A Problem Even The Wizard Can't Solve (I hope!)
It is a classic problem, and those who have arrived at the correct answer can be counted on one hand, one foot, one ear, and one nose. Even the original proponent, Sir Arthur Eddington, didn't get it right.
If A, B, C and D each speak the truth once in three times (independently), and A affirms that B denies that C declares that D is a liar, what is the probability that D was speaking the truth?
I hope DorothyGale will be the first to respond and show her mettle. Perhaps this will do the trick:
DorothyGale looked in her mirror and it broke!
She went to the hardware store and got a plastic mirror and put in on the wall.
She looked in it - it melted!
She went to the hardware store and got a mirror made of polished aluminum.
It would neither break nor melt. She put it on the wall.
She looked in it and the mirror cried "What the Hell is THAT, got down off the wall and kicked her in the ass.
Over to you, Dorothy.
Quote: statmanSir Arthur Eddington, didn't get it right.
Yes, he did.
Quote: statmanThis post is not intended as a challenge to the Wizard; it is an attempt to get at DorothyGale, one of my most outspoken admirers, without anyone else stepping in and spilling the beans, as they did in "How's The Weather in Nerdville?"
It is a classic problem, and those who have arrived at the correct answer can be counted on one hand, one foot, one ear, and one nose. Even the original proponent, Sir Arthur Eddington, didn't get it right.
If A, B, C and D each speak the truth once in three times (independently), and A affirms that B denies that C declares that D is a liar, what is the probability that D was speaking the truth?
If you want to communicate with DorothyGale in private, do so. This is a public forum and you posted a puzzler in the math section. You should expect replies.
I get a number less than 1/3 but more than 30%, conditional on D actually saying something to begin with (if D is silent, the probability is zero; EDIT, actually we have to assume that everyone says something, and furthermore that the statement of the problem is true 100% of the time.). I'm being intentionally vague so as not to give away the solution. But my first question to you is:
Can you solve it?
This statement is false.
Quote: dwheatleyThe answer(s) can be found in a number of books on probability. Apparently the good Knight made a certain assumption based on who is saying exactly what that led him to one answer, when a more straightforward assumption leads to another answer.
This statement is false.
Which one -- the last one or the one above it? :)
Quote: dwheatleyApparently the good Knight made a certain assumption based on who is saying exactly what that led him to one answer, when a more straightforward assumption leads to another answer.
Actually, Eddigton's assumption was fairly straightforward if you ask me, it was also the only one that agrees with the rules of the predicate calculus.
Also, don't forget, he is the one, who formulated the problem. If he says his assumption is correct, it is correct :)
I have to admit I am at once flattered by your attention and at the same time so very sad for you ...Quote: statmanThis post is ... is an attempt to get at DorothyGale
--Ms. D.
Quote: weaselmanActually, Eddigton's assumption was fairly straightforward if you ask me, it was also the only one that agrees with the rules of the predicate calculus.
Also, don't forget, he is the one, who formulated the problem. If he says his assumption is correct, it is correct :)
Having just read up on it, the answer does indeed depend on the assumptions you make. Specifically, if you assume each party makes a relevant statement (as I did), you end up with a different answer than if you make no assumptions about who spoke and/or about what. However, in the latter case it would seem Eddington didn't go far enough: as I said earlier, if D has said nothing, the probability of D having spoken the truth is zero.
Quote: MoscaThey each speak the truth one in three times. I think that's the answer right there.
It would be if the statements were independent, but the question is actually asking a conditional probability: what is the probability that D was truthful given that A said what he said.
Its somewhat like saying "The probability of each number 1 through 6 appearing on a die is 1/6. Two dice are rolled: one die shows a four and one die does not show a six. What is the chance the total is seven?"
The answer isn't 6/36 -- the unconditional probability of rolling a 7 -- because there are conditions involved. (For a very small amount of extra credit, what's the answer?)
Looks to me as if there is zero chance that the total is seven. The one die that does not show a six has already been revealed to show a four. Thus, we are specifically told that the dice show 6 and 4 for a total of 10.Quote: MathExtremist"The probability of each number 1 through 6 appearing on a die is 1/6. Two dice are rolled: one die shows a four and one die does not show a six. What is the chance the total is seven?"
I was thinking about this and here's a smaller example, so let me see if I have this right :
C declares D is lying.
Lets say D makes a statement "The last car we saw was red". Lets call that statement x.
x is either true or false.
Assuming C KNOWS whether x is true or false, they can make a statement about D's statement x.
So, I think this ends up in some sort of Combinatory Logic, which I've tried to udnerstand several times by reading Raymond Smullyan's "To Mock a Mockingbird", but after we leave the island of Knights and Knaves, the bird forests make my head spin.
There's four possibilities :
x is true, D states it to be true and C is lying about D lying.
x is true, D states it to be false and C is telling the truth about the lie D made
x is false, D states it to be true and C is telling the truth about the lie D made
x is false, D states it to be false, and C is lying about D lying.
D's only telling the truth when C lies about it. C lies 2/3rds of the time. Thus D is telling the truth 2/3rds of the time?
Hmm. Seems wrong to me....
Quote: DorothyGaleI have to admit I am at once flattered by your attention and at the same time so very sad for you ...
--Ms. D.
I like this. +1
Everyone should have a stalker!
Assuming that the question is not being worded in an exceptionally misleading way (for instance, the statement, "Two dice are rolled: one die shows a four and one die does not show a six." could be claiming the obvious, that the one die showing a four also does not show a six and referring twice to the same die) I would say that the odds would be 1/5.Quote: MathExtremist
Its somewhat like saying "The probability of each number 1 through 6 appearing on a die is 1/6. Two dice are rolled: one die shows a four and one die does not show a six. What is the chance the total is seven?"
The answer isn't 6/36 -- the unconditional probability of rolling a 7 -- because there are conditions involved. (For a very small amount of extra credit, what's the answer?)
Although the problem is briefly stated, my solution is rather long because it is broken down into very small steps in order to deal with the twisted logic one twist at a time. "A affirms that B denies that C declares..." takes you on a nice roller coaster of logic. Eddington submitted it to the world in his book New Pathways in Science, and that is the only book in which it is discussed. If ever I do win the Nobel Prize (nyuk, nyuk) I'll invite Dr. Olofsson to come along. Stockholm is his home town.
DorothyGale: Do not be flattered. I am out to bring you down for the extremely vicious remarks you made about me in the thread "New Probability Tables for Roulette IV." Buzzpaff now is lying in a back alley, but you did not give me as much ammunition as he did, and the worst I can do for you is to humiliate you before the forum. Here's another story about you:
DorothyGale booked a flight to Las Vegas in economy class. At 8,000 feet she called the stewardess over and said: "I'm not comfortable here. There are people on either side of me, I can't see out the window, and the coffee is lukewarm. I know there are empty seats in first class. Couldn't you seat me there?" "I'll call the captain," the stewardess said.
In a minute the co-pilot appeared. "Ms. Gale, I understand you're not happy with your seat and want to be put in first class." "Yes," she replied. "I'm a first class person and I think I deserve it." "I'm sorry, Ms. Gale," the co-pilot replied, "But on this flight first class doesn't go to Las Vegas." "Oh," said Dorothy, "I didn't realize that."
Quote: statman
Eddington submitted it to the world in his book New Pathways in Science, and that is the only book in which it is discussed.
The only one you heard of, perhaps?
I have seen at least three other books that deal with it. One is "Elementary Probability" by David Stirzaker (unlike you, it presents a well-defined statement of the problem btw, that does not contain ambiguity).
There is also a discussion in one of the books of Martin Gardner, but I can't remember which one. The third one I saw it in was called "Introduction to Information Theory", but I can't remember the author.
Quote: TheNightflyAssuming that the question is not being worded in an exceptionally misleading way (for instance, the statement, "Two dice are rolled: one die shows a four and one die does not show a six." could be claiming the obvious, that the one die showing a four also does not show a six and referring twice to the same die) I would say that the odds would be 1/5.
If one die does not show a six, it means that the other one does :) The total is 10.
I'm not sure what Dorothy did to get your feathers ruffled, but posting stupid shit like this make you look even more like the idiot that we all know you to be.Quote: statmanDorothyGale booked a flight to Las Vegas in economy class. At 8,000 feet she called the stewardess over and said: "I'm not comfortable here. There are people on either side of me, I can't see out the window, and the coffee is lukewarm. I know there are empty seats in first class. Couldn't you seat me there?" "I'll call the captain," the stewardess said.
In a minute the co-pilot appeared. "Ms. Gale, I understand you're not happy with your seat and want to be put in first class." "Yes," she replied. "I'm a first class person and I think I deserve it." "I'm sorry, Ms. Gale," the co-pilot replied, "But on this flight first class doesn't go to Las Vegas." "Oh," said Dorothy, "I didn't realize that."
Quote: weaselman
If one die does not show a six, it means that the other one does :) The total is 10.
Yes, it's a little like the puzzle "I have 2 coins in my pocket that total 15 cents. One of the coins is not a dime - what 2 coins have I got?"
a) Dorothy solves the problem correctly;
b) She decides this nonsense is not worth her time, and
c) Dorothy cant solve it, or does so incorrectly.
1. Given event a occurs, what is the probability that statman would become able to handle criticism any better? The probability that his ego would become any less fragile?
2. Given event b occurs, what is the probability that anyone gives a shit? What is the expected length of time before statman runs out of old magazines to copy and paste from and will have to find a new hobby?
3. Given event c occurs - the result statman dreams of day and night - what is the incremental free time he will have, due to his colleagues and annoyed grandchildren avoiding him to get out of yet another 'once I told off this guy on the internet' victory speeches, which he will fill the void by copy and pasting even more old magazines from. (careful, this will affect 2b!)
For bonus marks, assign probabilities to a, b and c that will win statman the most respect he can get out of his challenge. Now that's something even the wizard can't solve.
Gog -Quote: gog...
For bonus marks, assign probabilities to a, b and c that will win statman the most respect he can get out of his challenge. Now that's something even the wizard can't solve.
For a member that doesn't post much, that was priceless. Hell, the entire post was excellent.
Bravo!
Quote: MathExtremistIt would be if the statements were independent, but the question is actually asking a conditional probability: what is the probability that D was truthful given that A said what he said.
Its somewhat like saying "The probability of each number 1 through 6 appearing on a die is 1/6. Two dice are rolled: one die shows a four and one die does not show a six. What is the chance the total is seven?"
The answer isn't 6/36 -- the unconditional probability of rolling a 7 -- because there are conditions involved. (For a very small amount of extra credit, what's the answer?)
You know, I saw that. What happened was that I went and looked up the original problem, and Eddington had originally stated it as "what is the conditional probability...."
But statman left out the word conditional. So to me, I figured that that meant the problem simplified to the original declaration; what A, B, and C say are irrelevant.
Quote: statman
DorothyGale: Do not be flattered. I am out to bring you down for the extremely vicious remarks you made about me in the thread "New Probability Tables for Roulette IV." Buzzpaff now is lying in a back alley, but you did not give me as much ammunition as he did, and the worst I can do for you is to humiliate you before the forum.
You are delusional.
My advice would be to ignore things you don't like to hear.
Do you watch Big Bang Theory. One of my favorite lines is when Sheldon is bowling against his nemesis, Wil Wheaton (ST:TNG's Wesley Crusher).
Quote: Big Bang Theory, The Wheaton Recurrence
Wil Wheaton: You're not still sore that I beat you at that card tournament, are you?
Sheldon Cooper: I am the proud owner of ihatewilwheaton.com, .net and .org. What does that tell you?
Wil Wheaton: It tells me that I am living rent-free right here. - Points at Sheldon's head
Quote: boymimbo
Do you watch Big Bang Theory. One of my favorite lines is when Sheldon is bowling against his nemesis, Wil Wheaton (ST:TNG's Wesley Crusher).
Quote: Big Bang Theory, The Wheaton Recurrence
Wil Wheaton: You're not still sore that I beat you at that card tournament, are you?
Sheldon Cooper: I am the proud owner of ihatewilwheaton.com, .net and .org. What does that tell you?
Wil Wheaton: It tells me that I am living rent-free right here. - Points at Sheldon's head
I think about that episode a lot when I read conversations here. But I'm thinking more of:
Quote: SheldonDo you people even hear yourselves? It’s not the Wesley Crushers. It’s not the “Wesley” Crushers. It’s the Wesley “Crushers.”
Quote: TheNightflyI would say the odds (of the total being 7) would be 1/5.
Quote: weaselmanIf one die does not show a six, it means that the other one does :) The total is 10. (implying that the probability of a total of 7 is zero)
Neither 1/5 nor 0 is the answer I had in mind when I wrote the problem. This is a perfect example of how even a simple probability question has an answer that depends on the assumptions you make about the language. It goes to show that identifying what remains unsaid in the statement of a problem is just as important, if not more so, than the problem statement itself. If there are 3 different interpretations for something as simple as my dice problem, it is not surprising that there are multiple interpretations of the Eddington problem.
Quote: statmanMathExtremist: Yes, I managed to solve it,
What's your answer?
Quote: statmanDorothyGale booked a flight to Las Vegas in economy class. At 8,000 feet she called the stewardess over and said: "I'm not comfortable here. There are people on either side of me, I can't see out the window, and the coffee is lukewarm. I know there are empty seats in first class. Couldn't you seat me there?" "I'll call the captain," the stewardess said.
Everyone knows that you don't get your coffee at 8,000 feet.
Quote: MathExtremist
Everyone knows that you don't get your coffee at 8,000 feet.
My first thought, too. The flight attendants aren't even out of their seats at that time.
The 8000' foot issue was already raised, the terminology "stewardess" is no longer used, and it is no longer possible for a pilot or co-pilot to leave the flight deck to visit a passenger. And I am a brunette, the original joke was about a blonde. Original joke. That's four errors ...Quote: statmanAt 8,000 feet ... stewardess ... In a minute the co-pilot appeared ...
--Ms. D.
You gotta think outside of the box when it comes to the statman.
But do those planes have a separate first class section and serve coffee at 8000'? Also, the original joke had the flight attendant calling the captain, but the co-pilot came out instead. So, under your interpretation, the flight attendant called the captain, who tossed it back to the flight attendant = co-pilot, who under the new personality ... OMG ... Sheldon, is that you up there?Quote: boymimboActually, there's still a few old passenger planes where the co-pilot IS the flight attendant.
--Ms. D.
Quote: statman... snip some faff about talking to people about a solution and a bunch of rubbish about it's uniqueness ...
... attack on DorothyGale ...
Well done sir, you've managed to act like a complete douche bag. Or in the vernacular of your time, a cad, a rotter and downright poopy-pants.
Hell, I even defended your posting of your magical roulette tables, to give you a chance. Looks like you didn't deserve it.
Second, this is probably wrong, and I'm not even sure I understand the question, but my answer is 20%.
Quote: statmanMathExtremist: If ever I do win the Nobel Prize (nyuk, nyuk) I'll invite Dr. Olofsson to come along. Stockholm is his home town.
DorothyGale: Do not be flattered. I am out to bring you down for the extremely vicious remarks you made about me in the thread "New Probability Tables for Roulette IV."
Get back on your Meds Statman! Dorothy didn't fillet you because she is mean, but because you came off sounding like an idiot. But there is hope- they gave a Nobel prize to the guys who screwed up Long Term Capital Mgmt... to this day they are still saying that their was "nothing wrong with their models, the market was wrong....?" Maybe you'll get yours (nyuk, nyuk).
I voted you should stay around when you put up the poll, little did I know you would be so entertaining.
Especially when we've seen what happened here, here, and here. I just can't get over the third one...a self-described instructor of college level mathematics thinks that a no-zero roulette wheel would still have a 2.78% house advantage? Really??
So then you gotta solve for the situation in the problem.
A affirms that B denies that C declares that D is a liar. What is the probability that D is speaking the truth?
If B denies that C declares that D is a liar, then B is affirming that C declares that D is telling the truth. Therefore, we have the following situations where either A, B, C are telling the truth, or there are two lies among A, B, and C. D is always telling the truth.
By my calculation, I get (1 + 4 + 4 + 4) / 81 = 13/81, but only if we assume that everyone said something. Eddington got 25/71. Another got 1/3rd. Another got 13/41.
Which is right? Who cares?
I interpret "A affirms that B denies that C declares that D is a liar" as:
D: (something)
C: D is lying
B: C did not declare that D is lying
A: B said that C did not declare that D is lying
The probabilities of A and B making their responses are the same regardless of whether or not D told the truth, since B's statement would not change - B didn't say, "C is the one who is lying", but said, "I deny that C said that D is lying."
The range of possibilities is now (a) D is lying, and, as a result, C is telling the truth, and (b) D is telling the truth, and, as a result, C is lying.
(a) has a probability of 1/3 x 2/3 = 4/9; (b) has a probability of 2/3 x 1/3 = 4/9.
(a) and (b) have equal probability, and as they represent the entire range of results, each is 1/2.
Only (b) has D telling the truth.
Quote: boymimboBy my calculation, I get (1 + 4 + 4 + 4) / 81 = 13/81, but only if we assume that everyone said something.
That's the probability that A's statement was true *and* D is speaking the truth. You're looking for the probability that D is speaking the truth *conditioned upon* A's statement being true, p(D | A), which by Bayes equals p(D & A) / p(A)
p(D & A) = 13/81
p(A) = 41/81 (any even number of liars will do, including those in which D is lying: 4c0 + 4c2 + 4c4 = 41)
p(D | A) = 13/41
Again, this assumes that not only did everyone say something, but that it was either an affirmation or denial as indicated. A affirms what B says, but if A is lying, we still need to assume that B spoke about what C said instead of something unrelated like the weather.
Edit: to clarify, I mean that the statement "A affirms that B denies ..." is true, not necessarily that A was speaking the truth.
Quote: MathExtremistThat's the probability that A's statement was true *and* D is speaking the truth. You're looking for the probability that D is speaking the truth *conditioned upon* A's statement being true, p(D | A)
Are you? We just know that A said that B denied ... not that he spoke the truth, don't we?
Why am I looking for the probability that D is speaking the truth conditioned on A's statement being true. The question asks for the probability that D speaks the truth. Or did I read the question wrong?
If A, B, C and D each speak the truth once in three times (independently), and A affirms that B denies that C declares that D is a liar, what is the probability that D was speaking the truth?
Oh, okay, the question asks for the probability that D is speaking the truth give that A affirmed so on and so forth. I get it.
Quote: weaselmanAre you? We just know that A said that B denied ... not that he spoke the truth, don't we?
A's statement turns out to be true more often than the situations when A is telling the truth. If A is lying and B is lying, A's statement is still true (double negative) etc. Maybe I should have said "the proposition that A said that B denied..." is true, and if I muddled things I apologize.
Quote: MathExtremistIf A is lying ... , A's statement is still true
Really? :)
And that's why even talking about these sorts of problems is a pain...Quote: weaselmanReally? :)
Quote: WizardYou guys will have to figure it out yourselves, statman was just permanently banned.
What'd he do (just now) that got him banned? Or was it just the previous bizarre attacks on DorothyGale?
Quote: MathExtremistWhat'd he do (just now) that got him banned? Or was it just the previous bizarre attacks on DorothyGale?
Give me a day or so to answer that question. There is more to it than just his posts.
