Thread about an unsolvable riddle? August 20, 2014 10:29 AM Subscribe
I can't seem to locate a thread I read here once about an impossible riddle or logic puzzle. I don't recall the riddle itself except that it was math-related, and the answers were multiple choice series of percentages: 50%, 25%, 10%, etc. The ensuing metafilter meltdown over the answer was highly entertaining, but I foolishly didn't mark it as a favorite, and now my searches are coming up empty. Can anyone help?
If it's the one I remember, the question was something like "What is the probability of answering this question correctly?". It creates a feedback between what you choose and what the right answer is. I'll look around for it.
posted by benito.strauss at 10:40 AM on August 20, 2014 [2 favorites]
posted by benito.strauss at 10:40 AM on August 20, 2014 [2 favorites]
Sniped. Grrrrrr.
posted by benito.strauss at 10:41 AM on August 20, 2014 [1 favorite]
posted by benito.strauss at 10:41 AM on August 20, 2014 [1 favorite]
Benito.strauss and zamboni have it! And thanks to Cortex, I have two other new threads for lunch break reading. Thanks so much!
posted by backwards compatible at 10:44 AM on August 20, 2014 [1 favorite]
posted by backwards compatible at 10:44 AM on August 20, 2014 [1 favorite]
5 minutes ago I would have said the chance of me taking some ibuprofen was 50%. Now it's definitely 100%.
posted by Marie Mon Dieu at 11:15 AM on August 20, 2014 [3 favorites]
posted by Marie Mon Dieu at 11:15 AM on August 20, 2014 [3 favorites]
If enough people think about this we end up destroying the Matrix so keep circulating those tapes!
posted by Potomac Avenue at 12:51 PM on August 20, 2014 [2 favorites]
posted by Potomac Avenue at 12:51 PM on August 20, 2014 [2 favorites]
I love this problem, because the answer is so obviously "those were the times that I carried you."
posted by gauche at 12:58 PM on August 20, 2014 [17 favorites]
posted by gauche at 12:58 PM on August 20, 2014 [17 favorites]
The ultimate answer to any and every question is it depends on what you mean by [question].
For example, the idea that we should rely on the lettered options provided is not indicated by the question itself. In their absence, our answer must be...
...
...
...
Nevermind.
The abyss just gazed into me, and now I have a headache.
posted by The Confessor at 1:24 PM on August 20, 2014 [4 favorites]
For example, the idea that we should rely on the lettered options provided is not indicated by the question itself. In their absence, our answer must be...
...
...
...
Nevermind.
The abyss just gazed into me, and now I have a headache.
posted by The Confessor at 1:24 PM on August 20, 2014 [4 favorites]
I think the question is improved if c) 60 is deleted and d) none of the above is added, making the choices a) 25 b) 50 c) 25 d) none of the above.
As originally posed, I can buy into the "0% theory", since whichever of the enumerated answers you choose leads to an inconsistent situation. But with "none of the above" as an alternative, even the "0% theory" doesn't liberate you from choosing an answer which isn't present to give some claim of a consistent system.
posted by jepler at 2:45 PM on August 20, 2014 [2 favorites]
As originally posed, I can buy into the "0% theory", since whichever of the enumerated answers you choose leads to an inconsistent situation. But with "none of the above" as an alternative, even the "0% theory" doesn't liberate you from choosing an answer which isn't present to give some claim of a consistent system.
posted by jepler at 2:45 PM on August 20, 2014 [2 favorites]
> I love this problem, because the answer is so obviously "those were the times that I carried you."
Well, it's a math problem, so it's more likely the answer is "those were the time I carried two".
posted by benito.strauss at 3:30 PM on August 20, 2014 [14 favorites]
Well, it's a math problem, so it's more likely the answer is "those were the time I carried two".
posted by benito.strauss at 3:30 PM on August 20, 2014 [14 favorites]
I think the question is improved if c) 60 is deleted and d) none of the above is added, making the choices a) 25 b) 50 c) 25 d) none of the above.
But if you mean that "none of the above" would be right, then the answer is 25%. But then A and C are also right, so the answer is 75%. But if the answer is 75%, then A, B, and C are wrong, so the answer is "none of the above." But then the answer is 25% ...
posted by John Cohen at 7:35 PM on August 20, 2014 [1 favorite]
But if you mean that "none of the above" would be right, then the answer is 25%. But then A and C are also right, so the answer is 75%. But if the answer is 75%, then A, B, and C are wrong, so the answer is "none of the above." But then the answer is 25% ...
posted by John Cohen at 7:35 PM on August 20, 2014 [1 favorite]
If you pick 50% or 60% then the answer is 25%. If you pick 25% then the answer is 50%.
So it's impossible to pick a right answer.
posted by Chocolate Pickle at 7:40 PM on August 20, 2014
So it's impossible to pick a right answer.
posted by Chocolate Pickle at 7:40 PM on August 20, 2014
Jepler is saying that "none of the above" itself includes "0%", which would make it the correct answer, except then if 1/4 of the answers is correct then 25% of random guesses will get it, et cetera.
posted by teremala at 8:05 PM on August 20, 2014
posted by teremala at 8:05 PM on August 20, 2014
Jepler is saying that "none of the above" itself includes "0%", which would make it the correct answer, except then if 1/4 of the answers is correct then 25% of random guesses will get it, et cetera.
That's why I'm saying that changing "60%" to "none of the above" makes it no less paradoxical.
posted by John Cohen at 9:55 PM on August 20, 2014
That's why I'm saying that changing "60%" to "none of the above" makes it no less paradoxical.
posted by John Cohen at 9:55 PM on August 20, 2014
That's why I'm saying that changing "60%" to "none of the above" makes it no less paradoxical.
The point is that it currently isn't paradoxical, the answer simply isn't a-d, it's secret option e, 0%. If 60% is replaced with 0% then there is now no correct answer.
posted by Cannon Fodder at 12:25 AM on August 21, 2014 [1 favorite]
The point is that it currently isn't paradoxical, the answer simply isn't a-d, it's secret option e, 0%. If 60% is replaced with 0% then there is now no correct answer.
posted by Cannon Fodder at 12:25 AM on August 21, 2014 [1 favorite]
I'm not quite convinced, jepler. Once you answer "none of the above", you're home free. You've answered the question "what is the chance you will be correct?" And you close the book.
"But if the answer is D, then the answer is really 25%, and that leads to an inconsistency!"
Nope, you don't have to go there. Once you've answered "none of the above", you're done. It's not an affirmative chance statement that affects the other answers.
posted by naju at 1:30 AM on August 21, 2014
"But if the answer is D, then the answer is really 25%, and that leads to an inconsistency!"
Nope, you don't have to go there. Once you've answered "none of the above", you're done. It's not an affirmative chance statement that affects the other answers.
posted by naju at 1:30 AM on August 21, 2014
(To be clear, I think it ultimately changes the question into a philosophical one on language and logic, i.e. "can something be an answer, a non-answer and a denial of the possibility of an answer at the same time?" It can. It's just not satisfying, but we don't need satisfying. Talk to me further and I'll explain why the impossibility of God is actually proof of God, in between massive bong rips.)
posted by naju at 1:53 AM on August 21, 2014
posted by naju at 1:53 AM on August 21, 2014
The World Famous: "The abyss just gazed into me, and now I have a headache.
I haven't given a moment's thought to The Simpsons in longer than I can recall, but that's got to be a Ralph Wiggum quote, right?"
I couldn't tell if item was making another referential joke that I'm not familiar with, or if he actually thought the "that's where I'm a viking!" thing originated on mefi...
posted by Grither at 4:27 AM on August 21, 2014
I haven't given a moment's thought to The Simpsons in longer than I can recall, but that's got to be a Ralph Wiggum quote, right?"
I couldn't tell if item was making another referential joke that I'm not familiar with, or if he actually thought the "that's where I'm a viking!" thing originated on mefi...
posted by Grither at 4:27 AM on August 21, 2014
The correct answer is
a) None of the above.
posted by Obscure Reference at 5:49 AM on August 21, 2014 [1 favorite]
a) None of the above.
posted by Obscure Reference at 5:49 AM on August 21, 2014 [1 favorite]
I love this problem, because the answer is so obviously "those were the times that I carried you."
Friend of mine had the "Footprints" poster on his way, but he'd used Photoshop wizardry to change it, so his read, "But Lord, at my worst times I looked down and there were only one set of footprints in the sand," and the Lord said, "Yeah, those were the times when you were a real downer to be around and an asshole and I can't abide assholes." He had it on his wall for years and says I was the only one to ever notice. Everyone else just read the first line and knew what it was and moved on.
posted by cjorgensen at 6:26 AM on August 21, 2014 [6 favorites]
Friend of mine had the "Footprints" poster on his way, but he'd used Photoshop wizardry to change it, so his read, "But Lord, at my worst times I looked down and there were only one set of footprints in the sand," and the Lord said, "Yeah, those were the times when you were a real downer to be around and an asshole and I can't abide assholes." He had it on his wall for years and says I was the only one to ever notice. Everyone else just read the first line and knew what it was and moved on.
posted by cjorgensen at 6:26 AM on August 21, 2014 [6 favorites]
The point is that it currently isn't paradoxical, the answer simply isn't a-d, it's secret option e, 0%. If 60% is replaced with 0% then there is now no correct answer.
Sure it's paradoxical. The implication is that there's a right answer. I know: there just is no right answer. But if you don't think that's a paradox, I'd suggest looking into the meaning of the word "paradox." It might be subtler than you think.
posted by John Cohen at 6:30 AM on August 21, 2014
Sure it's paradoxical. The implication is that there's a right answer. I know: there just is no right answer. But if you don't think that's a paradox, I'd suggest looking into the meaning of the word "paradox." It might be subtler than you think.
posted by John Cohen at 6:30 AM on August 21, 2014
(If you interpret the word "paradox" narrowly and strictly enough — two things that are true but mutually incompatible — there's no such thing as a paradox!)
posted by John Cohen at 6:31 AM on August 21, 2014
posted by John Cohen at 6:31 AM on August 21, 2014
Nope, just a reference to Nietzsche, and a rephrase of Marie Mon Dieu's ibuprofen comment.
posted by The Confessor at 7:30 AM on August 21, 2014
posted by The Confessor at 7:30 AM on August 21, 2014
Well, it's a math problem, so it's more likely the answer is "those were the time I carried two".
I wish I could favorite that twice, benito.strauss. That is the best pun I have heard in an indeterminate period of time.
posted by maryr at 8:03 AM on August 21, 2014 [2 favorites]
I wish I could favorite that twice, benito.strauss. That is the best pun I have heard in an indeterminate period of time.
posted by maryr at 8:03 AM on August 21, 2014 [2 favorites]
We've quips and quibbles heard in flocks, but none to beat this paradox!
posted by Kabanos at 9:28 AM on August 21, 2014
posted by Kabanos at 9:28 AM on August 21, 2014
cjorgensen, the best detournement of the Footprints poster I've even seen has Christ respond "Oh, that's when I was surfing."
If your friend's poster looks something like this, they could also photoshop in a small image on the distant waves, long hair an robe flapping in the wind.
posted by benito.strauss at 10:19 AM on August 21, 2014 [2 favorites]
If your friend's poster looks something like this, they could also photoshop in a small image on the distant waves, long hair an robe flapping in the wind.
posted by benito.strauss at 10:19 AM on August 21, 2014 [2 favorites]
If you interpret the word "paradox" narrowly and strictly enough — two things that are true but mutually incompatible — there's no such thing as a paradox!
Israel has a right to defend itself.
Palestine has a right to exist free of occupation by a foreign power.
posted by flabdablet at 10:22 AM on August 21, 2014
Israel has a right to defend itself.
Palestine has a right to exist free of occupation by a foreign power.
posted by flabdablet at 10:22 AM on August 21, 2014
Something something Godwin something
posted by flabdablet at 10:24 AM on August 21, 2014
posted by flabdablet at 10:24 AM on August 21, 2014
Chocolate Pickle: "If you pick 50% or 60% then the answer is 25%. If you pick 25% then the answer is 50%.
So it's impossible to pick a right answer."
You're not picking 25 percent, though. You're picking A or C. Don't confuse the key for the value.
posted by boo_radley at 10:58 AM on August 21, 2014 [3 favorites]
So it's impossible to pick a right answer."
You're not picking 25 percent, though. You're picking A or C. Don't confuse the key for the value.
posted by boo_radley at 10:58 AM on August 21, 2014 [3 favorites]
The thing you have to remember about "Footprints" is that we are not the intended audience. It was written for the sort of Christian so married to their faith that they cannot even conceive of the reality that the circumstances of our lives are the result of human action or inaction upon an environment that is essentially random for its complexity.
I was on the verge of calling it a "new parable" when it hit me: most of the parables of Jesus have a moral lesson that applies even beyond the context of Christian faith.
The sole purpose of "Footprints" is to say "Hey, you! Struggling with your faith? Had a few recent setbacks? Here's a facile fantasy to help you square your unfortunate experiences with the existence of a God worthy of worship."
posted by The Confessor at 10:58 AM on August 21, 2014 [2 favorites]
I was on the verge of calling it a "new parable" when it hit me: most of the parables of Jesus have a moral lesson that applies even beyond the context of Christian faith.
The sole purpose of "Footprints" is to say "Hey, you! Struggling with your faith? Had a few recent setbacks? Here's a facile fantasy to help you square your unfortunate experiences with the existence of a God worthy of worship."
posted by The Confessor at 10:58 AM on August 21, 2014 [2 favorites]
This whole debate reminds me of this: Whoah, that's a hell of a question.
posted by maryr at 3:02 PM on August 21, 2014 [5 favorites]
posted by maryr at 3:02 PM on August 21, 2014 [5 favorites]
benito.strauss: ... photoshop in a small image on the distant waves, long hair an robe flapping in the wind
Cowabunga, dude!
posted by filthy light thief at 3:14 PM on August 21, 2014 [1 favorite]
Cowabunga, dude!
posted by filthy light thief at 3:14 PM on August 21, 2014 [1 favorite]
(At least nobody's claiming every true or false statement has, by definition, a 50/50 chance of being one or the other. I skimmed through the original thread, started getting...agitated when that theme showed up, and then -- finally -- remembered I'd commented in that thread however many years ago, and that I was maybe a bit of a jerk about it.)
posted by nobody at 5:31 PM on August 21, 2014
posted by nobody at 5:31 PM on August 21, 2014
So did the person who created it say there was an answer and if so what is it?
posted by Carillon at 6:01 PM on August 21, 2014
posted by Carillon at 6:01 PM on August 21, 2014
Here's a paradox for the Monty Hall problem, which I've been thinking about since cortex mentioned it.
You have a 2/3 chance of being right if you switch, and a 1/3 chance of being right if you stay.
What if you obliterate this information? Either pick a door based on the flip of a coin, or have someone who doesn't know which door you picked, pick one of the remaining doors? Is the chance 50/50, or does it remain 2/3 to 1/3?
posted by graymouser at 7:57 AM on August 22, 2014
You have a 2/3 chance of being right if you switch, and a 1/3 chance of being right if you stay.
What if you obliterate this information? Either pick a door based on the flip of a coin, or have someone who doesn't know which door you picked, pick one of the remaining doors? Is the chance 50/50, or does it remain 2/3 to 1/3?
posted by graymouser at 7:57 AM on August 22, 2014
What's important is that Monty Hall has two doors (2/3 chance of the car) and always narrows it down to one door by showing you a goat. Thus you only get the better odds if you know which door Monty Hall has left unopened. If you don't know which door is Monty Hall's remainder and which was the original choice, your odds are 1/2, the same as if you begin the game with just two doors.
posted by Thing at 9:23 AM on August 22, 2014
posted by Thing at 9:23 AM on August 22, 2014
The key factor is whether the door the person who chooses a door to open is based on knowledge of what's behind the doors; the thing Monty will never, ever do is pick a door with the prize behind it to open, because he'd be losing the game for you. So any replacement system has to either (a) maintain that function of avoiding-the-winning-door or (b) fundamentally changes the setup.
Both your proposed mechanics break the game's assumptions taken at face value:
- Flip a coin when the player's chosen a non-winner and you'll reveal a winner half the time. As the player will initially choose a non-winner in two thirds of all cases, that means .666 * .5 = .333 of all games will end with a coin flip opening the winning door and you're presumably not allowed to switch to the winning door once it's revealed, so you've lost. The remaining two thirds of cases are split evenly between the other two doors, which each have equal chances of being the winner, so your chances of winning by either sticking or switching are even at .5 vs .5. That's .5 chance of being correct to stick or switch * .666 of games = .333 chance of winning all in all. One third of the time you second-guess right and win, one third of the time you second-guess wrong and lose, and one third of the time you don't a chance to second guess because the coinflip opens the winner door and loses the game out from under you. p = .333 and the game develops a new weird instant-loser possibility.
- Have hapless, inattentive Milton Hall, who was distracted by a bright light on the wall and so doesn't know what door you chose, pick at random a door to choose and one third of the time he'll choose the same door you chose. Given that it was your first choice, let's suppose that if it opens on a prize that means you win immediately because you'd be an idiot to switch to another door; if it does not open on a prize, it's a useful hint and you get to make your second guess with the knowledge that you should definitely switch away from that door. So: in one third of cases, Milton picks the same as you; in two thirds he picks other doors. In the case where he picks the same as you, one third of those will be winners, so that's .333 * .333 = .111 total wins. In two thirds of cases, Milton chooses differently from you; one third of those will be Milton choosing the the winning door by chance, losing you the game, so that's .666 * .333 = .222 total losses. The remaining cases, Milton chooses the same non-winner door as you and Milton chooses a non-winning different door than you, all reduce to a .5 vs .5 guess as to whether your original or the remaining door is correct. So that's the remaining .666 of cases with .5 chance of winning, .666 * .5 = .333 winning on second guess results. Add to that the .111 of same-guess instant win results and you've got a .444 chance or four ninths odds of winning. And, again, a somewhat weirder flow of the game.
Both of those depend on some just-so handwaving on my part about how to deal with the hint revealing a winner or loser; you could tweak them so that e.g. the hint door revealing a chosen loser or revealing an unchosen is not an instant loss but instead a really great hint and that would up the odds of winning in both cases significantly. But would also be weird.
Not that the aesthetics of the game show should matter for a probability story problem, but, hey, there's a certain elegance all else aside to the Monty Hall proposition.
posted by cortex (staff) at 9:30 AM on August 22, 2014
Both your proposed mechanics break the game's assumptions taken at face value:
- Flip a coin when the player's chosen a non-winner and you'll reveal a winner half the time. As the player will initially choose a non-winner in two thirds of all cases, that means .666 * .5 = .333 of all games will end with a coin flip opening the winning door and you're presumably not allowed to switch to the winning door once it's revealed, so you've lost. The remaining two thirds of cases are split evenly between the other two doors, which each have equal chances of being the winner, so your chances of winning by either sticking or switching are even at .5 vs .5. That's .5 chance of being correct to stick or switch * .666 of games = .333 chance of winning all in all. One third of the time you second-guess right and win, one third of the time you second-guess wrong and lose, and one third of the time you don't a chance to second guess because the coinflip opens the winner door and loses the game out from under you. p = .333 and the game develops a new weird instant-loser possibility.
- Have hapless, inattentive Milton Hall, who was distracted by a bright light on the wall and so doesn't know what door you chose, pick at random a door to choose and one third of the time he'll choose the same door you chose. Given that it was your first choice, let's suppose that if it opens on a prize that means you win immediately because you'd be an idiot to switch to another door; if it does not open on a prize, it's a useful hint and you get to make your second guess with the knowledge that you should definitely switch away from that door. So: in one third of cases, Milton picks the same as you; in two thirds he picks other doors. In the case where he picks the same as you, one third of those will be winners, so that's .333 * .333 = .111 total wins. In two thirds of cases, Milton chooses differently from you; one third of those will be Milton choosing the the winning door by chance, losing you the game, so that's .666 * .333 = .222 total losses. The remaining cases, Milton chooses the same non-winner door as you and Milton chooses a non-winning different door than you, all reduce to a .5 vs .5 guess as to whether your original or the remaining door is correct. So that's the remaining .666 of cases with .5 chance of winning, .666 * .5 = .333 winning on second guess results. Add to that the .111 of same-guess instant win results and you've got a .444 chance or four ninths odds of winning. And, again, a somewhat weirder flow of the game.
Both of those depend on some just-so handwaving on my part about how to deal with the hint revealing a winner or loser; you could tweak them so that e.g. the hint door revealing a chosen loser or revealing an unchosen is not an instant loss but instead a really great hint and that would up the odds of winning in both cases significantly. But would also be weird.
Not that the aesthetics of the game show should matter for a probability story problem, but, hey, there's a certain elegance all else aside to the Monty Hall proposition.
posted by cortex (staff) at 9:30 AM on August 22, 2014
Leonard Mlodinow's The Drunkard's Walk explains the Monty Hall Problem very thoroughly...
I first saw this puzzle on Reddit and copied what I took to be the most thorough analysis
None of the choices are self-consistent.
If A or D is correct, that means that the probability of getting the correct answer is both 25% (because that's what A and D say) and 50% (because you have a 50% chance of choosing A or D. That's a contradiction. It can't be both at once, so A and D are both wrong.
If B is correct, then the correct answer is both 50% (what it says) and 25% (the odds of choosing B). Same deal. B is self-contradictory, therefore wrong.
Likewise, if C is correct, then the answer is both 60% and 25%.
So none of the choices are right, and the actual probability of getting the right answer by choosing among them (by any means) is 0%. No paradox, no ambiguity, just a multiple-choice question that doesn't give any right answers, like "What is the longest river in Europe? A) The Nile B) The Mississippi C) CowboyNeal"
The tricky part is that if you did include 0% among the answers -- replacing C, let's say -- it would stop being right. It would become subject to the same problem as the current C does: it would imply both a 0% and a 25% chance of getting the correct answer. In that case, you wouldn't just have a multiple-choice question that omits the correct answer, but one that doesn't have a correct answer, like "Is the correct answer to this question 'no'?" But paradoxes of this sort are well-understood and don't faze logicians.
The variant I find more interesting is when there's more than one self-consistent answer. Like, what if the choices were: A) 50% B) 25% C) 60% D) 50%? You could answer A/D without contradiction, but you could also answer B without contradiction. You could even say "None of the answers are correct; the probability is 0%" without contradiction. But questions of the form "What is the probability of event X?" aren't the sort of thing that can have more than one correct answer. You just have no basis for choosing one answer over another. Well, that's not unusual: it's easy to come up with probability questions where you just don't have enough information to answer. (For example: "I have ten marbles in a bag. Some are black, some are white. If I draw one out, what is the probability that it is black?") But this variant isn't quite like that: it's impossible that more information will allow us to choose one answer over another. It's as unanswerable as the paradoxical variant, but without the paradox.
posted by lazycomputerkids at 10:10 AM on August 22, 2014 [1 favorite]
I first saw this puzzle on Reddit and copied what I took to be the most thorough analysis
None of the choices are self-consistent.
If A or D is correct, that means that the probability of getting the correct answer is both 25% (because that's what A and D say) and 50% (because you have a 50% chance of choosing A or D. That's a contradiction. It can't be both at once, so A and D are both wrong.
If B is correct, then the correct answer is both 50% (what it says) and 25% (the odds of choosing B). Same deal. B is self-contradictory, therefore wrong.
Likewise, if C is correct, then the answer is both 60% and 25%.
So none of the choices are right, and the actual probability of getting the right answer by choosing among them (by any means) is 0%. No paradox, no ambiguity, just a multiple-choice question that doesn't give any right answers, like "What is the longest river in Europe? A) The Nile B) The Mississippi C) CowboyNeal"
The tricky part is that if you did include 0% among the answers -- replacing C, let's say -- it would stop being right. It would become subject to the same problem as the current C does: it would imply both a 0% and a 25% chance of getting the correct answer. In that case, you wouldn't just have a multiple-choice question that omits the correct answer, but one that doesn't have a correct answer, like "Is the correct answer to this question 'no'?" But paradoxes of this sort are well-understood and don't faze logicians.
The variant I find more interesting is when there's more than one self-consistent answer. Like, what if the choices were: A) 50% B) 25% C) 60% D) 50%? You could answer A/D without contradiction, but you could also answer B without contradiction. You could even say "None of the answers are correct; the probability is 0%" without contradiction. But questions of the form "What is the probability of event X?" aren't the sort of thing that can have more than one correct answer. You just have no basis for choosing one answer over another. Well, that's not unusual: it's easy to come up with probability questions where you just don't have enough information to answer. (For example: "I have ten marbles in a bag. Some are black, some are white. If I draw one out, what is the probability that it is black?") But this variant isn't quite like that: it's impossible that more information will allow us to choose one answer over another. It's as unanswerable as the paradoxical variant, but without the paradox.
posted by lazycomputerkids at 10:10 AM on August 22, 2014 [1 favorite]
What if you obliterate this information? Either pick a door based on the flip of a coin, or have someone who doesn't know which door you picked, pick one of the remaining doors? Is the chance 50/50, or does it remain 2/3 to 1/3?
I take it that by "remaining doors" you mean "doors Monty didn't open", and you're proposing to alter the game as follows:
1. Contestant nominates a door.
2. Monty chooses a different door that has a goat behind it, and opens it.
3. Contestant flips a coin. If the coin turns up heads the contestant opens the originally nominated door. If it turns up tails the contestant opens the other unopened door.
After the completion of step 2, it is certain that the car is behind either the originally chosen door or the the other unopened door. Picking one of those on a coin flip gives you a winning probability of 1/2.
However, after step 2 it is also the case that the probability of the car being behind the originally nominated door is 1/3; behind the other unopened door, 2/3.
So picking your final door on a coin flip is a better idea than always staying with the original door, but a worse idea than always switching to the other unopened door.
If you want to break it down into cases, here are the cases:
1. Coin turns up heads (p = 1/2) and car is behind original door (p = 1/3); win (p = 1/6).
2. Coin turns up heads (p = 1/2) and car is behind other door (p = 2/3); lose (p = 1/3).
3. Coin turns up tails (p = 1/2) and car is behind original door (p = 1/3); lose (p = 1/6).
4. Coin turns up tails (p = 1/2) and car is behind other door (p = 2/3); win (p = 1/3).
The win and loss probabilities add up to 1/2 each.
posted by flabdablet at 12:05 PM on August 22, 2014
I take it that by "remaining doors" you mean "doors Monty didn't open", and you're proposing to alter the game as follows:
1. Contestant nominates a door.
2. Monty chooses a different door that has a goat behind it, and opens it.
3. Contestant flips a coin. If the coin turns up heads the contestant opens the originally nominated door. If it turns up tails the contestant opens the other unopened door.
After the completion of step 2, it is certain that the car is behind either the originally chosen door or the the other unopened door. Picking one of those on a coin flip gives you a winning probability of 1/2.
However, after step 2 it is also the case that the probability of the car being behind the originally nominated door is 1/3; behind the other unopened door, 2/3.
So picking your final door on a coin flip is a better idea than always staying with the original door, but a worse idea than always switching to the other unopened door.
If you want to break it down into cases, here are the cases:
1. Coin turns up heads (p = 1/2) and car is behind original door (p = 1/3); win (p = 1/6).
2. Coin turns up heads (p = 1/2) and car is behind other door (p = 2/3); lose (p = 1/3).
3. Coin turns up tails (p = 1/2) and car is behind original door (p = 1/3); lose (p = 1/6).
4. Coin turns up tails (p = 1/2) and car is behind other door (p = 2/3); win (p = 1/3).
The win and loss probabilities add up to 1/2 each.
posted by flabdablet at 12:05 PM on August 22, 2014
I love this problem, because the answer is so obviously "those were the times that I carried you."
The Onion is on it.
excerpt:
The Onion is on it.
excerpt:
Sure, the sandal footprints came back when I got that big job promotion, but right at the point where my son Tommy died, they veer off again. Actually, now that I look again, it seems like there's an unusually large distance between each of the sandal-wearer's footprints around the time of my son's death, as if the person were actually running away.posted by el io at 2:23 PM on August 22, 2014 [3 favorites]
A rabbi, a priest, and a guru walk into a bar. The conditional probabilities of each one having a drink are 50%, 25%, and 10%. What is the probability all of them will get drunk?
Voted Second Prize for "Worst Joke" at the International Festival of Bad Humor.
posted by twoleftfeet at 7:09 PM on August 22, 2014
Voted Second Prize for "Worst Joke" at the International Festival of Bad Humor.
posted by twoleftfeet at 7:09 PM on August 22, 2014
Right, but who cares about friggin probability.
DOES THE @#%#$ PLANE TAKE OFF OR NOT ?
posted by k5.user at 10:46 AM on August 25, 2014
DOES THE @#%#$ PLANE TAKE OFF OR NOT ?
posted by k5.user at 10:46 AM on August 25, 2014
Here's an insoluble logic puzzle for you: Does the set of all possible sets contain itself?
posted by clarknova at 5:45 PM on August 25, 2014
posted by clarknova at 5:45 PM on August 25, 2014
Look, Kurt, I told you to stay out of this tavern when I threw you out last time for making Bertrand cry.
posted by cortex (staff) at 6:04 PM on August 25, 2014 [1 favorite]
posted by cortex (staff) at 6:04 PM on August 25, 2014 [1 favorite]
Does the set of all possible sets contain itself?
If you say so.
posted by flabdablet at 8:51 PM on August 25, 2014
If you say so.
posted by flabdablet at 8:51 PM on August 25, 2014
The complex number that electrical engineers conventionally refer to as j has long been assumed to be the same one that mathematicians conventionally refer to as i. But what if that's not true? What if j is actually -i? Is there any way to be sure?
posted by flabdablet at 10:15 PM on August 25, 2014
posted by flabdablet at 10:15 PM on August 25, 2014
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posted by cortex (staff) at 10:31 AM on August 20, 2014 [2 favorites]