Three doors puzzle
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November 30th, 2017 at 1:04:59 PM
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There are three doors numbered 1, 2, and 3. Behind each, placed randomly, are red, blue, and green cars.
There are three contestants one each wearing a red, blue, and green hat.
Each may open two doors, one at a time. If each player sees the car of the same color as his hat, then all three will win cars. If at least one doesn't see the car of his own color, then all three lose.
The three players must go one at a time and may not communicate once they start. Each player must leave all three doors closed after his turn. They may communicate before the first player's turn.
What is the greatest chance they can achieve of winning the three cars and how should they do it?
For example, if each picked randomly, their chances of winning would (2/3)^3. But can that strategy be improved upon?
This is not a trick question like peeking under the door or anything like that.
Free beer to the first satisfactory answer and solution. The question for the poll is what is the answer?
There are three contestants one each wearing a red, blue, and green hat.
Each may open two doors, one at a time. If each player sees the car of the same color as his hat, then all three will win cars. If at least one doesn't see the car of his own color, then all three lose.
The three players must go one at a time and may not communicate once they start. Each player must leave all three doors closed after his turn. They may communicate before the first player's turn.
What is the greatest chance they can achieve of winning the three cars and how should they do it?
For example, if each picked randomly, their chances of winning would (2/3)^3. But can that strategy be improved upon?
This is not a trick question like peeking under the door or anything like that.
Free beer to the first satisfactory answer and solution. The question for the poll is what is the answer?
Last edited by: Wizard on Nov 30, 2017
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
November 30th, 2017 at 1:29:49 PM
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Two players should agree to open the same two doors. They have a 1/3 chance of choosing the right 2 door combination. The third player should select the remaining door and either of the first two doors. If the first two players were successful, then it is guaranteed that the third player is successful. This increases their odds from 8/27 to 1/3.
I heart Crystal Math.
November 30th, 2017 at 2:12:10 PM
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This sounds right to me. I briefly posted a different answer, but I had misread one of the rules.
If the House lost every hand, they wouldn't deal the game.
November 30th, 2017 at 4:12:33 PM
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They all look at door one. The player who matches chooses no more doors. Beforehand, it was decided which of the non-matching players would take door two. Say, by the lowest frequency of the colors of hats remaining. Lowest goes first.
Never make a bet that you wouldn't take, yourself.
November 30th, 2017 at 4:20:48 PM
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Quote: CrystalMath
Two players should agree to open the same two doors. They have a 1/3 chance of choosing the right 2 door combination. The third player should select the remaining door and either of the first two doors. If the first two players were successful, then it is guaranteed that the third player is successful. This increases their odds from 8/27 to 1/3.
The probability of success can be improved upon that.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
November 30th, 2017 at 4:43:26 PM
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Huh?Quote: WizardEach may open two doors, open at a time.
November 30th, 2017 at 5:00:42 PM
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My shot:
Player one picks doors one and two. He has a 2/3 chance of being right. Player two picks door two and three. If player one was correct, player two now has a 3/4 chance of being right. Which means that through the first two players, they have a 1/2 chance of still having a chance. Player three then picks doors one and three. If the first two players were right, player three has a 3/4 chance of being right, which gives the team a 3/8 chance of winning.
Player one picks doors one and two. He has a 2/3 chance of being right. Player two picks door two and three. If player one was correct, player two now has a 3/4 chance of being right. Which means that through the first two players, they have a 1/2 chance of still having a chance. Player three then picks doors one and three. If the first two players were right, player three has a 3/4 chance of being right, which gives the team a 3/8 chance of winning.
November 30th, 2017 at 5:17:29 PM
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66.66% + (1/2)*33.33% = 83.3333%
Red person checks door 1. If his colored car is there, he stays behind the door. If it's not there, he returns to his group. Then blue person goes, he checks door 2. If his colored car is there, he stays behind the door. Otherwise, he returns. Green dude checks door 3. If it's not green, he returns.
The probability none of them get the correct door on the first try is 1/3 or 33%. If one of them gets the correct door (66% of the time), then they're guaranteed to win, because there are only 2 options available, and the remaining contestants either have already looked behind one door and they know that's not or they still have 2 "tries" remaining.
Of the 33% times they are all wrong on the initial door guess, they now have a 50% chance of winning. Red guy goes into door #2. If the car is red, he stays behind the door. The others now know door #2 is taken, so blue checks door 3 and green checks door 1 (since door 1 can't be blue, because if it was, then door 3 would have to be green, but the green guy already checked door 3).
Red person checks door 1. If his colored car is there, he stays behind the door. If it's not there, he returns to his group. Then blue person goes, he checks door 2. If his colored car is there, he stays behind the door. Otherwise, he returns. Green dude checks door 3. If it's not green, he returns.
The probability none of them get the correct door on the first try is 1/3 or 33%. If one of them gets the correct door (66% of the time), then they're guaranteed to win, because there are only 2 options available, and the remaining contestants either have already looked behind one door and they know that's not or they still have 2 "tries" remaining.
Of the 33% times they are all wrong on the initial door guess, they now have a 50% chance of winning. Red guy goes into door #2. If the car is red, he stays behind the door. The others now know door #2 is taken, so blue checks door 3 and green checks door 1 (since door 1 can't be blue, because if it was, then door 3 would have to be green, but the green guy already checked door 3).
November 30th, 2017 at 5:23:45 PM
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Quote: RSHuh?
I meant one at a time.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
November 30th, 2017 at 5:25:04 PM
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Quote: TomGMy shot:
Player one picks doors one and two. He has a 2/3 chance of being right. Player two picks door two and three. If player one was correct, player two now has a 3/4 chance of being right. Which means that through the first two players, they have a 1/2 chance of still having a chance. Player three then picks doors one and three. If the first two players were right, player three has a 3/4 chance of being right, which gives the team a 3/8 chance of winning.
You can improve upon a 3/8 chance
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
November 30th, 2017 at 5:26:59 PM
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Quote: RS66.66% + (1/2)*33.33% = 83.3333%
Red person checks door 1. If his colored car is there, he stays behind the door. If it's not there, he returns to his group. Then blue person goes, he checks door 2. If his colored car is there, he stays behind the door. Otherwise, he returns. Green dude checks door 3. If it's not green, he returns.
The probability none of them get the correct door on the first try is 1/3 or 33%. If one of them gets the correct door (66% of the time), then they're guaranteed to win, because there are only 2 options available, and the remaining contestants either have already looked behind one door and they know that's not or they still have 2 "tries" remaining.
Of the 33% times they are all wrong on the initial door guess, they now have a 50% chance of winning. Red guy goes into door #2. If the car is red, he stays behind the door. The others now know door #2 is taken, so blue checks door 3 and green checks door 1 (since door 1 can't be blue, because if it was, then door 3 would have to be green, but the green guy already checked door 3).
Let me clarify the rules. After each player opens one or two doors, he must return to wherever he started from. You can't hide behind doors. Imagine the game show host saying "I can confirm the last man is done, you may proceed with your turn" once the person before you finishes.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
November 30th, 2017 at 5:44:59 PM
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Ah, you gotta use the ol' chain method, as they say.
2/3 or 66.66%
A is assigned to 1, B is assigned to 2, C is assigned to 3.
A opens door 1. If it's "A", then he doesn't need to check another door. If it's not his corresponding car color, then he goes to door 2 if it's the "B" color or to door 3 if it's the "C" color.
B checks door 2, if it's not his, then he either goes to door 1 if it's the "A" color or door 3 if it's the "C" color.
C checks door 3, if it's not his, then he either goes to door 1 if it's the "A" color or 2 if it's the "B" color.
Possible combos:
ABC
ACB
BAC
BCA
CAB
CBA
ABC - win - easy to see how
ACB - win - A picks on first try. B sees 'C' in his door so he checks door 3. C checks door 3, sees it's B, so he goes to 2.
BAC - win - Same as above, except it's A and B swapping
BCA - lose - A sees 'B' in his door, so he checks door 2 and sees 'C'. He used his 2 tries and failed (miserably).
CAB - lose - A sees 'C' in his door, so he checks door 3 and and sees 'B'. He used his 2 tries and failed again, wow, he sucks.
CBA - win - same as the second and third example, and A and C are swapping.
2/3 or 66.66%
A is assigned to 1, B is assigned to 2, C is assigned to 3.
A opens door 1. If it's "A", then he doesn't need to check another door. If it's not his corresponding car color, then he goes to door 2 if it's the "B" color or to door 3 if it's the "C" color.
B checks door 2, if it's not his, then he either goes to door 1 if it's the "A" color or door 3 if it's the "C" color.
C checks door 3, if it's not his, then he either goes to door 1 if it's the "A" color or 2 if it's the "B" color.
Possible combos:
ABC
ACB
BAC
BCA
CAB
CBA
ABC - win - easy to see how
ACB - win - A picks on first try. B sees 'C' in his door so he checks door 3. C checks door 3, sees it's B, so he goes to 2.
BAC - win - Same as above, except it's A and B swapping
BCA - lose - A sees 'B' in his door, so he checks door 2 and sees 'C'. He used his 2 tries and failed (miserably).
CAB - lose - A sees 'C' in his door, so he checks door 3 and and sees 'B'. He used his 2 tries and failed again, wow, he sucks.
CBA - win - same as the second and third example, and A and C are swapping.
Instead of a beer, I'll be humble and just take 1/4 of a WizCoin. =)
November 30th, 2017 at 6:20:07 PM
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Quote: RSAh, you gotta use the ol' chain method, as they say.
2/3 or 66.66%
A is assigned to 1, B is assigned to 2, C is assigned to 3.
A opens door 1. If it's "A", then he doesn't need to check another door. If it's not his corresponding car color, then he goes to door 2 if it's the "B" color or to door 3 if it's the "C" color.
B checks door 2, if it's not his, then he either goes to door 1 if it's the "A" color or door 3 if it's the "C" color.
C checks door 3, if it's not his, then he either goes to door 1 if it's the "A" color or 2 if it's the "B" color.
Possible combos:
ABC
ACB
BAC
BCA
CAB
CBA
ABC - win - easy to see how
ACB - win - A picks on first try. B sees 'C' in his door so he checks door 3. C checks door 3, sees it's B, so he goes to 2.
BAC - win - Same as above, except it's A and B swapping
BCA - lose - A sees 'B' in his door, so he checks door 2 and sees 'C'. He used his 2 tries and failed (miserably).
CAB - lose - A sees 'C' in his door, so he checks door 3 and and sees 'B'. He used his 2 tries and failed again, wow, he sucks.
CBA - win - same as the second and third example, and A and C are swapping.
Congratulations! That looks right. Nice to see a new winner for a change. I'll let you swap the beer for 0.01% of a Wizcoin.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
November 30th, 2017 at 6:28:38 PM
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Quote: WizardQuote: RSAh, you gotta use the ol' chain method, as they say.
2/3 or 66.66%
A is assigned to 1, B is assigned to 2, C is assigned to 3.
A opens door 1. If it's "A", then he doesn't need to check another door. If it's not his corresponding car color, then he goes to door 2 if it's the "B" color or to door 3 if it's the "C" color.
B checks door 2, if it's not his, then he either goes to door 1 if it's the "A" color or door 3 if it's the "C" color.
C checks door 3, if it's not his, then he either goes to door 1 if it's the "A" color or 2 if it's the "B" color.
Possible combos:
ABC
ACB
BAC
BCA
CAB
CBA
ABC - win - easy to see how
ACB - win - A picks on first try. B sees 'C' in his door so he checks door 3. C checks door 3, sees it's B, so he goes to 2.
BAC - win - Same as above, except it's A and B swapping
BCA - lose - A sees 'B' in his door, so he checks door 2 and sees 'C'. He used his 2 tries and failed (miserably).
CAB - lose - A sees 'C' in his door, so he checks door 3 and and sees 'B'. He used his 2 tries and failed again, wow, he sucks.
CBA - win - same as the second and third example, and A and C are swapping.
Congratulations! That looks right. Nice to see a new winner for a change. I'll let you swap the beer for 0.01% of a Wizcoin.
Thanks and thanks. I'll accept the 0.01% of a wizcoin.
November 30th, 2017 at 7:07:48 PM
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Quote: WizardQuote: RSAh, you gotta use the ol' chain method, as they say.
2/3 or 66.66%
A is assigned to 1, B is assigned to 2, C is assigned to 3.
A opens door 1. If it's "A", then he doesn't need to check another door. If it's not his corresponding car color, then he goes to door 2 if it's the "B" color or to door 3 if it's the "C" color.
B checks door 2, if it's not his, then he either goes to door 1 if it's the "A" color or door 3 if it's the "C" color.
C checks door 3, if it's not his, then he either goes to door 1 if it's the "A" color or 2 if it's the "B" color.
Possible combos:
ABC
ACB
BAC
BCA
CAB
CBA
ABC - win - easy to see how
ACB - win - A picks on first try. B sees 'C' in his door so he checks door 3. C checks door 3, sees it's B, so he goes to 2.
BAC - win - Same as above, except it's A and B swapping
BCA - lose - A sees 'B' in his door, so he checks door 2 and sees 'C'. He used his 2 tries and failed (miserably).
CAB - lose - A sees 'C' in his door, so he checks door 3 and and sees 'B'. He used his 2 tries and failed again, wow, he sucks.
CBA - win - same as the second and third example, and A and C are swapping.
Congratulations! That looks right.
Except that involves communication between them after the players begin to check the doors. It's not a set plan from the start. More to not communicating than just not closing doors afterward.
Never make a bet that you wouldn't take, yourself.
November 30th, 2017 at 7:10:48 PM
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Quote: 1MatterToMotionQuote: WizardQuote: RSAh, you gotta use the ol' chain method, as they say.
2/3 or 66.66%
A is assigned to 1, B is assigned to 2, C is assigned to 3.
A opens door 1. If it's "A", then he doesn't need to check another door. If it's not his corresponding car color, then he goes to door 2 if it's the "B" color or to door 3 if it's the "C" color.
B checks door 2, if it's not his, then he either goes to door 1 if it's the "A" color or door 3 if it's the "C" color.
C checks door 3, if it's not his, then he either goes to door 1 if it's the "A" color or 2 if it's the "B" color.
Possible combos:
ABC
ACB
BAC
BCA
CAB
CBA
ABC - win - easy to see how
ACB - win - A picks on first try. B sees 'C' in his door so he checks door 3. C checks door 3, sees it's B, so he goes to 2.
BAC - win - Same as above, except it's A and B swapping
BCA - lose - A sees 'B' in his door, so he checks door 2 and sees 'C'. He used his 2 tries and failed (miserably).
CAB - lose - A sees 'C' in his door, so he checks door 3 and and sees 'B'. He used his 2 tries and failed again, wow, he sucks.
CBA - win - same as the second and third example, and A and C are swapping.
Congratulations! That looks right.
Except that involves communication between them after the players begin to check the doors. It's not a set plan from the start. More to not communicating than just not closing doors afterward.
Where's the communication?
Person "A" goes in thinking, "I'm going to check door 1, if it's blue then I'll check door 2, if it's geeen I'll check door 3, and if it's red, it doesn't matter, yolo."
December 1st, 2017 at 8:50:10 AM
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Hi. Player A's actions tell the others what's in door one. Even if the others don't use that information, we can't know they don't. The operative words in the original question are "... and may not communicate once they start." May not, but not can not. This means they are not specifically kept in "sound proof booths" out of sight of everyone else throughout the players' turns, however, as one might expect, anyway.
Never make a bet that you wouldn't take, yourself.
December 1st, 2017 at 9:42:18 AM
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I still say RS's answer is good. I don't see any communication after the first player took his turn.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
December 1st, 2017 at 9:59:43 AM
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Quote: 1MatterToMotionHi. Player A's actions tell the others what's in door one. Even if the others don't use that information, we can't know they don't. The operative words in the original question are "... and may not communicate once they start." May not, but not can not. This means they are not specifically kept in "sound proof booths" out of sight of everyone else throughout the players' turns, however, as one might expect, anyway.
I think a clarification is needed.
First, I assume that each player knows the color of each hat, including his own.
Each one follows the same procedure:
Open "your' door (the one wearing the red hat opens door 1, the green hat opens door 2, and the blue hat opens door 3).
If you see a red car, open door 1 as your second door.
If you see a green car, open door 2 as your second door.
If you see a blue car, open door 3 as your second door.
The possibilities for the locations of the cars are:
R, G, B - each one finds his car with his first door; they win
R, B, G - Red sees his car with his first door; Green sees a blue car, so he opens door 3, which is his car; Blue sees a green car, so he opens door 2, which is his car; they win
G, R, B - Blue sees his car with his first door; Red sees a green car, so he opens door 2, which is his car; Green sees a red car, so he opens door 1, which is his car; they win
G, B, R - Red sees a red car, so he opens door 2, which has a blue car; they lose
B, R, G - Red sees a blue car, so he opens door 3, which has a green car; they lose
B, G, R - Green sees his car with his first door; Red sees a blue car, so he opens door 3, which is his car; Blue sees a red car, so he opens door 1, which is his car; they win
Not only is there no communication once they start, but the order in which they take their turns is irrelevant as well.
December 1st, 2017 at 10:16:57 AM
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This answer gives them a 2/3 chance of winning and is simpler than other posted answers:
They agree that the first player will begin by opening door#1. If door #1 is NOT his own color, player 1 will open door #2. If Door #1 is his own color, he will open door #3.
If player 1 opens door#1 and #3, they realize that #1 was player 1's color, so players 2 and 3 will simply open doors #2 and #3.
On the other hand, if player 1 opens doors #1 and #2, the other players know that door#1 was not player 1's color. Therefore, the other two players must assume that Door#2 was Player 1's color, because otherwise they cannot win. Therefore, the players 2 and 3 must pick doors #1 and #3.
They agree that the first player will begin by opening door#1. If door #1 is NOT his own color, player 1 will open door #2. If Door #1 is his own color, he will open door #3.
If player 1 opens door#1 and #3, they realize that #1 was player 1's color, so players 2 and 3 will simply open doors #2 and #3.
On the other hand, if player 1 opens doors #1 and #2, the other players know that door#1 was not player 1's color. Therefore, the other two players must assume that Door#2 was Player 1's color, because otherwise they cannot win. Therefore, the players 2 and 3 must pick doors #1 and #3.
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
December 1st, 2017 at 10:22:24 AM
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Quote: gordonm888This answer gives them a 2/3 chance of winning and is simpler than other posted answers:
They agree that the first player will begin by opening door#1. If door #1 is NOT his own color, player 1 will open door #2. If Door #1 is his own color, he will open door #3.
If player 1 opens door#1 and #3, they realize that #1 was player 1's color, so players 2 and 3 will simply open doors #2 and #3.
But this version implies that the doors remain open after each choice. The problem as stated says that the doors are closed after each choice.
December 1st, 2017 at 10:42:55 AM
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Quote: ThatDonGuy
Not only is there no communication once they start, but the order in which they take their turns is irrelevant as well.
Further communication isn't necessary to achieve the 2/3 result, but it happens regardless when B and C see what A, who goes first, does. B and C know A's decision making process, but what A ultimately does is new information to them. Unless all the players act simultaneously or in isolation. Maybe I am missing something, as who knows.
Never make a bet that you wouldn't take, yourself.
December 1st, 2017 at 11:22:05 AM
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Remember that communication is allowed before the first player acts. There are lots of "devise a strategy" riddles like this.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
December 2nd, 2017 at 2:39:36 AM
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I think that there are a number of also math errors in this thread, beginning with the players' success rate if each chooses randomly (on his own).Quote: 1MatterToMotionMaybe I am missing something, as who knows.
Never make a bet that you wouldn't take, yourself.
December 3rd, 2017 at 12:49:56 PM
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"There are three doors numbered 1, 2, and 3. Behind each, placed randomly, are red, blue, and green cars." The only way to do this truly randomly is to somewhat account for or counter the possibility of also the 2/3 probabilistic strategy. The cars aren't arranged truly randomly as long as the players can improve upon a random strategy using probability to determine which car is behind which door.
Secondly, suppose if the player with the red hat goes into door 1, and next toward door 2 (when car 1 isn't red), then the player with the blue hat who has seen the red car at door 3 hurries into door 2 to force the player with the red hat into door 3, unless the player with the green hat saw the green car in door 2, and hurries back even faster into door 2 to force the player with the blue hat into door 1. Each player just instinctively moves for his own benefit as much as the group's, without any attempt to communicate anything to the others. The player with the red hat doesn't necessarily communicate what is in door 1 by his possibly later moving to door 2 or door 3. And, the time to look into a door isn't the same as standing beside it after looking, as it blocks another player from entering as long as they are taking turns. Then, their chance of success is p = 1.
Secondly, suppose if the player with the red hat goes into door 1, and next toward door 2 (when car 1 isn't red), then the player with the blue hat who has seen the red car at door 3 hurries into door 2 to force the player with the red hat into door 3, unless the player with the green hat saw the green car in door 2, and hurries back even faster into door 2 to force the player with the blue hat into door 1. Each player just instinctively moves for his own benefit as much as the group's, without any attempt to communicate anything to the others. The player with the red hat doesn't necessarily communicate what is in door 1 by his possibly later moving to door 2 or door 3. And, the time to look into a door isn't the same as standing beside it after looking, as it blocks another player from entering as long as they are taking turns. Then, their chance of success is p = 1.
Last edited by: 1MatterToMotion on Dec 3, 2017
Never make a bet that you wouldn't take, yourself.
December 3rd, 2017 at 1:51:37 PM
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Quote: 1MatterToMotionI think that there are a number of also math errors in this thread, beginning with the players' success rate if each chooses randomly (on his own).
And what would be the probability of success of every player picked randomly, if not (2/3)3 = 8/27?
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
December 3rd, 2017 at 1:52:52 PM
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Quote: 1MatterToMotion...then the player with the blue hat who has seen the red car at door 3 hurries into door 2 to force the player with the red hat into door 3, ...
The players go one at a time. Furthermore, your answer amounts to communication during the game.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
December 5th, 2017 at 9:27:47 AM
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The players open the doors one at a time, regardless of them, one at a time, going to or deciding on, a door. And, if acting in isolation, they set their watches for when to go to the doors, and "meet up" at the doors such that they never see each other, and their freedom to check or take turns in doors remains otherwise unrestricted.Quote: WizardThe players go one at a time. How they get to a door seems to be an open matter. As does the order of turns, and all selections in one turn? Furthermore, your answer amounts to communication during the game.
Isn't any move a form of communication if not done in complete isolation? And the additional information from a move that is disallowed? If A says exactly what is in door 1 by going to door 2 or 3, that is additional information. It's okay for A with a red hat to say door 1 is a red car by going back to it, or by not opening an other door, but not for A to say door 1 is specifically a green or blue car by opening door 2 or 3 respectively. So, it's okay for A to say that door 1 is not a red car. Player A can "tell" B not to go to a particular door (that isn't B's hat color) by making that door his choice or move in a way that involves only the opening of the doors in some sequence, such as by already looking in a door at the spit second B arrives there; and B would then, logically or otherwise, go on to an other door to continue to maximize also his own chance of success. B isn't allowed to stand beside the door, and shouldn't give up on looking further. Nor can he know that A is already in the door if they are acting in isolation. Both players are merely doing or trying to do what has been set out. B could only try to open the door that holds another player looking at it, and the other player isn't obliged to show it to him. Is there a rule here that we always get what we want in life, and that all doors will open to us?
In any event, off hand, I think that this strategy works without blocking or interrupting a player's approach to a door when the players can see each other. Would have to think this over a bit. But, then, players could just make their lengths of times in a door an indication of what's up? Or would that constitute more communication after the fact? As people don't, normally, walk and think to at the same rate. Nor do they look at or process things at once. Some of us slow down, or change our minds in midstream. Elegant solutions are for perfect people and worlds.
Incidentally, I listed only the strategy for the player with the green hat when he sees a green car at door 2 on his first try. When door 2 is blue or red, he acts last behind the other players.
As in the above strategy, a player can and may go back to the same door, especially if he has already matched his hat there. And, to spread out selections of doors among also the individual players isn't entirely random. What if one of the players can't fathom any strategy, and wanders around a couple of times without settling on any door, or all three doors? Furthermore, who's to say that the players won't randomly pick an overall, better strategy (to look at doors) instead of a different one (in the form of a simple selection) each time? Maybe, one of the players already has a list of lucky numbers. But, there's always a chance that they all independently settle on even the 2/3 strategy by accident to use each time. Even a broken clock is right twice a day. I'm sure that others things could happen too. People are the most unpredictable, at times.Quote: WizardAnd what would be the probability of success of every player picked randomly, if not (2/3)3 = 8/27?
Never make a bet that you wouldn't take, yourself.
December 5th, 2017 at 10:55:42 AM
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Quote: WizardThe three players must go one at a time and may not communicate once they start.
What is so hard to understand about this rule of no communication? I didn't mean just no verbal communication. I meant no hiding behind doors, winking, Morse code, opening and closing doors with timed precision, or standing in another player's way. Do I have to list every possible way communication may happen? I mean NO COMMUNICATION AT ALL!!!.
*sigh*
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
December 5th, 2017 at 12:23:34 PM
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Quote: 1MatterToMotionThe players open the doors one at a time, regardless of them, one at a time, going to or deciding on, a door. And, if acting in isolation, they set their watches for when to go to the doors, and "meet up" at the doors such that they never see each other, and their freedom to check or take turns in doors remains otherwise unrestricted.
Isn't any move a form of communication if not done in complete isolation? And the additional information from a move that is disallowed? If A says exactly what is in door 1 by going to door 2 or 3, that is additional information. It's okay for A with a red hat to say door 1 is a red car by going back to it, or by not opening an other door, but not for A to say door 1 is specifically a green or blue car by opening door 2 or 3 respectively. So, it's okay for A to say that door 1 is not a red car. Player A can "tell" B not to go to a particular door (that isn't B's hat color) by making that door his choice or move in a way that involves only the opening of the doors in some sequence, such as by already looking in a door at the spit second B arrives there; and B would then, logically or otherwise, go on to an other door to continue to maximize also his own chance of success. B isn't allowed to stand beside the door, and shouldn't give up on looking further. Nor can he know that A is already in the door if they are acting in isolation. Both players are merely doing or trying to do what has been set out. B could only try to open the door that holds another player looking at it, and the other player isn't obliged to show it to him. Is there a rule here that we always get what we want in life, and that all doors will open to us?
In any event, off hand, I think that this strategy works without blocking or interrupting a player's approach to a door when the players can see each other. Would have to think this over a bit. But, then, players could just make their lengths of times in a door an indication of what's up? Or would that constitute more communication after the fact? As people don't, normally, walk and think to at the same rate. Nor do they look at or process things at once. Some of us slow down, or change our minds in midstream. Elegant solutions are for perfect people and worlds.
Incidentally, I listed only the strategy for the player with the green hat when he sees a green car at door 2 on his first try. When door 2 is blue or red, he acts last behind the other players.As in the above strategy, a player can and may go back to the same door, especially if he has already matched his hat there. And, to spread out selections of doors among also the individual players isn't entirely random. What if one of the players can't fathom any strategy, and wanders around a couple of times without settling on any door, or all three doors? Furthermore, who's to say that the players won't randomly pick an overall, better strategy (to look at doors) instead of a different one (in the form of a simple selection) each time? Maybe, one of the players already has a list of lucky numbers. But, there's always a chance that they all independently settle on even the 2/3 strategy by accident to use each time. Even a broken clock is right twice a day. I'm sure that others things could happen too. People are the most unpredictable, at times.Quote: WizardAnd what would be the probability of success of every player picked randomly, if not (2/3)3 = 8/27?
