If the 34 people are indeed in the bedroom, are they there BEFORE you entered? If so, there are 35 after. Are “you” one of the people who was already in the bedroom and part of the 34? If yes, then killing 30 people in the bedroom would leave four plus the killer. As I said, it just isn’t clear.
Better riddle. Three people are lined up in a line so that the person in the back can see the two in front of him, the middle person can only see the one person ahead of him, and the person in front cannot see anybody. They are each made to wear a hat: green or red. They all know that there are three green hats total and two red hats total in the pile of five. They cannot see what color hat they are wearing, but they can see the colors of the hats of the people IN FRONT of them.
The person in the back is asked to guess the color of his hat and he says that he doesn’t know. The person in the middle is asked to guess the color of his hat and he too says that he doesn’t know. The person in front says that he knows the color of his own hat. What color is it and what is the conclusive reasoning used by the person in front to deduce the color of his own hat?
If the 34 people are indeed in the bedroom, are they there BEFORE you entered? If so, there are 35 after. Are “you” one of the people who was already in the bedroom and part of the 34? If yes, then killing 30 people in the bedroom would leave four plus the killer. As I said, it just isn’t clear.
Better riddle. Three people are lined up in a line so that the person in the back can see the two in front of him, the middle person can only see the one person ahead of him, and the person in front cannot see anybody. They are each made to wear a hat: green or red. They all know that there are three green hats total and two red hats total in the pile of five. They cannot see what color hat they are wearing, but they can see the colors of the hats of the people IN FRONT of them.
The person in the back is asked to guess the color of his hat and he says that he doesn’t know. The person in the middle is asked to guess the color of his hat and he too says that he doesn’t know. The person in front says that he knows the color of his own hat. What color is it and what is the conclusive reasoning used by the person in front to deduce the color of his own hat?
Also, on my riddle, they can each hear the answers given by the previous person/people.
If the 34 people are indeed in the bedroom, are they there BEFORE you entered? If so, there are 35 after. Are “you” one of the people who was already in the bedroom and part of the 34? If yes, then killing 30 people in the bedroom would leave four plus the killer. As I said, it just isn’t clear.
Better riddle. Three people are lined up in a line so that the person in the back can see the two in front of him, the middle person can only see the one person ahead of him, and the person in front cannot see anybody. They are each made to wear a hat: green or red. They all know that there are three green hats total and two red hats total in the pile of five. They cannot see what color hat they are wearing, but they can see the colors of the hats of the people IN FRONT of them.
The person in the back is asked to guess the color of his hat and he says that he doesn’t know. The person in the middle is asked to guess the color of his hat and he too says that he doesn’t know. The person in front says that he knows the color of his own hat. What color is it and what is the conclusive reasoning used by the person in front to deduce the color of his own hat?
Since he entered before saying there are 34 I would say there are 34 people. I would say 34 people are in the room. Even if they are dead the dead bodies are still in the room.
Sure. I can see that if the people are in the room in the first place, although this might get into a philosophical debate of what is a “person.” Is a dead body without any life still a person?
If the 34 people are indeed in the bedroom, are they there BEFORE you entered? If so, there are 35 after. Are “you” one of the people who was already in the bedroom and part of the 34? If yes, then killing 30 people in the bedroom would leave four plus the killer. As I said, it just isn’t clear.
Better riddle. Three people are lined up in a line so that the person in the back can see the two in front of him, the middle person can only see the one person ahead of him, and the person in front cannot see anybody. They are each made to wear a hat: green or red. They all know that there are three green hats total and two red hats total in the pile of five. They cannot see what color hat they are wearing, but they can see the colors of the hats of the people IN FRONT of them.
The person in the back is asked to guess the color of his hat and he says that he doesn’t know. The person in the middle is asked to guess the color of his hat and he too says that he doesn’t know. The person in front says that he knows the color of his own hat. What color is it and what is the conclusive reasoning used by the person in front to deduce the color of his own hat?
Well, I’m going to leave it up for a little while to see if anyone has the brains to figure it out. I can assure all of you that there are no silly tricks to my riddle. It’s a legitimate answer based on reasoning.
Well, I’m going to leave it up for a little while to see if anyone has the brains to figure it out. I can assure all of you that there are no silly tricks to my riddle. It’s a legitimate answer based on reasoning.
Better question is the Monte Hall one:
Quote:
The Monty Hall problem is a brain teaser, in the form of a probability puzzle, loosely based on the American television game show Let's Make a Deal and named after its original host, Monty Hall. The problem was originally posed (and solved) in a letter by Steve Selvin to the American Statistician in 1975 (Selvin 1975a), (Selvin 1975b). It became famous as a question from a reader's letter quoted in Marilyn vos Savant's "Ask Marilyn" column in Parade magazine in 1990 (vos Savant 1990a):
Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?
Well, I’m going to leave it up for a little while to see if anyone has the brains to figure it out. I can assure all of you that there are no silly tricks to my riddle. It’s a legitimate answer based on reasoning.
Better question is the Monte Hall one:
Quote:
The Monty Hall problem is a brain teaser, in the form of a probability puzzle, loosely based on the American television game show Let's Make a Deal and named after its original host, Monty Hall. The problem was originally posed (and solved) in a letter by Steve Selvin to the American Statistician in 1975 (Selvin 1975a), (Selvin 1975b). It became famous as a question from a reader's letter quoted in Marilyn vos Savant's "Ask Marilyn" column in Parade magazine in 1990 (vos Savant 1990a):
Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?
Link - ( New Window )
ALWAYS switch. Your initial probability is 1/3 to get the car. The fact that the host, WHO KNOWS WHATS BEHIND THE DOORS, opens a door and shows you a goat doesn’t change your probability from 1/3 because he will always open the door for the goat.
But given that you have NEW INFORMATION, you have a 1/2 probability of getting the car by switching because it’s a 50-50 between the two unopened doors. The key to this puzzle is understanding that the probability on your initial door does NOT change to 1/2 after the host opens the door showing the goat because he will always open a losing door. If the host had no clue and could have revealed a door with the car, then it’s true that it would be a 50-50 either way, But since the host’s decision is made based on inside knowledge, the probability on your initial choice remains 1/3. So 1/2>1/3 means always switch.
Btw, most people would stick with the door they initially chose for the irrational reason that it’s far more painful to switch away from a winning door that to not switch to the winning door. This is true despite the odds being clearly in favor of a switch. Psychology vs. logic.
Sorry. Meant to write that the probably on switching is 2/3, which is greater than 1/3 (not 1/2). Same reasoning for everything else. So always switch.
If the 34 people are indeed in the bedroom, are they there BEFORE you entered? If so, there are 35 after. Are “you” one of the people who was already in the bedroom and part of the 34? If yes, then killing 30 people in the bedroom would leave four plus the killer. As I said, it just isn’t clear.
Better riddle. Three people are lined up in a line so that the person in the back can see the two in front of him, the middle person can only see the one person ahead of him, and the person in front cannot see anybody. They are each made to wear a hat: green or red. They all know that there are three green hats total and two red hats total in the pile of five. They cannot see what color hat they are wearing, but they can see the colors of the hats of the people IN FRONT of them.
The person in the back is asked to guess the color of his hat and he says that he doesn’t know. The person in the middle is asked to guess the color of his hat and he too says that he doesn’t know. The person in front says that he knows the color of his own hat. What color is it and what is the conclusive reasoning used by the person in front to deduce the color of his own hat?
Also, on my riddle, they can each hear the answers given by the previous person/people.
I think the guy in the front knows he has a green hat on.
If the guy in the back saw two red hats, he would know his was green since there were only two red hats. So he must have seen either two green hats, or one red and one green. The guy in the middle knows this, so if he saw a red hat, he would know his was green, but he didn't know either, so he must have seen a green hat.
If the 34 people are indeed in the bedroom, are they there BEFORE you entered? If so, there are 35 after. Are “you” one of the people who was already in the bedroom and part of the 34? If yes, then killing 30 people in the bedroom would leave four plus the killer. As I said, it just isn’t clear.
Better riddle. Three people are lined up in a line so that the person in the back can see the two in front of him, the middle person can only see the one person ahead of him, and the person in front cannot see anybody. They are each made to wear a hat: green or red. They all know that there are three green hats total and two red hats total in the pile of five. They cannot see what color hat they are wearing, but they can see the colors of the hats of the people IN FRONT of them.
The person in the back is asked to guess the color of his hat and he says that he doesn’t know. The person in the middle is asked to guess the color of his hat and he too says that he doesn’t know. The person in front says that he knows the color of his own hat. What color is it and what is the conclusive reasoning used by the person in front to deduce the color of his own hat?
Also, on my riddle, they can each hear the answers given by the previous person/people.
I think the guy in the front knows he has a green hat on.
If the guy in the back saw two red hats, he would know his was green since there were only two red hats. So he must have seen either two green hats, or one red and one green. The guy in the middle knows this, so if he saw a red hat, he would know his was green, but he didn't know either, so he must have seen a green hat.
Sorry. Meant to write that the probably on switching is 2/3, which is greater than 1/3 (not 1/2). Same reasoning for everything else. So always switch.
Correct. It's all about information. If Monte pulled someone off the street not seeing what had unfolded in the game HIS odds would be 50-50. Think of it this way: Say there were 100 doors instead of 3. And Monte opened 98 empty doors. Would one switch then? Of course!
OTOH, with 3 doors, the contestant knew Monte was going to open a door with a goat BEFORE he picked no matter what he picked. So when Monte opened a goat door that was already known information. So what information was really gained?
That what makes this so interesting.
More for Wiki:
Quote:
Vos Savant's response was that the contestant should switch to the other door (vos Savant 1990a). Under the standard assumptions, contestants who switch have a 2/3 chance of winning the car, while contestants who stick to their initial choice have only a 1/3 chance.
The given probabilities depend on specific assumptions about how the host and contestant choose their doors. A key insight is that, under these standard conditions, there is more information about doors 2 and 3 than was available at the beginning of the game when door 1 was chosen by the player: the host's deliberate action adds value to the door he did not choose to eliminate, but not to the one chosen by the contestant originally. Another insight is that switching doors is a different action than choosing between the two remaining doors at random, as the first action uses the previous information and the latter does not. Other possible behaviors than the one described can reveal different additional information, or none at all, and yield different probabilities. Yet another insight is that your chance of winning by switching doors is directly related to your chance of choosing the winning door in the first place: if you choose the correct door on your first try, then switching loses; if you choose a wrong door on your first try, then switching wins; your chance of choosing the correct door on your first try is 1/3, and the chance of choosing a wrong door is 2/3.
Many readers of vos Savant's column refused to believe switching is beneficial despite her explanation. After the problem appeared in Parade, approximately 10,000 readers, including nearly 1,000 with PhDs, wrote to the magazine, most of them claiming vos Savant was wrong (Tierney 1991). Even when given explanations, simulations, and formal mathematical proofs, many people still do not accept that switching is the best strategy (vos Savant 1991a). Paul Erdős, one of the most prolific mathematicians in history, remained unconvinced until he was shown a computer simulation demonstrating vos Savant's predicted result (Vazsonyi 1999).
The problem is a paradox of the veridical type, because the correct choice (that one should switch doors) is so counterintuitive it can seem absurd, but is nevertheless demonstrably true.
See, you assumed the question in the last line was referring to the same bedroom mentioned in the first line, when it was actually referring to the one next door, where Mr. & Mrs. Jones were sleeping.
That's the problem with it being so ambiguous (and not in a good riddle kind of way) - there are plenty of answers that are as equally unsatisfying as the one you were looking for.
Sorry. Meant to write that the probably on switching is 2/3, which is greater than 1/3 (not 1/2). Same reasoning for everything else. So always switch.
Correct. It's all about information. If Monte pulled someone off the street not seeing what had unfolded in the game HIS odds would be 50-50. Think of it this way: Say there were 100 doors instead of 3. And Monte opened 98 empty doors. Would one switch then? Of course!
OTOH, with 3 doors, the contestant knew Monte was going to open a door with a goat BEFORE he picked no matter what he picked. So when Monte opened a goat door that was already known information. So what information was really gained?
That what makes this so interesting.
More for Wiki:
Quote:
Vos Savant's response was that the contestant should switch to the other door (vos Savant 1990a). Under the standard assumptions, contestants who switch have a 2/3 chance of winning the car, while contestants who stick to their initial choice have only a 1/3 chance.
The given probabilities depend on specific assumptions about how the host and contestant choose their doors. A key insight is that, under these standard conditions, there is more information about doors 2 and 3 than was available at the beginning of the game when door 1 was chosen by the player: the host's deliberate action adds value to the door he did not choose to eliminate, but not to the one chosen by the contestant originally. Another insight is that switching doors is a different action than choosing between the two remaining doors at random, as the first action uses the previous information and the latter does not. Other possible behaviors than the one described can reveal different additional information, or none at all, and yield different probabilities. Yet another insight is that your chance of winning by switching doors is directly related to your chance of choosing the winning door in the first place: if you choose the correct door on your first try, then switching loses; if you choose a wrong door on your first try, then switching wins; your chance of choosing the correct door on your first try is 1/3, and the chance of choosing a wrong door is 2/3.
Many readers of vos Savant's column refused to believe switching is beneficial despite her explanation. After the problem appeared in Parade, approximately 10,000 readers, including nearly 1,000 with PhDs, wrote to the magazine, most of them claiming vos Savant was wrong (Tierney 1991). Even when given explanations, simulations, and formal mathematical proofs, many people still do not accept that switching is the best strategy (vos Savant 1991a). Paul Erdős, one of the most prolific mathematicians in history, remained unconvinced until he was shown a computer simulation demonstrating vos Savant's predicted result (Vazsonyi 1999).
The problem is a paradox of the veridical type, because the correct choice (that one should switch doors) is so counterintuitive it can seem absurd, but is nevertheless demonstrably true.
this is going to mess with me all day, because I'm ready to pound the table and say that you're no better off switching in spite of all of the citations you laid out (which I definitely won't go find/read.
So I'm in the fun position of believing strongly that I am right while also KNOWING that I am not right.
I've seen it ask at job interviews. The is an unknown number of each color hat, they could all be red or all be green etc. Make the riddle much harder. The are given a night to come up with a strategy to save as many as possible. The hats will be placed on their hards the following morning. Starting in the back they will be asked what their color it, if wrong they die. They can only say red or green....
I've seen it ask at job interviews. The is an unknown number of each color hat, they could all be red or all be green etc. Make the riddle much harder. The are given a night to come up with a strategy to save as many as possible. The hats will be placed on their hards the following morning. Starting in the back they will be asked what their color it, if wrong they die. They can only say red or green....
??? I didn’t understand at all what your getting at. The riddle I presented isn’t “wrong.” Maybe you know a different version of it, but the one I gave is certainly a valid version of it that can be properly deduced through reason.
because you are assuming that the people in line can piece it out logically, when most would want to stick to door 1, or not understand who is in the bedroom killing people or counting myself or dead people as people.
I thought the person in front was looking into a mirror...
Sorry. Meant to write that the probably on switching is 2/3, which is greater than 1/3 (not 1/2). Same reasoning for everything else. So always switch.
Correct. It's all about information. If Monte pulled someone off the street not seeing what had unfolded in the game HIS odds would be 50-50. Think of it this way: Say there were 100 doors instead of 3. And Monte opened 98 empty doors. Would one switch then? Of course!
OTOH, with 3 doors, the contestant knew Monte was going to open a door with a goat BEFORE he picked no matter what he picked. So when Monte opened a goat door that was already known information. So what information was really gained?
That what makes this so interesting.
More for Wiki:
Quote:
Vos Savant's response was that the contestant should switch to the other door (vos Savant 1990a). Under the standard assumptions, contestants who switch have a 2/3 chance of winning the car, while contestants who stick to their initial choice have only a 1/3 chance.
The given probabilities depend on specific assumptions about how the host and contestant choose their doors. A key insight is that, under these standard conditions, there is more information about doors 2 and 3 than was available at the beginning of the game when door 1 was chosen by the player: the host's deliberate action adds value to the door he did not choose to eliminate, but not to the one chosen by the contestant originally. Another insight is that switching doors is a different action than choosing between the two remaining doors at random, as the first action uses the previous information and the latter does not. Other possible behaviors than the one described can reveal different additional information, or none at all, and yield different probabilities. Yet another insight is that your chance of winning by switching doors is directly related to your chance of choosing the winning door in the first place: if you choose the correct door on your first try, then switching loses; if you choose a wrong door on your first try, then switching wins; your chance of choosing the correct door on your first try is 1/3, and the chance of choosing a wrong door is 2/3.
Many readers of vos Savant's column refused to believe switching is beneficial despite her explanation. After the problem appeared in Parade, approximately 10,000 readers, including nearly 1,000 with PhDs, wrote to the magazine, most of them claiming vos Savant was wrong (Tierney 1991). Even when given explanations, simulations, and formal mathematical proofs, many people still do not accept that switching is the best strategy (vos Savant 1991a). Paul Erdős, one of the most prolific mathematicians in history, remained unconvinced until he was shown a computer simulation demonstrating vos Savant's predicted result (Vazsonyi 1999).
The problem is a paradox of the veridical type, because the correct choice (that one should switch doors) is so counterintuitive it can seem absurd, but is nevertheless demonstrably true.
this is going to mess with me all day, because I'm ready to pound the table and say that you're no better off switching in spite of all of the citations you laid out (which I definitely won't go find/read.
So I'm in the fun position of believing strongly that I am right while also KNOWING that I am not right.
DRATS.
Bigbluehoya, the easiest way to demonstrate this is with a deck of cards.
The key to this is that Monty Hall KNOWS which door has the prize and so he is intentionally showing you a door that has nothing . He will always be able to do this because at least 1 of the 2 doors will always have nothing.
But again, try it with a deck of cards and someone else. Have them pick a single card form the deck and not look at it, leave it face down. Off the In this case, the prize is the Ace of Spades. The person's goal is the pick the Ace of Spades.
When they do this, they will have about a 2% of picking the Ace of spades.
Scenario A: they pick the Ace of spades, you have 51 "junk" cards (2%)
Scenario B: they pick a junk card, you have 50 junk cards and the Ace of Spacdes. (98%)
Now, you take the remaining stack of 51 cards, look through them all, and remove 50 of those 51 cards that aren't the Ace of Spades. Show person A all of those junk cards and then separate them from the pile, leaving one card in front of you face down.
Ask that person if they want to switch.
The catch is that the person thinks that because you removed 50 cards that aren't the Ace of Spades, that their odds went up that they picked the Ace of Spades.
But the reality is that whether that person picked the AoS or not, you will always be able to show that 50 of your cards are NOT the AoS. And because you are looking through the cards beforehand, this is not at random. So the person who picked the single card never increases his odds from 2% that he picked the Ace of Spades. Since you are selecting the 50 junk cards to remove, the remaining card you have still represents the other 51 cards. Because again, whether the person picks the Ace of Spades or not, you are able to show him 50 junk cards.
This is different from a game like Deal or no Deal where the suitcases are removed blindly.
So go try it. You don't even need another person. Take one card from a deck and leave it face down. See how many times that pile of 1 card is the Ace of Spades vs. the pile of 51 cards remaining.
Once that starts to make sense with a larger number like 51 cards instead of 3 doors, the same logic applies to 3 doors. Monty Hall will always have a junk door to show the contestant, and so long as he knows beforehand which doors are junk, it's all just smoke and mirrors to trick the contestant.
If the 34 people are indeed in the bedroom, are they there BEFORE you entered? If so, there are 35 after. Are “you” one of the people who was already in the bedroom and part of the 34? If yes, then killing 30 people in the bedroom would leave four plus the killer. As I said, it just isn’t clear.
Better riddle. Three people are lined up in a line so that the person in the back can see the two in front of him, the middle person can only see the one person ahead of him, and the person in front cannot see anybody. They are each made to wear a hat: green or red. They all know that there are three green hats total and two red hats total in the pile of five. They cannot see what color hat they are wearing, but they can see the colors of the hats of the people IN FRONT of them.
The person in the back is asked to guess the color of his hat and he says that he doesn’t know. The person in the middle is asked to guess the color of his hat and he too says that he doesn’t know. The person in front says that he knows the color of his own hat. What color is it and what is the conclusive reasoning used by the person in front to deduce the color of his own hat?
34 is my guess for the first one but not sure.
Didn’t look at any comment after this one but think I have it ...
If guy in back sees 2 reds he knows he has green. If he sees 2 green or 1 and 1 he doesn’t know. So middle guy can assume it’s NOT 2 red since back guy doesn’t know.
If next guy assumes not 2 red it’ must be 1 and 1 or 2 green. If he sees 1 red he knows he has green. If he sees green he wouldn’t know. So he must see green. Person in front can assume this.
Better riddle. Three people are lined up in a line so that the person in the back can see the two in front of him, the middle person can only see the one person ahead of him, and the person in front cannot see anybody. They are each made to wear a hat: green or red. They all know that there are three green hats total and two red hats total in the pile of five. They cannot see what color hat they are wearing, but they can see the colors of the hats of the people IN FRONT of them.
The person in the back is asked to guess the color of his hat and he says that he doesn’t know. The person in the middle is asked to guess the color of his hat and he too says that he doesn’t know. The person in front says that he knows the color of his own hat. What color is it and what is the conclusive reasoning used by the person in front to deduce the color of his own hat?
Better riddle. Three people are lined up in a line so that the person in the back can see the two in front of him, the middle person can only see the one person ahead of him, and the person in front cannot see anybody. They are each made to wear a hat: green or red. They all know that there are three green hats total and two red hats total in the pile of five. They cannot see what color hat they are wearing, but they can see the colors of the hats of the people IN FRONT of them.
The person in the back is asked to guess the color of his hat and he says that he doesn’t know. The person in the middle is asked to guess the color of his hat and he too says that he doesn’t know. The person in front says that he knows the color of his own hat. What color is it and what is the conclusive reasoning used by the person in front to deduce the color of his own hat?
Also, on my riddle, they can each hear the answers given by the previous person/people.
Better riddle. Three people are lined up in a line so that the person in the back can see the two in front of him, the middle person can only see the one person ahead of him, and the person in front cannot see anybody. They are each made to wear a hat: green or red. They all know that there are three green hats total and two red hats total in the pile of five. They cannot see what color hat they are wearing, but they can see the colors of the hats of the people IN FRONT of them.
The person in the back is asked to guess the color of his hat and he says that he doesn’t know. The person in the middle is asked to guess the color of his hat and he too says that he doesn’t know. The person in front says that he knows the color of his own hat. What color is it and what is the conclusive reasoning used by the person in front to deduce the color of his own hat?
Since he entered before saying there are 34 I would say there are 34 people. I would say 34 people are in the room. Even if they are dead the dead bodies are still in the room.
Sure. I can see that if the people are in the room in the first place, although this might get into a philosophical debate of what is a “person.” Is a dead body without any life still a person?
Better riddle. Three people are lined up in a line so that the person in the back can see the two in front of him, the middle person can only see the one person ahead of him, and the person in front cannot see anybody. They are each made to wear a hat: green or red. They all know that there are three green hats total and two red hats total in the pile of five. They cannot see what color hat they are wearing, but they can see the colors of the hats of the people IN FRONT of them.
The person in the back is asked to guess the color of his hat and he says that he doesn’t know. The person in the middle is asked to guess the color of his hat and he too says that he doesn’t know. The person in front says that he knows the color of his own hat. What color is it and what is the conclusive reasoning used by the person in front to deduce the color of his own hat?
WAY to complicated. Mine was simple.
Better question is the Monte Hall one:
Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?
Link - ( New Window )
Quote:
Well, I’m going to leave it up for a little while to see if anyone has the brains to figure it out. I can assure all of you that there are no silly tricks to my riddle. It’s a legitimate answer based on reasoning.
Better question is the Monte Hall one:
Quote:
The Monty Hall problem is a brain teaser, in the form of a probability puzzle, loosely based on the American television game show Let's Make a Deal and named after its original host, Monty Hall. The problem was originally posed (and solved) in a letter by Steve Selvin to the American Statistician in 1975 (Selvin 1975a), (Selvin 1975b). It became famous as a question from a reader's letter quoted in Marilyn vos Savant's "Ask Marilyn" column in Parade magazine in 1990 (vos Savant 1990a):
Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?
Link - ( New Window )
ALWAYS switch. Your initial probability is 1/3 to get the car. The fact that the host, WHO KNOWS WHATS BEHIND THE DOORS, opens a door and shows you a goat doesn’t change your probability from 1/3 because he will always open the door for the goat.
But given that you have NEW INFORMATION, you have a 1/2 probability of getting the car by switching because it’s a 50-50 between the two unopened doors. The key to this puzzle is understanding that the probability on your initial door does NOT change to 1/2 after the host opens the door showing the goat because he will always open a losing door. If the host had no clue and could have revealed a door with the car, then it’s true that it would be a 50-50 either way, But since the host’s decision is made based on inside knowledge, the probability on your initial choice remains 1/3. So 1/2>1/3 means always switch.
Btw, most people would stick with the door they initially chose for the irrational reason that it’s far more painful to switch away from a winning door that to not switch to the winning door. This is true despite the odds being clearly in favor of a switch. Psychology vs. logic.
I never fucked anybody over in my life who didn’t have it coming to them.
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If the 34 people are indeed in the bedroom, are they there BEFORE you entered? If so, there are 35 after. Are “you” one of the people who was already in the bedroom and part of the 34? If yes, then killing 30 people in the bedroom would leave four plus the killer. As I said, it just isn’t clear.
Better riddle. Three people are lined up in a line so that the person in the back can see the two in front of him, the middle person can only see the one person ahead of him, and the person in front cannot see anybody. They are each made to wear a hat: green or red. They all know that there are three green hats total and two red hats total in the pile of five. They cannot see what color hat they are wearing, but they can see the colors of the hats of the people IN FRONT of them.
The person in the back is asked to guess the color of his hat and he says that he doesn’t know. The person in the middle is asked to guess the color of his hat and he too says that he doesn’t know. The person in front says that he knows the color of his own hat. What color is it and what is the conclusive reasoning used by the person in front to deduce the color of his own hat?
Also, on my riddle, they can each hear the answers given by the previous person/people.
I think the guy in the front knows he has a green hat on.
If the guy in the back saw two red hats, he would know his was green since there were only two red hats. So he must have seen either two green hats, or one red and one green. The guy in the middle knows this, so if he saw a red hat, he would know his was green, but he didn't know either, so he must have seen a green hat.
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In comment 14949241 Dukie Dimes said:
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If the 34 people are indeed in the bedroom, are they there BEFORE you entered? If so, there are 35 after. Are “you” one of the people who was already in the bedroom and part of the 34? If yes, then killing 30 people in the bedroom would leave four plus the killer. As I said, it just isn’t clear.
Better riddle. Three people are lined up in a line so that the person in the back can see the two in front of him, the middle person can only see the one person ahead of him, and the person in front cannot see anybody. They are each made to wear a hat: green or red. They all know that there are three green hats total and two red hats total in the pile of five. They cannot see what color hat they are wearing, but they can see the colors of the hats of the people IN FRONT of them.
The person in the back is asked to guess the color of his hat and he says that he doesn’t know. The person in the middle is asked to guess the color of his hat and he too says that he doesn’t know. The person in front says that he knows the color of his own hat. What color is it and what is the conclusive reasoning used by the person in front to deduce the color of his own hat?
Also, on my riddle, they can each hear the answers given by the previous person/people.
I think the guy in the front knows he has a green hat on.
If the guy in the back saw two red hats, he would know his was green since there were only two red hats. So he must have seen either two green hats, or one red and one green. The guy in the middle knows this, so if he saw a red hat, he would know his was green, but he didn't know either, so he must have seen a green hat.
Well done.
Correct. It's all about information. If Monte pulled someone off the street not seeing what had unfolded in the game HIS odds would be 50-50. Think of it this way: Say there were 100 doors instead of 3. And Monte opened 98 empty doors. Would one switch then? Of course!
OTOH, with 3 doors, the contestant knew Monte was going to open a door with a goat BEFORE he picked no matter what he picked. So when Monte opened a goat door that was already known information. So what information was really gained?
That what makes this so interesting.
More for Wiki:
The given probabilities depend on specific assumptions about how the host and contestant choose their doors. A key insight is that, under these standard conditions, there is more information about doors 2 and 3 than was available at the beginning of the game when door 1 was chosen by the player: the host's deliberate action adds value to the door he did not choose to eliminate, but not to the one chosen by the contestant originally. Another insight is that switching doors is a different action than choosing between the two remaining doors at random, as the first action uses the previous information and the latter does not. Other possible behaviors than the one described can reveal different additional information, or none at all, and yield different probabilities. Yet another insight is that your chance of winning by switching doors is directly related to your chance of choosing the winning door in the first place: if you choose the correct door on your first try, then switching loses; if you choose a wrong door on your first try, then switching wins; your chance of choosing the correct door on your first try is 1/3, and the chance of choosing a wrong door is 2/3.
Many readers of vos Savant's column refused to believe switching is beneficial despite her explanation. After the problem appeared in Parade, approximately 10,000 readers, including nearly 1,000 with PhDs, wrote to the magazine, most of them claiming vos Savant was wrong (Tierney 1991). Even when given explanations, simulations, and formal mathematical proofs, many people still do not accept that switching is the best strategy (vos Savant 1991a). Paul Erdős, one of the most prolific mathematicians in history, remained unconvinced until he was shown a computer simulation demonstrating vos Savant's predicted result (Vazsonyi 1999).
The problem is a paradox of the veridical type, because the correct choice (that one should switch doors) is so counterintuitive it can seem absurd, but is nevertheless demonstrably true.
Unless you are saying dead people are still people.
"You kill 30" - there is no indication that the people killed nor the one who did the killing are in the bedroom
"I enter the bedroom" - there is only 1 person who we are certain entered the bedroom
18>15.
"You kill 30" - there is no indication that the people killed nor the one who did the killing are in the bedroom
"I enter the bedroom" - there is only 1 person who we are certain entered the bedroom
Yes, Exact explanation. Credit to others who said 1.
I liked the simplicity of the question.
I didn’t look at the answer yet or scroll all the way down. I’ll say 3.
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"there are 34 people" -you never say those 34 people are also in the bedroom.
"You kill 30" - there is no indication that the people killed nor the one who did the killing are in the bedroom
"I enter the bedroom" - there is only 1 person who we are certain entered the bedroom
Yes, Exact explanation. Credit to others who said 1.
I liked the simplicity of the question.
Is this Momento?
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Got it wrong.
I didn’t look at the answer yet or scroll all the way down. I’ll say 3.
Most people 'from where I found this' not BBI. #Reading comprehension.
That's the problem with it being so ambiguous (and not in a good riddle kind of way) - there are plenty of answers that are as equally unsatisfying as the one you were looking for.
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Sorry. Meant to write that the probably on switching is 2/3, which is greater than 1/3 (not 1/2). Same reasoning for everything else. So always switch.
Correct. It's all about information. If Monte pulled someone off the street not seeing what had unfolded in the game HIS odds would be 50-50. Think of it this way: Say there were 100 doors instead of 3. And Monte opened 98 empty doors. Would one switch then? Of course!
OTOH, with 3 doors, the contestant knew Monte was going to open a door with a goat BEFORE he picked no matter what he picked. So when Monte opened a goat door that was already known information. So what information was really gained?
That what makes this so interesting.
More for Wiki:
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Vos Savant's response was that the contestant should switch to the other door (vos Savant 1990a). Under the standard assumptions, contestants who switch have a 2/3 chance of winning the car, while contestants who stick to their initial choice have only a 1/3 chance.
The given probabilities depend on specific assumptions about how the host and contestant choose their doors. A key insight is that, under these standard conditions, there is more information about doors 2 and 3 than was available at the beginning of the game when door 1 was chosen by the player: the host's deliberate action adds value to the door he did not choose to eliminate, but not to the one chosen by the contestant originally. Another insight is that switching doors is a different action than choosing between the two remaining doors at random, as the first action uses the previous information and the latter does not. Other possible behaviors than the one described can reveal different additional information, or none at all, and yield different probabilities. Yet another insight is that your chance of winning by switching doors is directly related to your chance of choosing the winning door in the first place: if you choose the correct door on your first try, then switching loses; if you choose a wrong door on your first try, then switching wins; your chance of choosing the correct door on your first try is 1/3, and the chance of choosing a wrong door is 2/3.
Many readers of vos Savant's column refused to believe switching is beneficial despite her explanation. After the problem appeared in Parade, approximately 10,000 readers, including nearly 1,000 with PhDs, wrote to the magazine, most of them claiming vos Savant was wrong (Tierney 1991). Even when given explanations, simulations, and formal mathematical proofs, many people still do not accept that switching is the best strategy (vos Savant 1991a). Paul Erdős, one of the most prolific mathematicians in history, remained unconvinced until he was shown a computer simulation demonstrating vos Savant's predicted result (Vazsonyi 1999).
The problem is a paradox of the veridical type, because the correct choice (that one should switch doors) is so counterintuitive it can seem absurd, but is nevertheless demonstrably true.
this is going to mess with me all day, because I'm ready to pound the table and say that you're no better off switching in spite of all of the citations you laid out (which I definitely won't go find/read.
So I'm in the fun position of believing strongly that I am right while also KNOWING that I am not right.
DRATS.
??? I didn’t understand at all what your getting at. The riddle I presented isn’t “wrong.” Maybe you know a different version of it, but the one I gave is certainly a valid version of it that can be properly deduced through reason.
Just wondering...
I thought the person in front was looking into a mirror...
Just wondering...
Yup. Read above.
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In comment 14949269 Dukie Dimes said:
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Sorry. Meant to write that the probably on switching is 2/3, which is greater than 1/3 (not 1/2). Same reasoning for everything else. So always switch.
Correct. It's all about information. If Monte pulled someone off the street not seeing what had unfolded in the game HIS odds would be 50-50. Think of it this way: Say there were 100 doors instead of 3. And Monte opened 98 empty doors. Would one switch then? Of course!
OTOH, with 3 doors, the contestant knew Monte was going to open a door with a goat BEFORE he picked no matter what he picked. So when Monte opened a goat door that was already known information. So what information was really gained?
That what makes this so interesting.
More for Wiki:
Quote:
Vos Savant's response was that the contestant should switch to the other door (vos Savant 1990a). Under the standard assumptions, contestants who switch have a 2/3 chance of winning the car, while contestants who stick to their initial choice have only a 1/3 chance.
The given probabilities depend on specific assumptions about how the host and contestant choose their doors. A key insight is that, under these standard conditions, there is more information about doors 2 and 3 than was available at the beginning of the game when door 1 was chosen by the player: the host's deliberate action adds value to the door he did not choose to eliminate, but not to the one chosen by the contestant originally. Another insight is that switching doors is a different action than choosing between the two remaining doors at random, as the first action uses the previous information and the latter does not. Other possible behaviors than the one described can reveal different additional information, or none at all, and yield different probabilities. Yet another insight is that your chance of winning by switching doors is directly related to your chance of choosing the winning door in the first place: if you choose the correct door on your first try, then switching loses; if you choose a wrong door on your first try, then switching wins; your chance of choosing the correct door on your first try is 1/3, and the chance of choosing a wrong door is 2/3.
Many readers of vos Savant's column refused to believe switching is beneficial despite her explanation. After the problem appeared in Parade, approximately 10,000 readers, including nearly 1,000 with PhDs, wrote to the magazine, most of them claiming vos Savant was wrong (Tierney 1991). Even when given explanations, simulations, and formal mathematical proofs, many people still do not accept that switching is the best strategy (vos Savant 1991a). Paul Erdős, one of the most prolific mathematicians in history, remained unconvinced until he was shown a computer simulation demonstrating vos Savant's predicted result (Vazsonyi 1999).
The problem is a paradox of the veridical type, because the correct choice (that one should switch doors) is so counterintuitive it can seem absurd, but is nevertheless demonstrably true.
this is going to mess with me all day, because I'm ready to pound the table and say that you're no better off switching in spite of all of the citations you laid out (which I definitely won't go find/read.
So I'm in the fun position of believing strongly that I am right while also KNOWING that I am not right.
DRATS.
Bigbluehoya, the easiest way to demonstrate this is with a deck of cards.
The key to this is that Monty Hall KNOWS which door has the prize and so he is intentionally showing you a door that has nothing . He will always be able to do this because at least 1 of the 2 doors will always have nothing.
But again, try it with a deck of cards and someone else. Have them pick a single card form the deck and not look at it, leave it face down. Off the In this case, the prize is the Ace of Spades. The person's goal is the pick the Ace of Spades.
When they do this, they will have about a 2% of picking the Ace of spades.
Scenario A: they pick the Ace of spades, you have 51 "junk" cards (2%)
Scenario B: they pick a junk card, you have 50 junk cards and the Ace of Spacdes. (98%)
Now, you take the remaining stack of 51 cards, look through them all, and remove 50 of those 51 cards that aren't the Ace of Spades. Show person A all of those junk cards and then separate them from the pile, leaving one card in front of you face down.
Ask that person if they want to switch.
The catch is that the person thinks that because you removed 50 cards that aren't the Ace of Spades, that their odds went up that they picked the Ace of Spades.
But the reality is that whether that person picked the AoS or not, you will always be able to show that 50 of your cards are NOT the AoS. And because you are looking through the cards beforehand, this is not at random. So the person who picked the single card never increases his odds from 2% that he picked the Ace of Spades. Since you are selecting the 50 junk cards to remove, the remaining card you have still represents the other 51 cards. Because again, whether the person picks the Ace of Spades or not, you are able to show him 50 junk cards.
This is different from a game like Deal or no Deal where the suitcases are removed blindly.
So go try it. You don't even need another person. Take one card from a deck and leave it face down. See how many times that pile of 1 card is the Ace of Spades vs. the pile of 51 cards remaining.
Once that starts to make sense with a larger number like 51 cards instead of 3 doors, the same logic applies to 3 doors. Monty Hall will always have a junk door to show the contestant, and so long as he knows beforehand which doors are junk, it's all just smoke and mirrors to trick the contestant.
1)Car Goat-a Goat-b
2)Car Goat-a Goat-b
3)Car Goat-a Goat-b
3 possible outcomes:
In #1 you pick car, Monte picks goat-a, you switch and get goat-b.
In #2 you pick goat-a, Monte picks goat-b, you switch and get car.
In #3 you pick goat-b, Monte picks goat-a, you switch and get car.
2/3 chance of getting car when switching.
Better riddle. Three people are lined up in a line so that the person in the back can see the two in front of him, the middle person can only see the one person ahead of him, and the person in front cannot see anybody. They are each made to wear a hat: green or red. They all know that there are three green hats total and two red hats total in the pile of five. They cannot see what color hat they are wearing, but they can see the colors of the hats of the people IN FRONT of them.
The person in the back is asked to guess the color of his hat and he says that he doesn’t know. The person in the middle is asked to guess the color of his hat and he too says that he doesn’t know. The person in front says that he knows the color of his own hat. What color is it and what is the conclusive reasoning used by the person in front to deduce the color of his own hat?
34 is my guess for the first one but not sure.
Didn’t look at any comment after this one but think I have it ...
If guy in back sees 2 reds he knows he has green. If he sees 2 green or 1 and 1 he doesn’t know. So middle guy can assume it’s NOT 2 red since back guy doesn’t know.
If next guy assumes not 2 red it’ must be 1 and 1 or 2 green. If he sees 1 red he knows he has green. If he sees green he wouldn’t know. So he must see green. Person in front can assume this.
Therefore person in front is wearing green.