay
661
answered 7 hours ago
JayJay
661
661
New contributor
Jay is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Jay is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
Ooo, I like it! I'll have to think more on it (Sorry, my discrete math's a bit rough around the edges)
$endgroup$
– Jay
7 hours ago
2
$begingroup$
This seems far too simple to be the answer, but I can't think of a better one.
$endgroup$
– Christian
7 hours ago
$begingroup$
I think you can do it in less if you do it in pairs. I'll try to organize this idea and write up an answer
$endgroup$
– João Bravo
6 hours ago
$begingroup$
seems i did something wrong with my assumptions and failed with question, this answer seems quite right to me now... :/ sorry...
$endgroup$
– Oray
6 hours ago
add a comment |
$begingroup$
Ooo, I like it! I'll have to think more on it (Sorry, my discrete math's a bit rough around the edges)
$endgroup$
– Jay
7 hours ago
2
$begingroup$
This seems far too simple to be the answer, but I can't think of a better one.
$endgroup$
– Christian
7 hours ago
$begingroup$
I think you can do it in less if you do it in pairs. I'll try to organize this idea and write up an answer
$endgroup$
– João Bravo
6 hours ago
$begingroup$
seems i did something wrong with my assumptions and failed with question, this answer seems quite right to me now... :/ sorry...
$endgroup$
– Oray
6 hours ago
$begingroup$
Ooo, I like it! I'll have to think more on it (Sorry, my discrete math's a bit rough around the edges)
$endgroup$
– Jay
7 hours ago
$begingroup$
Ooo, I like it! I'll have to think more on it (Sorry, my discrete math's a bit rough around the edges)
$endgroup$
– Jay
7 hours ago
2
2
$begingroup$
This seems far too simple to be the answer, but I can't think of a better one.
$endgroup$
– Christian
7 hours ago
$begingroup$
This seems far too simple to be the answer, but I can't think of a better one.
$endgroup$
– Christian
7 hours ago
$begingroup$
I think you can do it in less if you do it in pairs. I'll try to organize this idea and write up an answer
$endgroup$
– João Bravo
6 hours ago
$begingroup$
I think you can do it in less if you do it in pairs. I'll try to organize this idea and write up an answer
$endgroup$
– João Bravo
6 hours ago
$begingroup$
seems i did something wrong with my assumptions and failed with question, this answer seems quite right to me now... :/ sorry...
$endgroup$
– Oray
6 hours ago
$begingroup$
seems i did something wrong with my assumptions and failed with question, this answer seems quite right to me now... :/ sorry...
$endgroup$
– Oray
6 hours ago
add a comment |
$begingroup$
This is like a reverse puzzle to the well-known
"How often do you have to break a chocolate apart to end up with 24 pieces?" - The answer is 23 because every act of breaking - no matter how - increases the number of pieces by exactly one
Here in reverse:
Any failed attempt with two cards in effect only tells you that these two cards have the same colour. As they are henceforth equivalent, you can as well dispose one of them - or to stay in the spirit of the chocolate puzzle, glue them together to a larger monochromatic card cluster.
The only move that guarantees that the door opens is when you would upon failure produce a card cluster of 67 or more cards (as a cluster of this size cannot be monochromatic). But producing a cluster of size 67 is easily possible in many different ways with 66 moves, but definitely not with less than 66 moves (cf. chocolate puzzle)
A remark as a thinking outside the box solution:
Most cases of colour blindness are red-green blindness and as far as I know can distinguish blue from green ;)
answered 6 hours ago
Hagen von EitzenHagen von Eitzen
40627
$endgroup$
1
$begingroup$
As a colorblind person myself, I can confirm that yes, I can generally distinguish blue from green. That said, if the green is slightly blue-tinted or vice versa, the two colors would look a lot more similar to me than they would for a normal-sighted person. In general, for a colorblind person, all colors look a bit more like other colors than they should (with the exception of red and green, which look a lot more like each other than they should). So it's possible that the cards are shades of blue and green that would stump only colorblind people.
$endgroup$
– Gilad M
5 hours ago
add a comment |
$begingroup$
This is like a reverse puzzle to the well-known
"How often do you have to break a chocolate apart to end up with 24 pieces?" - The answer is 23 because every act of breaking - no matter how - increases the number of pieces by exactly one
Here in reverse:
Any failed attempt with two cards in effect only tells you that these two cards have the same colour. As they are henceforth equivalent, you can as well dispose one of them - or to stay in the spirit of the chocolate puzzle, glue them together to a larger monochromatic card cluster.
The only move that guarantees that the door opens is when you would upon failure produce a card cluster of 67 or more cards (as a cluster of this size cannot be monochromatic). But producing a cluster of size 67 is easily possible in many different ways with 66 moves, but definitely not with less than 66 moves (cf. chocolate puzzle)
A remark as a thinking outside the box solution:
Most cases of colour blindness are red-green blindness and as far as I know can distinguish blue from green ;)
answered 6 hours ago
Hagen von EitzenHagen von Eitzen
40627
$endgroup$
1
$begingroup$
As a colorblind person myself, I can confirm that yes, I can generally distinguish blue from green. That said, if the green is slightly blue-tinted or vice versa, the two colors would look a lot more similar to me than they would for a normal-sighted person. In general, for a colorblind person, all colors look a bit more like other colors than they should (with the exception of red and green, which look a lot more like each other than they should). So it's possible that the cards are shades of blue and green that would stump only colorblind people.
$endgroup$
– Gilad M
5 hours ago
add a comment |
$begingroup$
This is like a reverse puzzle to the well-known
"How often do you have to break a chocolate apart to end up with 24 pieces?" - The answer is 23 because every act of breaking - no matter how - increases the number of pieces by exactly one
Here in reverse:
Any failed attempt with two cards in effect only tells you that these two cards have the same colour. As they are henceforth equivalent, you can as well dispose one of them - or to stay in the spirit of the chocolate puzzle, glue them together to a larger monochromatic card cluster.
The only move that guarantees that the door opens is when you would upon failure produce a card cluster of 67 or more cards (as a cluster of this size cannot be monochromatic). But producing a cluster of size 67 is easily possible in many different ways with 66 moves, but definitely not with less than 66 moves (cf. chocolate puzzle)
A remark as a thinking outside the box solution:
Most cases of colour blindness are red-green blindness and as far as I know can distinguish blue from green ;)
answered 6 hours ago
Hagen von EitzenHagen von Eitzen
40627
$endgroup$
This is like a reverse puzzle to the well-known
"How often do you have to break a chocolate apart to end up with 24 pieces?" - The answer is 23 because every act of breaking - no matter how - increases the number of pieces by exactly one
Here in reverse:
Any failed attempt with two cards in effect only tells you that these two cards have the same colour. As they are henceforth equivalent, you can as well dispose one of them - or to stay in the spirit of the chocolate puzzle, glue them together to a larger monochromatic card cluster.
The only move that guarantees that the door opens is when you would upon failure produce a card cluster of 67 or more cards (as a cluster of this size cannot be monochromatic). But producing a cluster of size 67 is easily possible in many different ways with 66 moves, but definitely not with less than 66 moves (cf. chocolate puzzle)
A remark as a thinking outside the box solution:
Most cases of colour blindness are red-green blindness and as far as I know can distinguish blue from green ;)
answered 6 hours ago
Hagen von EitzenHagen von Eitzen
40627
edited 6 hours ago
answered 6 hours ago
Hagen von EitzenHagen von Eitzen
40627
answered 6 hours ago
Hagen von EitzenHagen von Eitzen
40627
answered 6 hours ago
Hagen von EitzenHagen von Eitzen
40627
40627
1
$begingroup$
As a colorblind person myself, I can confirm that yes, I can generally distinguish blue from green. That said, if the green is slightly blue-tinted or vice versa, the two colors would look a lot more similar to me than they would for a normal-sighted person. In general, for a colorblind person, all colors look a bit more like other colors than they should (with the exception of red and green, which look a lot more like each other than they should). So it's possible that the cards are shades of blue and green that would stump only colorblind people.
$endgroup$
– Gilad M
5 hours ago
add a comment |
1
$begingroup$
As a colorblind person myself, I can confirm that yes, I can generally distinguish blue from green. That said, if the green is slightly blue-tinted or vice versa, the two colors would look a lot more similar to me than they would for a normal-sighted person. In general, for a colorblind person, all colors look a bit more like other colors than they should (with the exception of red and green, which look a lot more like each other than they should). So it's possible that the cards are shades of blue and green that would stump only colorblind people.
$endgroup$
– Gilad M
5 hours ago
1
1
$begingroup$
As a colorblind person myself, I can confirm that yes, I can generally distinguish blue from green. That said, if the green is slightly blue-tinted or vice versa, the two colors would look a lot more similar to me than they would for a normal-sighted person. In general, for a colorblind person, all colors look a bit more like other colors than they should (with the exception of red and green, which look a lot more like each other than they should). So it's possible that the cards are shades of blue and green that would stump only colorblind people.
$endgroup$
– Gilad M
5 hours ago
$begingroup$
As a colorblind person myself, I can confirm that yes, I can generally distinguish blue from green. That said, if the green is slightly blue-tinted or vice versa, the two colors would look a lot more similar to me than they would for a normal-sighted person. In general, for a colorblind person, all colors look a bit more like other colors than they should (with the exception of red and green, which look a lot more like each other than they should). So it's possible that the cards are shades of blue and green that would stump only colorblind people.
$endgroup$
– Gilad M
5 hours ago
add a comment |
$begingroup$
If I can get the cards back after an attempt
the minimum number of tries to open the door is 3. Pick 3 cards. Worst case scenario is: you pick green and blue for the first attempt; then you pick one of the first and test it against the third, which fails. Then, you know that the third card will match the card unpicked on the second attempt and you can open the door.
Otherwise,
It is not guaranteed that you can open the door, because there is a chance that you will always pick different colors for the 66 attempts.
answered 6 hours ago
cinicocinico
27416
$endgroup$
3
$begingroup$
This is the right answer, but to a different question (the left and right sock problem). Read again, what condition opens the door.
$endgroup$
– Hagen von Eitzen
6 hours ago
$begingroup$
I'm must be slower than usual today, because I don't understand what I misunderstood from the conditions
$endgroup$
– cinico
5 hours ago
1
$begingroup$
You are thinking that equal colors will open the door. They must be different
$endgroup$
– João Bravo
5 hours ago
add a comment |
$begingroup$
If I can get the cards back after an attempt
the minimum number of tries to open the door is 3. Pick 3 cards. Worst case scenario is: you pick green and blue for the first attempt; then you pick one of the first and test it against the third, which fails. Then, you know that the third card will match the card unpicked on the second attempt and you can open the door.
Otherwise,
It is not guaranteed that you can open the door, because there is a chance that you will always pick different colors for the 66 attempts.
answered 6 hours ago
cinicocinico
27416
$endgroup$
3
$begingroup$
This is the right answer, but to a different question (the left and right sock problem). Read again, what condition opens the door.
$endgroup$
– Hagen von Eitzen
6 hours ago
$begingroup$
I'm must be slower than usual today, because I don't understand what I misunderstood from the conditions
$endgroup$
– cinico
5 hours ago
1
$begingroup$
You are thinking that equal colors will open the door. They must be different
$endgroup$
– João Bravo
5 hours ago
add a comment |
$begingroup$
If I can get the cards back after an attempt
the minimum number of tries to open the door is 3. Pick 3 cards. Worst case scenario is: you pick green and blue for the first attempt; then you pick one of the first and test it against the third, which fails. Then, you know that the third card will match the card unpicked on the second attempt and you can open the door.
Otherwise,
It is not guaranteed that you can open the door, because there is a chance that you will always pick different colors for the 66 attempts.
answered 6 hours ago
cinicocinico
27416
$endgroup$
If I can get the cards back after an attempt
the minimum number of tries to open the door is 3. Pick 3 cards. Worst case scenario is: you pick green and blue for the first attempt; then you pick one of the first and test it against the third, which fails. Then, you know that the third card will match the card unpicked on the second attempt and you can open the door.
Otherwise,
It is not guaranteed that you can open the door, because there is a chance that you will always pick different colors for the 66 attempts.
answered 6 hours ago
cinicocinico
27416
answered 6 hours ago
cinicocinico
27416
answered 6 hours ago
cinicocinico
27416
answered 6 hours ago
cinicocinico
27416
27416
3
$begingroup$
This is the right answer, but to a different question (the left and right sock problem). Read again, what condition opens the door.
$endgroup$
– Hagen von Eitzen
6 hours ago
$begingroup$
I'm must be slower than usual today, because I don't understand what I misunderstood from the conditions
$endgroup$
– cinico
5 hours ago
1
$begingroup$
You are thinking that equal colors will open the door. They must be different
$endgroup$
– João Bravo
5 hours ago
add a comment |
3
$begingroup$
This is the right answer, but to a different question (the left and right sock problem). Read again, what condition opens the door.
$endgroup$
– Hagen von Eitzen
6 hours ago
$begingroup$
I'm must be slower than usual today, because I don't understand what I misunderstood from the conditions
$endgroup$
– cinico
5 hours ago
1
$begingroup$
You are thinking that equal colors will open the door. They must be different
$endgroup$
– João Bravo
5 hours ago
3
3
$begingroup$
This is the right answer, but to a different question (the left and right sock problem). Read again, what condition opens the door.
$endgroup$
– Hagen von Eitzen
6 hours ago
$begingroup$
This is the right answer, but to a different question (the left and right sock problem). Read again, what condition opens the door.
$endgroup$
– Hagen von Eitzen
6 hours ago
$begingroup$
I'm must be slower than usual today, because I don't understand what I misunderstood from the conditions
$endgroup$
– cinico
5 hours ago
$begingroup$
I'm must be slower than usual today, because I don't understand what I misunderstood from the conditions
$endgroup$
– cinico
5 hours ago
1
1
$begingroup$
You are thinking that equal colors will open the door. They must be different
$endgroup$
– João Bravo
5 hours ago
$begingroup$
You are thinking that equal colors will open the door. They must be different
$endgroup$
– João Bravo
5 hours ago
add a comment |
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$begingroup$
Can I get the cards back after an attempt?
$endgroup$
– cinico
6 hours ago
1
$begingroup$
@cinico of course
$endgroup$
– Oray
6 hours ago