Conundrums, Puzzles, Logic Problems (1 Viewer)

[ELM]

******* spoilers below ********when it comes to arguments about infinity. logic breaks downIt simply is not logical that Bob could come up witha strategy that will yeild him better than 50%Infinity - 10 trillion is still infinity.If you don't buy that, then lets's calculate probabilitiesthere are 4 scenariosBoth Al's numbers are greater than Bob's n+.5 = 25 percent of the time - Bob will guess lower and Win half of the time - 12.5%One is high and one is lower bob will Win always - 50%Both Al's numbers are less than Bob's n+.5 = 25 percent of the time - Bob will guess higher and Win half of the time = 25 percent of the time - win half 12.5%Bob will win 75% of the time rightI'm sure somebody can use this to prove that .75 = .5 at which point all Rational Mathematics breaks down.********************************wrap your head around that one...

Logic doesn't break down when we start talking about infinite sets.Common sense breaks down.I certainly agree that you can take any finite number of elements out of an infinite set and the infinite set won't become finite. You can't thin an infinite set.If Al and Bob choose 3 different integers comletely at random, Bob will get the highest number 1/3 of the time, the middle number 1/3 of the time, and the lowest number 1/3 of the time - it's not 25%, 50%, and 25%. Finally, none of the solution relies on funky math. It's flies in the face of common sense, but given that we're talking about infinite sets, the fact that the solution doesn't pass the smell test doesn't mean anything. All sorts of facts about infinite sets fly in the face of common sense (e.g. that some infinite sets are strictly larger than other infinite sets - proved by Cantor a while ago).As I said, if Bob's edge is supposed to reduce to zero, then you'd have to show that his probability of generating a number that falls between Al's numbers has to hit zero. If someone could show that, the solution would be invalidated.I guess what I'm asking is, is this reasoning incorrect? If not, what's the problem with the solution?

you have dialed yourself in to a false truth. You divide it into three subsets when there are ONLY two. If you pick a number that happens to be 7.5 for N+.5 and then you pick als right hand there are only two sets of numbers for what is in his left hand. There is either higher or lower than his right hand. That is it. End of story. So if N+.5 is 6.5 and the first number chosen is 100. Then he will say lower. There are only two sets of integers here. If the next number is lower than 100 he is right. If the next number is below 6.5 he is still right, therefore in this scenario your fictitious third scenario is just a subset. It is still only part of the 50% of the numbers that are lower than 100.

 

[TinHat]

******* spoilers below ********when it comes to arguments about infinity. logic breaks downIt simply is not logical that Bob could come up witha strategy that will yeild him better than 50%Infinity - 10 trillion is still infinity.If you don't buy that, then lets's calculate probabilitiesthere are 4 scenariosBoth Al's numbers are greater than Bob's n+.5 = 25 percent of the time - Bob will guess lower and Win half of the time - 12.5%One is high and one is lower bob will Win always - 50%Both Al's numbers are less than Bob's n+.5 = 25 percent of the time - Bob will guess higher and Win half of the time = 25 percent of the time - win half 12.5%Bob will win 75% of the time rightI'm sure somebody can use this to prove that .75 = .5 at which point all Rational Mathematics breaks down.********************************wrap your head around that one...

Logic doesn't break down when we start talking about infinite sets.Common sense breaks down.I certainly agree that you can take any finite number of elements out of an infinite set and the infinite set won't become finite. You can't thin an infinite set.If Al and Bob choose 3 different integers comletely at random, Bob will get the highest number 1/3 of the time, the middle number 1/3 of the time, and the lowest number 1/3 of the time - it's not 25%, 50%, and 25%. Finally, none of the solution relies on funky math. It's flies in the face of common sense, but given that we're talking about infinite sets, the fact that the solution doesn't pass the smell test doesn't mean anything. All sorts of facts about infinite sets fly in the face of common sense (e.g. that some infinite sets are strictly larger than other infinite sets - proved by Cantor a while ago).As I said, if Bob's edge is supposed to reduce to zero, then you'd have to show that his probability of generating a number that falls between Al's numbers has to hit zero. If someone could show that, the solution would be invalidated.I guess what I'm asking is, is this reasoning incorrect? If not, what's the problem with the solution?

This is why things break down with probability on infinite setsIf you out the three random numbers in a hat, the odds that Bob's number is the highest one is 1/3YETThe odds that Bobs' number higher than the Al's first number pulled out of the hat is 1/2The odds that Bobs' number higher than the Al's second number pulled out of the hat is 1/2combine these and we get that the odds that he will pick a number higher than Al's numbers is 1/41/4 = 1/3

 

[Aabye]

******* spoilers below ********when it comes to arguments about infinity. logic breaks downIt simply is not logical that Bob could come up witha strategy that will yeild him better than 50%Infinity - 10 trillion is still infinity.If you don't buy that, then lets's calculate probabilitiesthere are 4 scenariosBoth Al's numbers are greater than Bob's n+.5 = 25 percent of the time - Bob will guess lower and Win half of the time - 12.5%One is high and one is lower bob will Win always - 50%Both Al's numbers are less than Bob's n+.5 = 25 percent of the time - Bob will guess higher and Win half of the time = 25 percent of the time - win half 12.5%Bob will win 75% of the time rightI'm sure somebody can use this to prove that .75 = .5 at which point all Rational Mathematics breaks down.********************************wrap your head around that one...

Logic doesn't break down when we start talking about infinite sets.Common sense breaks down.I certainly agree that you can take any finite number of elements out of an infinite set and the infinite set won't become finite. You can't thin an infinite set.If Al and Bob choose 3 different integers comletely at random, Bob will get the highest number 1/3 of the time, the middle number 1/3 of the time, and the lowest number 1/3 of the time - it's not 25%, 50%, and 25%. Finally, none of the solution relies on funky math. It's flies in the face of common sense, but given that we're talking about infinite sets, the fact that the solution doesn't pass the smell test doesn't mean anything. All sorts of facts about infinite sets fly in the face of common sense (e.g. that some infinite sets are strictly larger than other infinite sets - proved by Cantor a while ago).As I said, if Bob's edge is supposed to reduce to zero, then you'd have to show that his probability of generating a number that falls between Al's numbers has to hit zero. If someone could show that, the solution would be invalidated.I guess what I'm asking is, is this reasoning incorrect? If not, what's the problem with the solution?

This is why things break down with probability on infinite setsIf you out the three random numbers in a hat, the odds that Bob's number is the highest one is 1/3YETThe odds that Bobs' number higher than the Al's first number pulled out of the hat is 1/2The odds that Bobs' number higher than the Al's second number pulled out of the hat is 1/2combine these and we get that the odds that he will pick a number higher than Al's numbers is 1/41/4 = 1/3

OK, if that's the case, we can play a game.I'll generate a random integer and you generate two random integers by the same procedure. They can be any integers you like. When my number falls between your numbers, I'll give you a dollar. When my number is either the highest or the lowest of the three, you give me a dollar.Do you see that I will win this game?You're treating this as if it's two discrete cases - my number versus your first number and my number versus your second number. It's not. It's one scenario.It also has nothing to do with infinite sets. If we limited the possibilities to numbers 1, 2, and 3 and I randomly pick one and you take the other two, what is the probability that my number is higher than your first number? 1/2. What is the probability that my number is higher than your second number? 1/2. So is the probability that my number is 3 (the number that is higher than either of your numbers) 1 in 4? Of course not. It's 1 in 3.Now, the above scenario is exactly like the case in which Al randomly chooses two different integers and Bob chooses an integer and adds 1/2, thereby producing a third number that is not identical to either of Al's numbers. The actual values of the numbers don't matter. We'll label the lowest number "1", the middle number "2", and the highest number "3". Now, what is the chance that Bob holds the highest number - "3"? It has to be 1 in 3.You're not doing the probabilities right. It doesn't have anything to do with infinite v. non-infinite sets. After the numbers are written down, one number has to be the largest, one number has to be the smallest, and one number has to be in the middle. Bob has a 1/3 chance of each, supposing that his procedure to generate random numbers is at least as good as Al's procedure.

 

[sartre]

It also has nothing to do with infinite sets.

The wording of the initial riddle has everything to do with infinite sets. The strategy simply will not work for a boundless pool of integers. Do you agree that

1/infinity = 22/infinity = 6000000000/infinity = 0 ?

If not, then I understand where you are coming from, though your conclusion remains mathematically flawed.

 

[Aabye]

******* spoilers below ********when it comes to arguments about infinity. logic breaks downIt simply is not logical that Bob could come up witha strategy that will yeild him better than 50%Infinity - 10 trillion is still infinity.If you don't buy that, then lets's calculate probabilitiesthere are 4 scenariosBoth Al's numbers are greater than Bob's n+.5 = 25 percent of the time - Bob will guess lower and Win half of the time - 12.5%One is high and one is lower bob will Win always - 50%Both Al's numbers are less than Bob's n+.5 = 25 percent of the time - Bob will guess higher and Win half of the time = 25 percent of the time - win half 12.5%Bob will win 75% of the time rightI'm sure somebody can use this to prove that .75 = .5 at which point all Rational Mathematics breaks down.********************************wrap your head around that one...

Logic doesn't break down when we start talking about infinite sets.Common sense breaks down.I certainly agree that you can take any finite number of elements out of an infinite set and the infinite set won't become finite. You can't thin an infinite set.If Al and Bob choose 3 different integers comletely at random, Bob will get the highest number 1/3 of the time, the middle number 1/3 of the time, and the lowest number 1/3 of the time - it's not 25%, 50%, and 25%. Finally, none of the solution relies on funky math. It's flies in the face of common sense, but given that we're talking about infinite sets, the fact that the solution doesn't pass the smell test doesn't mean anything. All sorts of facts about infinite sets fly in the face of common sense (e.g. that some infinite sets are strictly larger than other infinite sets - proved by Cantor a while ago).As I said, if Bob's edge is supposed to reduce to zero, then you'd have to show that his probability of generating a number that falls between Al's numbers has to hit zero. If someone could show that, the solution would be invalidated.I guess what I'm asking is, is this reasoning incorrect? If not, what's the problem with the solution?

you have dialed yourself in to a false truth. You divide it into three subsets when there are ONLY two. If you pick a number that happens to be 7.5 for N+.5 and then you pick als right hand there are only two sets of numbers for what is in his left hand. There is either higher or lower than his right hand. That is it. End of story. So if N+.5 is 6.5 and the first number chosen is 100. Then he will say lower. There are only two sets of integers here. If the next number is lower than 100 he is right. If the next number is below 6.5 he is still right, therefore in this scenario your fictitious third scenario is just a subset. It is still only part of the 50% of the numbers that are lower than 100.

It looks like you're treating this as two independent tests as well - Bob's number versus Al's right-hand number and Bob's number versus Al's left hand number and treating these two cases as completely independent. But they're not. It's a single scenario which includes three numbers of different values.Let's go back to the simplified case of just writing the numbers 1, 2, and 3 on slips of paper. Bob draws one and Al draws two.Bob does not look at his own number. Now Bob chooses either Al's right or left hand and a third observer, call him Carl, tells Bob whether the number in Al's hand is higher or lower than the number in Bob's hand. He doesn't tell him what the number is, just whether it is higher or lower than Bob's number.Now Bob has to guess whether the number is Al's opposite hand is larger or smaller than the number in the hand that Bob chose first.Bob should say "higher" if Carl tells him that Al's first number was smaller than Bob's number and Bob should say "lower" if Carl tells him that Al's first number was larger than Bob's number.Do you see why?There are three possibilities if Carl tells Bob that Al's first number is larger than Bob's number. They are: (1) Al has 3 and Bob has 1, (2) Al has 3 and Bob has 2 and (3) Al has 2 and Bob has 1. In two of these cases, the number in Al's other hand is lower than the number in the hand that Bob picked. So Bob should say "lower" and expect to win 2/3 of the time.Likewise, there are three possibilities if Carl tells Bob that Al's first number is smaller than Bob's number. They are: (1) Al has 1 and Bob has 2, (2) Al has 1 and Bob has 3 and (3) Al has 2 and Bob has 3. In two of these cases, the number in Al's other hand is higher than the number in the hand that Bob picked. So Bob should say "higher" and expect to win 2/3 of the time.That exact reasoning translates to the problem at hand - the one in which Al generates both of his numbers randomly.Let's replace the numbers 1, 2, and 3 with 3 random non-identical integers. Does this make any difference for the case? Of course not. It doesn't matter to Bob what the actual numbers are.Let the numbers be -60, 45, and 900. Carl tells Bob that his number is higher than Al's first number when (1) Al's first number is 900 and Bob's number is 45, (2) Al's first number is 900 and Bob's number is -60, and (3) Al's first number is 45 and Bob's number is -60. Again, Bob wins 2/3 of these cases if he guesses that Al's other number is lower.Same reasoning when Carl tells Bob that his number is lower than Al's first number.And it should be clear that it doesn't matter what numbers Al and Bob generate, the situation does not change.

 

[Aabye]

It also has nothing to do with infinite sets.

The wording of the initial riddle has everything to do with infinite sets. The strategy simply will not work for a boundless pool of integers. Do you agree that

1/infinity = 22/infinity = 6000000000/infinity = 0 ?

If not, then I understand where you are coming from, though your conclusion remains mathematically flawed.

I agree that 1/infinity = 22/infinity = 6000000000/infinity.But imagine that you choose a single number from the set of integers and ask me choose a single number from the set of integers as well. What are the chances that my number matches yours?

It can't be strictly zero. It's a limit approaching zero from the positive side.

Here's a link.

 

[videoguy505]

there's a magical island that is populated by 99 dragons. all of these dragons have red eyes, but they do not know this. if at any time it is possible for a dragon to deduce his eye color, the dragon will die the next day at noon. dragons, being the intelligent creatures they are, never discuss eye colors and they avoid looking in water where they might accidentally see their reflection. dragons are also extremely observant, so each dragon knows that the other 98 have red eyes. one day, an evil logician appears on the island and gathers the dragons together. he yells to them, "there is at least one dragon with red eyes on this island!" and suddenly vanishes. the dragons freak out as eye colors are not supposed to be discussed on the island, but eventually they settle down upon seeing many other dragons with red eyes. 99 days later, 99 dragons are dead at noon. why?

The logician didn't tell them anything they weren't already aware of.If the dragons died, it wasn't because of that, it was some unrelated dragon catastrophe.

you got an answer or what?

The answer is that you framed the riddle totally wrong.Here it is in its correct glory:

The Problem

Three dragons are living on an island. On this island, dragons either have red eyes or blue eyes. If a dragon with red eyes realizes that he has red eyes, then he will die at midnight that night. There are no reflective surfaces on the island (i.e. there is no way for a dragon to see his own eye-colour). The three dragons never speak or communicate to each other in any way, but they do see each other everyday. Thus each dragon knows the eye-colour of the other two dragons. It turns out that all three dragons on this island have red eyes.

Many years pass without a dragon dying, because no dragon ever finds out his own eye-colour. One day, a cackling wizard appears in front of all three dragons and makes the statement, "At least one of you has red eyes!", and then disappears. The dragons know that the wizard never lies.

On the n-th night since the wizard appeared, all three dragons died. What is n, and how did the dragons realize that they had red eyes?

And the answer:

Spoiler

The Solution

I could give the solution for three dragons only, by breaking the situation down into four cases and handling each case; but I will do it a better, more elegant way. I will generalize this to any number of dragons, some with red eyes and some with blue.

Let the island have n > 0 dragons with red eyes, and m dragons with blue eyes. I claim that all n dragons with red eyes will die on the n-th night after the wizard makes the statement "At least one of you has red eyes." (So for the original problem, the answer is the 3rd night). I prove this using mathematical induction on n.

If n = 1, then the lone dragon with red eyes will look around the island and see that all other dragons have blue eyes. He concludes that he is the only one with red eyes, and thus dies on the first night.

Suppose that for n = k-1, k > 1, that all k-1 red-eyed dragons die on the (k-1)-th night. Then for n = k, this is what happens: each of the k dragons will look around and see that there are k-1 dragons with red eyes on the island. Each dragon thinks, "I don't want to die, so let's assume that I have blue eyes." Then each dragon should expect the other k-1 dragons to die in the (k-1)-th night; so they wait. On the k-th day, each dragon sees that the other k-1 dragons are still alive, and thus their initial assumption that "I have blue eyes" must be false. Thus each of the k dragons realize that they have red eyes, and die on the k-th night.

This is exactly what I said. And I only used "blue" to mean "not red". The dragons could look around and see all sorts of other colors, but if any were non-red, then that's the same as blue for all intents and purposes of the puzzle.

 

[videoguy505]

So if I understand this correctly....the solution to the Dragon puzzle is that you make up a half-way intelligent sounding answer and hope everyone is too afraid to look stupid to question the BS?

see post 109

That doesn't follow though.At least 1 of them has red eyes. They all already knew that 98 of them had red eyes at least.

Each individual dragon knew, before the evil logician came to their island, that they saw 98 other pairs of red eyes. But after the logician left and 99 days passed--the dragons all realized that all the other dragons saw 98 pairs of red eyes, too. Since they never discussed eye color, they never knew that everyone else saw exactly what they saw.

 

[videoguy505]

I think ELM is right about the backwards thing. There are two numbers in Al's hands, one low and one high. There is exactly a 50% chance of either being uncovered first and therefore exactly a 50% chance Bob will be correct when he guesses higher or lower. Surely, the act of Bob "thinking of a number" before choosing a hand does not affect those percentages.

You've just been Monty Hall-ed.

 

[videoguy505]

But, some infinities are bigger than others.

 

[Aabye]

It also has nothing to do with infinite sets.

The wording of the initial riddle has everything to do with infinite sets. The strategy simply will not work for a boundless pool of integers. Do you agree that

1/infinity = 22/infinity = 6000000000/infinity = 0 ?

If not, then I understand where you are coming from, though your conclusion remains mathematically flawed.

I agree that 1/infinity = 22/infinity = 6000000000/infinity.But imagine that you choose a single number from the set of integers and ask me choose a single number from the set of integers as well. What are the chances that my number matches yours?

It can't be strictly zero. It's a limit approaching zero from the positive side.

Here's a link.

I should also add that I don't mean that the original problem has to do with infinite sets (specifically countably infinite sets). I meant that tinhat's criticism could be equally applied to finite sets, and so the concern he raised was not generated by the fact that the original problem deals with countably infinite sets. It was a separate issue.

 

[TinHat]

It also has nothing to do with infinite sets.

The wording of the initial riddle has everything to do with infinite sets. The strategy simply will not work for a boundless pool of integers. Do you agree that

1/infinity = 22/infinity = 6000000000/infinity = 0 ?

If not, then I understand where you are coming from, though your conclusion remains mathematically flawed.

I agree that 1/infinity = 22/infinity = 6000000000/infinity.But imagine that you choose a single number from the set of integers and ask me choose a single number from the set of integers as well. What are the chances that my number matches yours?

It can't be strictly zero. It's a limit approaching zero from the positive side.

Here's a link.

I should also add that I don't mean that the original problem has to do with infinite sets (specifically countably infinite sets). I meant that tinhat's criticism could be equally applied to finite sets, and so the concern he raised was not generated by the fact that the original problem deals with countably infinite sets. It was a separate issue.

my critism could specifically and categoriicaly not be applied to finite sets.none of the probabities calculated would be correct.

the case you're trying to make here is that (infinity + |a1-a2|)/2 > infinity/2.

This is not true

 

[TinHat]

******* spoilers below ********when it comes to arguments about infinity. logic breaks downIt simply is not logical that Bob could come up witha strategy that will yeild him better than 50%Infinity - 10 trillion is still infinity.If you don't buy that, then lets's calculate probabilitiesthere are 4 scenariosBoth Al's numbers are greater than Bob's n+.5 = 25 percent of the time - Bob will guess lower and Win half of the time - 12.5%One is high and one is lower bob will Win always - 50%Both Al's numbers are less than Bob's n+.5 = 25 percent of the time - Bob will guess higher and Win half of the time = 25 percent of the time - win half 12.5%Bob will win 75% of the time rightI'm sure somebody can use this to prove that .75 = .5 at which point all Rational Mathematics breaks down.********************************wrap your head around that one...

Logic doesn't break down when we start talking about infinite sets.Common sense breaks down.I certainly agree that you can take any finite number of elements out of an infinite set and the infinite set won't become finite. You can't thin an infinite set.If Al and Bob choose 3 different integers comletely at random, Bob will get the highest number 1/3 of the time, the middle number 1/3 of the time, and the lowest number 1/3 of the time - it's not 25%, 50%, and 25%. Finally, none of the solution relies on funky math. It's flies in the face of common sense, but given that we're talking about infinite sets, the fact that the solution doesn't pass the smell test doesn't mean anything. All sorts of facts about infinite sets fly in the face of common sense (e.g. that some infinite sets are strictly larger than other infinite sets - proved by Cantor a while ago).As I said, if Bob's edge is supposed to reduce to zero, then you'd have to show that his probability of generating a number that falls between Al's numbers has to hit zero. If someone could show that, the solution would be invalidated.I guess what I'm asking is, is this reasoning incorrect? If not, what's the problem with the solution?

This is why things break down with probability on infinite setsIf you out the three random numbers in a hat, the odds that Bob's number is the highest one is 1/3YETThe odds that Bobs' number higher than the Al's first number pulled out of the hat is 1/2The odds that Bobs' number higher than the Al's second number pulled out of the hat is 1/2combine these and we get that the odds that he will pick a number higher than Al's numbers is 1/41/4 = 1/3

OK, if that's the case, we can play a game.I'll generate a random integer and you generate two random integers by the same procedure. They can be any integers you like. When my number falls between your numbers, I'll give you a dollar. When my number is either the highest or the lowest of the three, you give me a dollar.Do you see that I will win this game?You're treating this as if it's two discrete cases - my number versus your first number and my number versus your second number. It's not. It's one scenario.It also has nothing to do with infinite sets. If we limited the possibilities to numbers 1, 2, and 3 and I randomly pick one and you take the other two, what is the probability that my number is higher than your first number? 1/2. What is the probability that my number is higher than your second number? 1/2. So is the probability that my number is 3 (the number that is higher than either of your numbers) 1 in 4? Of course not. It's 1 in 3.

Please read up on calculating probablities "with replacement" v. "without replacement"Big difference there. "With relacement" treats occurences as separate. The example you gave is "without replacement."if I have thre numbers A1, A2 and B1, is the probabilty that B1 > A1 dependent at all on the value of B2?

 

[Aabye]

******* spoilers below ********when it comes to arguments about infinity. logic breaks downIt simply is not logical that Bob could come up witha strategy that will yeild him better than 50%Infinity - 10 trillion is still infinity.If you don't buy that, then lets's calculate probabilitiesthere are 4 scenariosBoth Al's numbers are greater than Bob's n+.5 = 25 percent of the time - Bob will guess lower and Win half of the time - 12.5%One is high and one is lower bob will Win always - 50%Both Al's numbers are less than Bob's n+.5 = 25 percent of the time - Bob will guess higher and Win half of the time = 25 percent of the time - win half 12.5%Bob will win 75% of the time rightI'm sure somebody can use this to prove that .75 = .5 at which point all Rational Mathematics breaks down.********************************wrap your head around that one...

Logic doesn't break down when we start talking about infinite sets.Common sense breaks down.I certainly agree that you can take any finite number of elements out of an infinite set and the infinite set won't become finite. You can't thin an infinite set.If Al and Bob choose 3 different integers comletely at random, Bob will get the highest number 1/3 of the time, the middle number 1/3 of the time, and the lowest number 1/3 of the time - it's not 25%, 50%, and 25%. Finally, none of the solution relies on funky math. It's flies in the face of common sense, but given that we're talking about infinite sets, the fact that the solution doesn't pass the smell test doesn't mean anything. All sorts of facts about infinite sets fly in the face of common sense (e.g. that some infinite sets are strictly larger than other infinite sets - proved by Cantor a while ago).As I said, if Bob's edge is supposed to reduce to zero, then you'd have to show that his probability of generating a number that falls between Al's numbers has to hit zero. If someone could show that, the solution would be invalidated.I guess what I'm asking is, is this reasoning incorrect? If not, what's the problem with the solution?

This is why things break down with probability on infinite setsIf you out the three random numbers in a hat, the odds that Bob's number is the highest one is 1/3YETThe odds that Bobs' number higher than the Al's first number pulled out of the hat is 1/2The odds that Bobs' number higher than the Al's second number pulled out of the hat is 1/2combine these and we get that the odds that he will pick a number higher than Al's numbers is 1/41/4 = 1/3

OK, if that's the case, we can play a game.I'll generate a random integer and you generate two random integers by the same procedure. They can be any integers you like. When my number falls between your numbers, I'll give you a dollar. When my number is either the highest or the lowest of the three, you give me a dollar.Do you see that I will win this game?You're treating this as if it's two discrete cases - my number versus your first number and my number versus your second number. It's not. It's one scenario.It also has nothing to do with infinite sets. If we limited the possibilities to numbers 1, 2, and 3 and I randomly pick one and you take the other two, what is the probability that my number is higher than your first number? 1/2. What is the probability that my number is higher than your second number? 1/2. So is the probability that my number is 3 (the number that is higher than either of your numbers) 1 in 4? Of course not. It's 1 in 3.

Please read up on calculating probablities "with replacement" v. "without replacement"Big difference there. "With relacement" treats occurences as separate. The example you gave is "without replacement."if I have thre numbers A1, A2 and B1, is the probabilty that B1 > A1 dependent at all on the value of B2?

I'm familiar with the difference between probabilities with and without replacement. I'm not sure what the point is here though. I don't think it's germane to the conversation.In your example of A1, A2, and B1, the probabilty that B1 > A1 is not at all dependent at all on the value of A2 (you said B2, but I assume that you meant A2). I am in full agreement with you on that. I think that may be where the talk about with replacement v. without replacement is coming in. But I'm not disputing that.Let's use an infinite set - the integers.Here 3 non-identical integers are generated. One is (necessarily) the largest, one is (necessarily) the smallest, and one is (necessarily) smaller than one and larger than the other. Now those three numbers are distributed randomly - two to Al and one to Bob.We start from there. It doesn't matter at all what the integers happen to be, so long as none of them are identical.The question here is "given that A1 is smaller than B1, what is the probability that A2 is larger than A1?"You're saying 50/50.I'm saying that that is incorrect. If three non-identical integers are chosen randomly, then when we are given the information that A1 is smaller than B1, A2 will be larger than A1 67% of the time, not 50% of the time.Notice that I'm not saying that the value of A1, A2, and B1 are in any way dependent on one another. They are chosen independently and randomly. The only stipulation is that they can't be the same number.You don't need to use the numbers 1, 2, and 3. All you need are 3 non-indentical integers and they'll fulfill the same function as the 1, 2, and 3 of the example problem. Use -53,534,365, 1,646,365,376,556, and 3,453,453,451,262,346 if you like. They aren't dependent on one another. But the solution still works. If we are given that A1 is smaller than B1, there are 3 possibilities:1) A1 = -53,534,365 and B1 = 1,646,365,376,5562) A1 = -53,534,365 and B1 = 3,453,453,451,262,3463) A1 = 1,646,365,376,556 and B1 = 3,453,453,451,262,346All of those are equally probable.In case #1, A2 = 3,453,453,451,262,346, so A2 is larger than A1.In case #2, A2 = 1,646,365,376,556, so A2 is larger than A1.In case #3, A2 = -53,534,365, so A2 is smaller than A1.So A2 will be larger than A1 2/3 of the time.Nothing about this makes any of the values of A1, A2, or B1 dependent in any way on any of the others. And since we're picking from an infinite set, it's sampling with replacement. So I don't see what the criticism of the solution is.

 

[fatguyinalittlecoat]

If Bob's N+1/2 is above both of Al's numbers, then Bob always says "higher". So it doesn't matter which number he chooses. He'll randomly choose the higher number 50% of the time and the lower number 50% of the time. It's a total wash.

Are we sure this is right? When Bob says "higher," and it turns out both numbers are lower, there are a finite number of integers where he is correct and and an infinite number of integers where he's wrong. It shouldn't be a 50% chance.

 

[videoguy505]

If Bob's N+1/2 is above both of Al's numbers, then Bob always says "higher". So it doesn't matter which number he chooses. He'll randomly choose the higher number 50% of the time and the lower number 50% of the time. It's a total wash.

Are we sure this is right? When Bob says "higher," and it turns out both numbers are lower, there are a finite number of integers where he is correct and and an infinite number of integers where he's wrong. It shouldn't be a 50% chance.

It's right. N>A1 and N>A2. The odds of him picking A1 first or A2 first, whichever is the lower one, is 50-50. He'd reply "higher" to either one. If he picked A1 and A1<A2, then he gets it right. If he picked A1 and A1>A2, then he gets it wrong. If he picks A2 and A2<A1, he gets it right. If he picks A2 and A2>A1, he gets it wrong.We're only examining the case where we know that Bob's N is higher than both of Al's numbers. So it's 50-50 that he picks the lower of Al's numbers.When Bob picks an N that is not in between A1 and A2, it's squarely a 50-50 proposition that "higher" or "lower" will be the correct answer. When Bob picks an N that is in between A1 and A2, he'll always answer correctly.

 

[Aabye]

The odds that Bobs' number higher than the Al's first number pulled out of the hat is 1/2

The odds that Bobs' number higher than the Al's second number pulled out of the hat is 1/2

combine these and we get that the odds that he will pick a number higher than Al's numbers is 1/4

I honestly have no idea where you're getting the idea that the odds of my picking the higest number is 1 in 4.As I said before, write 1, 2, and 3 on three pieces of paper. Distribute them to Al and Bob. Nobody looks at their number(s).

The probability that A1 is larger than B1 is exactly 50%. - The possibilities are 1/2, 1/3, 2/3, 2/1, 3/1, and 3/2. They are all identically likely and three of the six cases are instances in which A1 is smaller than B1.

The probability that A2 is larger than B1 is exactly 50%. - The possibilities are 1/2, 1/3, 2/3/, 2/1, 3/1, and 3/2. They are all identically likely and three of the six cases are instances in which A2 is smaller than B1.

Combine these - 1/2 and 1/2. Do you get that Bob should have the highest number 1/4 of the time?

How is this reasoning different from your reasoning about infinite sets?

You might want to protest by saying that we're not talking about 1, 2, and 3. We're talking about an infinite set of integers, so the values of the first two numbers does not fix the value of the last number. But that's completely irrelevant. The three random numbers chosen by Al and Bob fall into the following categories: one is the largest, one is the smallest, and one is in the middle. That's all that needs to happen. It has nothing to do with the finitude of {1, 2, 3} as compared to a countably infinite set. All you need in the new set {A1, A2, B2} is that they are non-identical, meaning that one number will be the biggest, one will be the smallest, and one will be in the middle.

Perhaps this will put the nail in it:

By the quoted line of reasoning, the probability that Bob's number is the largest is 1/4. The probability that Al's first number is the higest is also 1/4 (just apply exactly the same argument and switch some terms around). And the probability that Al's second number is the highest is also 1/4. So the chance of either Bob's or one of Al's numbers being the highest is 1/4+1/4+1/4, or 3/4. So 1/4 of the time none of the three will be the largest number.

Also, by logical extension, the probabilty that B1 is the middle number is 1/2. The same argument applies to show that the probability that A1 is the middle number is 1/2 and also to show that the probability that A2 is the middle number is 1/2. So the chance that one of these three numbers will be the middle number is 3/2, which is impossible.

This is a mistake in calculating probabilities, not a problem with infinite sets.

 

[joey]

um....

anyone got a different logic problem to solve?

pretty please?

 

[fatguyinalittlecoat]

If Bob's N+1/2 is above both of Al's numbers, then Bob always says "higher". So it doesn't matter which number he chooses. He'll randomly choose the higher number 50% of the time and the lower number 50% of the time. It's a total wash.

Are we sure this is right? When Bob says "higher," and it turns out both numbers are lower, there are a finite number of integers where he is correct and and an infinite number of integers where he's wrong. It shouldn't be a 50% chance.

It's right. N>A1 and N>A2. The odds of him picking A1 first or A2 first, whichever is the lower one, is 50-50. He'd reply "higher" to either one. If he picked A1 and A1<A2, then he gets it right. If he picked A1 and A1>A2, then he gets it wrong. If he picks A2 and A2<A1, he gets it right. If he picks A2 and A2>A1, he gets it wrong.

I don't know if it's fair to characterize it that way. The numbers aren't all picked at the same time. At the time of the decision, he knows what N is and he knows what A1 is. When he says "higher," he is saying that, if A1 and A2 are both lower than N, A2 is between A1 and N. That's a finite number of integers.

 

[videoguy505]

If Bob's N+1/2 is above both of Al's numbers, then Bob always says "higher". So it doesn't matter which number he chooses. He'll randomly choose the higher number 50% of the time and the lower number 50% of the time. It's a total wash.

Are we sure this is right? When Bob says "higher," and it turns out both numbers are lower, there are a finite number of integers where he is correct and and an infinite number of integers where he's wrong. It shouldn't be a 50% chance.

It's right. N>A1 and N>A2. The odds of him picking A1 first or A2 first, whichever is the lower one, is 50-50. He'd reply "higher" to either one. If he picked A1 and A1<A2, then he gets it right. If he picked A1 and A1>A2, then he gets it wrong. If he picks A2 and A2<A1, he gets it right. If he picks A2 and A2>A1, he gets it wrong.

I don't know if it's fair to characterize it that way. The numbers aren't all picked at the same time. At the time of the decision, he knows what N is and he knows what A1 is. When he says "higher," he is saying that, if A1 and A2 are both lower than N, A2 is between A1 and N. That's a finite number of integers.

Huh?Al has his two numbers, A1 and A2. Bob picks one of them, but doesn't know which one. He picks the left hand or the right hand. It doesn't matter which one, since whichever one he picks, he's going to answer higher, because in this scenario A1 and A2 are both lower than Bob's N.So, he's either picking the lower number or the higher number of A1 and A2.Let me put it another way. They are picking from the subset of the three numbers 1, 2, and 3. Bob's got 3. So Al has 1 and 2 in his hands.Bob picks the left hand and says "the right hand number is higher."So he's going to be right half the time and wrong half the time. If the left hand number is 1, then he's right. If it's 2, then he's wrong.

 

[fatguyinalittlecoat]

If Bob's N+1/2 is above both of Al's numbers, then Bob always says "higher". So it doesn't matter which number he chooses. He'll randomly choose the higher number 50% of the time and the lower number 50% of the time. It's a total wash.

Are we sure this is right? When Bob says "higher," and it turns out both numbers are lower, there are a finite number of integers where he is correct and and an infinite number of integers where he's wrong. It shouldn't be a 50% chance.

It's right. N>A1 and N>A2. The odds of him picking A1 first or A2 first, whichever is the lower one, is 50-50. He'd reply "higher" to either one. If he picked A1 and A1<A2, then he gets it right. If he picked A1 and A1>A2, then he gets it wrong. If he picks A2 and A2<A1, he gets it right. If he picks A2 and A2>A1, he gets it wrong.

I don't know if it's fair to characterize it that way. The numbers aren't all picked at the same time. At the time of the decision, he knows what N is and he knows what A1 is. When he says "higher," he is saying that, if A1 and A2 are both lower than N, A2 is between A1 and N. That's a finite number of integers.

Huh?Al has his two numbers, A1 and A2. Bob picks one of them, but doesn't know which one. He picks the left hand or the right hand. It doesn't matter which one, since whichever one he picks, he's going to answer higher, because in this scenario A1 and A2 are both lower than Bob's N.So, he's either picking the lower number or the higher number of A1 and A2.

Sure, he's either picking the higher or lower number. But I'm focusing on Aabye's three categories. At the start of the game, the left hand and right hand are equally likely to be the higher number, right? Once the first number is revealed, the overall chances of having picked the higher number are still 50%. Although the overall chances are still 50% of having picked the higher number, we have learned some new information regarding Aabye's categories. We now know we are not in category 3 (both A1 and A2 are higher than N). We might be in category 1 (both A1 and A2 are lower than N) or we might be in category 2 (N is between A1 and A2). In every category 2 situation, the number in the unpicked hand is higher than the number in the picked hand. So out of the 50% situation, you are essentially taking a subset out in which the unpicked hand is always higher and assigning it to a different category. That means that the combinations remaining in the old category shouldn't stay the same -- there are now more "unpicked is lower" choices remaining.I think I am convinced that Aabye is wrong.

 

[fatguyinalittlecoat]

Let me put it another way. They are picking from the subset of the three numbers 1, 2, and 3. Bob's got 3. So Al has 1 and 2 in his hands.Bob picks the left hand and says "the right hand number is higher."So he's going to be right half the time and wrong half the time. If the left hand number is 1, then he's right. If it's 2, then he's wrong.

Are you assuming Bob knows that the subset is 1,2, and 3? Because that's not the same as the puzzle.

 

[Buckychudd]

No, it works with 4 as well. Here's the reasoning.

mytagid = Math.floor( Math.random() * 100 );document.write("

Call the 4 dragons A, B, C, and D.

It easiest, I think, to first suppose that A has blue eyes and then start with how B would reason

B sees one set of blue eyes and two sets of red eyes. So B supposes that his eyes are blue.

If this were the case, then C would see two sets of blue eyes and one set of red eyes. So C supposes that his eyes are blue.

If this were the case, then D would see three sets of blue eyes and conclude that his own eyes must be red.

So if A, B, and C have blue eyes and D has red eyes, D will die on the first day ("first" meaning the first day after the logician appears).

When D does not die on the first day, C realizes that his eyes cannot be blue. So C realizes that his own eyes must also be red. So C will die on the second day.

So if A and B have blue eyes and C and D have red eyes, C and D will die on the second day.

When C and D do not die on the second day, B realizes that his own eyes must also be red. So B will die on the third day.

So if A has blue eyes and B, C, and D all have red eyes, then B, C, and D will die on the third day.

Now, if no dragons die on the third day, A realizes that his own eyes must also be red. So A, B, C, and D will die on the fourth day.

Now look at all those cases. They cover all of the possible distributions of eye color.

If one dragon has red eyes, he will die on day 1.

If two dragons have red eyes, they will die on day 2.

If three dragons have red eyes, they will die on day 3.

If four dragons have red eyes, they will die on day 4.

So if no dragons die on day 3, each dragon realizes that the only possible distribution is 4 sets of red eyes, and so all dragons die on day 4.

*** SPOILER ALERT! Click this link to display the potential spoiler text in this box. ***");document.close();

It should be clear that this process can be repeated for any number of dragons.

I see. Dragons shouldn't think so hard about what color their eyes are though.

The saddest part is that if they weren't so rational, they wouldn't have to die. RIP Smartdragons.

If the dragons were so smart, one of them would have turned to another and said "You have red eyes" and immediately killed him. Thus saving himself and the other 97 dragons.

 

[ELM]

If Bob's N+1/2 is above both of Al's numbers, then Bob always says "higher". So it doesn't matter which number he chooses. He'll randomly choose the higher number 50% of the time and the lower number 50% of the time. It's a total wash.

Are we sure this is right? When Bob says "higher," and it turns out both numbers are lower, there are a finite number of integers where he is correct and and an infinite number of integers where he's wrong. It shouldn't be a 50% chance.

It's right. N>A1 and N>A2. The odds of him picking A1 first or A2 first, whichever is the lower one, is 50-50. He'd reply "higher" to either one. If he picked A1 and A1<A2, then he gets it right. If he picked A1 and A1>A2, then he gets it wrong. If he picks A2 and A2<A1, he gets it right. If he picks A2 and A2>A1, he gets it wrong.

I don't know if it's fair to characterize it that way. The numbers aren't all picked at the same time. At the time of the decision, he knows what N is and he knows what A1 is. When he says "higher," he is saying that, if A1 and A2 are both lower than N, A2 is between A1 and N. That's a finite number of integers.

Huh?Al has his two numbers, A1 and A2. Bob picks one of them, but doesn't know which one. He picks the left hand or the right hand. It doesn't matter which one, since whichever one he picks, he's going to answer higher, because in this scenario A1 and A2 are both lower than Bob's N.So, he's either picking the lower number or the higher number of A1 and A2.

Sure, he's either picking the higher or lower number. But I'm focusing on Aabye's three categories. At the start of the game, the left hand and right hand are equally likely to be the higher number, right? Once the first number is revealed, the overall chances of having picked the higher number are still 50%. Although the overall chances are still 50% of having picked the higher number, we have learned some new information regarding Aabye's categories. We now know we are not in category 3 (both A1 and A2 are higher than N). We might be in category 1 (both A1 and A2 are lower than N) or we might be in category 2 (N is between A1 and A2). In every category 2 situation, the number in the unpicked hand is higher than the number in the picked hand. So out of the 50% situation, you are essentially taking a subset out in which the unpicked hand is always higher and assigning it to a different category. That means that the combinations remaining in the old category shouldn't stay the same -- there are now more "unpicked is lower" choices remaining.I think I am convinced that Aabye is wrong.

i think you understand now what I was trying to say.Lets say Al's numbers are A and B. Lets say Bobs number is C. C equates to saying higher or lower. It actually has no mathmatical value.We can all agree that Bob has a 50/50 shot at A or B right?So if Bob picks A his odds are 50/50 that his number is higher.If Bob picks B his odds are 50/50 that his number is lower.If C>A bob will guess higher.if C<A bob will guess lower.If C<B he will say higherIf C>B he will say lowerA will be lower than B 50% of the time.A will be higher than B 50% of the time.If A is chosen first and C>A Bob says higher.there are now two sets of numbers available. B>A and B<A. If B is greater than A and C that is still the same group as B>A as obviously we have already established C>AIf A is chosen first and C<A bob will say lower. Again only Two sets available. B>A and B<A. B less than A and C is still the same group as B<A as we have already established C<A.Do this for B being chosen first and again there are always only two sets of numbers available.The fictitious third category is already part of the other two.

 

[munga30]

If A is chosen first and C>A Bob says higher.there are now two sets of numbers available. B>A and B<A.

This is wrong. Given the first sentence, there are three cases where C>A: B>C>A; C>A>B; C>B>ABob always gets the first one right, always gets the second one wrong, and always gets the third one right.

 

[goldenchild]

Here's one that has been called the the hardest recreational logical puzzle ever created.Three gods A , B , and C are called, in some order, True, False, and Random. True always speaks truly, False always speaks falsely, but whether Random speaks truly or falsely is a completely random matter. Your task is to determine the identities of A , B , and C by asking three yes-no questions; each question must be put to exactly one god. The gods understand English, but will answer in their own language, in which the words for yes and no are “da” and “ja”, in some order. You do not know which word means which.I don't know the answer and haven't really ever tried to work it out, but it is apparently solvable. Perhaps with our combined powers...

Can we get to this one now?

 

[ELM]

If A is chosen first and C>A Bob says higher.there are now two sets of numbers available. B>A and B<A.

This is wrong. Given the first sentence, there are three cases where C>A: B>C>A; C>A>B; C>B>ABob always gets the first one right, always gets the second one wrong, and always gets the third one right.

nope.the first and third sets you mentioned are in the same group. Both are members of B>A.

 

[The Man from Laramie]

What is the next row in this series:

1

11

21

1211

111221

312211

13112221

1113213211

?

 

[videoguy505]

What is the next row in this series:

1

11

21

1211

111221

312211

13112221

1113213211

?

mytagid = Math.floor( Math.random() * 100 );document.write("31131211131221

*** SPOILER ALERT! Click this link to display the potential spoiler text in this box. ***

");document.close();

 

[The Man from Laramie]

What is the next row in this series:

1

11

21

1211

111221

312211

13112221

1113213211

?

mytagid = Math.floor( Math.random() * 100 );document.write("31131211131221

*** SPOILER ALERT! Click this link to display the potential spoiler text in this box. ***

");document.close();

nice. You had seen this before, no?

 

[videoguy505]

Here's one that has been called the the hardest recreational logical puzzle ever created.

Three gods A , B , and C are called, in some order, True, False, and Random. True always speaks truly, False always speaks falsely, but whether Random speaks truly or falsely is a completely random matter. Your task is to determine the identities of A , B , and C by asking three yes-no questions; each question must be put to exactly one god. The gods understand English, but will answer in their own language, in which the words for yes and no are “da” and “ja”, in some order. You do not know which word means which.

I don't know the answer and haven't really ever tried to work it out, but it is apparently solvable. Perhaps with our combined powers...

Can we get to this one now?

http://people.ucsc.edu/~jburke/three_gods.pdf

 

[videoguy505]

What is the next row in this series:

1

11

21

1211

111221

312211

13112221

1113213211

?

mytagid = Math.floor( Math.random() * 100 );document.write("31131211131221

*** SPOILER ALERT! Click this link to display the potential spoiler text in this box. ***

");document.close();

nice. You had seen this before, no?

No, but when i read my lips move.

 

[Aabye]

Here's one that has been called the the hardest recreational logical puzzle ever created.

Three gods A , B , and C are called, in some order, True, False, and Random. True always speaks truly, False always speaks falsely, but whether Random speaks truly or falsely is a completely random matter. Your task is to determine the identities of A , B , and C by asking three yes-no questions; each question must be put to exactly one god. The gods understand English, but will answer in their own language, in which the words for yes and no are “da” and “ja”, in some order. You do not know which word means which.

I don't know the answer and haven't really ever tried to work it out, but it is apparently solvable. Perhaps with our combined powers...

Can we get to this one now?

http://people.ucsc.edu/~jburke/three_gods.pdf

Here's the solution I came up with. If someone wants to verify it, I'd be quite happy. I think it works.mytagid = Math.floor( Math.random() * 100 );document.write("

Since QUESTION 1 is the only confusing one, here's how it works:

The goal of the question is to determine whether God 2 is Random or not. If God 2 is Random, the QUESTIONS 2 and 3 are directed at God 3. If God 2 is not Random, QUESTIONS 2 and 3 are directed at God 2. It won't matter if God 1 is Random, since I'll be asking the next question to either God 2 or God 3, so I'll be sure to ask QUESTIONS 2 and 3 to a non-Random God.

QUESTION 1:

I ask God 1: "Is 'da' the answer to the question 'is it the case that exactly one of the following statements are true:

(1) You are False

(2) God 2 is Random?'"

So there are 8 cases to consider. Here they are, in order, w/ God 1's response and the reasoning behind it.

A. God 1 is True, "da" means "yes", and God 2 is Random.

(1) is false

(2) is true

So exactly one of the statements is true. So "da" is the answer to the question "is it the case that exactly one of these statements is true?"

So "yes" is the answer to the question "Is 'da' the answer to the question 'is it the case that exactly one of the following statements are true?". So God 1 tells the truth and says "da".

B. God 1 is True, "da" means "yes", and God 2 is False.

(1) is false

(2) is false

So neither statement is true. So "ja" is the answer to the question "is it the case that exactly one of these statements is true?"

So "no" is the anwer to the question "Is 'da' the answer to the question 'is it the case that exactly one of the following statements are true?". So God 1 tells the truth and says "ja".

C. God 1 is True, "da" means "no", and God 2 is Random.

(1) is false

(2) is true

So exactly one statement is true. So "ja" is the answer to the question "is it the case that exactly one of these statements is true?"

So "no" is the anwer to the question "Is 'da' the answer to the question 'is it the case that exactly one of the following statements are true?". So God 1 tells the truth and says "da".

D. God 1 is True, "da" means "no", and God 2 is False.

(1) is false

(2) is false

So neither statement is true. So "da" is the answer to the question "is it the case that exactly one of these statements is true?"

So "yes" is the answer to the question "Is 'da' the answer to the question 'is it the case that exactly one of the following statements are true?". So God 1 tells the truth and says "ja".

E. God 1 is False, "da" means "yes", and God 2 is Random

(1) is true

(2) is true

So both statements are true. So "ja" is the answer to the question "is it the case that exactly one of these statements is true?"

So "no" is the answer to the question "Is 'da' the answer to the question 'is it the case that exactly one of the following statements are true?". So God 1 lies and says "da".

F. God 1 is False, "da" means "yes", and God 2 is True

(1) is true

(2) is false

So exactly one statement is true. So "da" is the answer to the question "is it the case that exactly one of these statements is true?"

So "yes" is the answer to the question "Is 'da' the answer to the question 'is it the case that exactly one of the following statements are true?". So God 1 lies and says "ja".

G. God 1 is False, "da" means "no", and God 2 is Random

(1) is true

(2) is true

So both statements are true. So "da" is the answer to the question "is it the case that exactly one of these statements is true?"

So "yes" is the answer to the question "Is 'da' the answer to the question 'is it the case that exactly one of the following statements are true?". So God 1 lies and says "da".

H. God 1 is False, "da" means "no", and God 2 is True

(1) is true

(2) is false

So exactly one statement is true. So "ja" is the answer to the question "is it the case that exactly one of these statements is true?"

So "no" is the answer to the question "Is 'da' the answer to the question 'is it the case that exactly one of the following statements are true?". So God 1 lies and says "ja".

So in each case where God 2 is Random, God 1 answers "da" and in each case where God 2 is not Random, God 1 answers "ja".

QUESTION 2 (thanks to BostonFred, circa May 2006)

I now ask God 2 or 3: "Does 'da' mean 'yes'?"

If God 2 or 3 answers "da", then he is True. If God 2 or 3 answers "ja" then he is False.

QUESTION

I ask the same God: "Is 'da' the answer to the question "Is God 1 Random?'"

If I'm addressing True, "da" means that God 1 is Random and "ja" means that God 1 is False.

If I'm addressing False, "ja" means that God 1 is Random and "da" means that God 1 is True.

*** SPOILER ALERT! Click this link to display the potential spoiler text in this box. ***");document.close();

 

[Alias]

Here's one that has been called the the hardest recreational logical puzzle ever created.

Three gods A , B , and C are called, in some order, True, False, and Random. True always speaks truly, False always speaks falsely, but whether Random speaks truly or falsely is a completely random matter. Your task is to determine the identities of A , B , and C by asking three yes-no questions; each question must be put to exactly one god. The gods understand English, but will answer in their own language, in which the words for yes and no are “da” and “ja”, in some order. You do not know which word means which.

I don't know the answer and haven't really ever tried to work it out, but it is apparently solvable. Perhaps with our combined powers...

Can we get to this one now?

http://people.ucsc.edu/~jburke/three_gods.pdf

Here's the solution I came up with. If someone wants to verify it, I'd be quite happy. I think it works.mytagid = Math.floor( Math.random() * 100 );document.write("

Since QUESTION 1 is the only confusing one, here's how it works:

The goal of the question is to determine whether God 2 is Random or not. If God 2 is Random, the QUESTIONS 2 and 3 are directed at God 3. If God 2 is not Random, QUESTIONS 2 and 3 are directed at God 2. It won't matter if God 1 is Random, since I'll be asking the next question to either God 2 or God 3, so I'll be sure to ask QUESTIONS 2 and 3 to a non-Random God.

QUESTION 1:

I ask God 1: "Is 'da' the answer to the question 'is it the case that exactly one of the following statements are true:

(1) You are False

(2) God 2 is Random?'"

So there are 8 cases to consider. Here they are, in order, w/ God 1's response and the reasoning behind it.

A. God 1 is True, "da" means "yes", and God 2 is Random.

(1) is false

(2) is true

So exactly one of the statements is true. So "da" is the answer to the question "is it the case that exactly one of these statements is true?"

So "yes" is the answer to the question "Is 'da' the answer to the question 'is it the case that exactly one of the following statements are true?". So God 1 tells the truth and says "da".

B. God 1 is True, "da" means "yes", and God 2 is False.

(1) is false

(2) is false

So neither statement is true. So "ja" is the answer to the question "is it the case that exactly one of these statements is true?"

So "no" is the anwer to the question "Is 'da' the answer to the question 'is it the case that exactly one of the following statements are true?". So God 1 tells the truth and says "ja".

C. God 1 is True, "da" means "no", and God 2 is Random.

(1) is false

(2) is true

So exactly one statement is true. So "ja" is the answer to the question "is it the case that exactly one of these statements is true?"

So "no" is the anwer to the question "Is 'da' the answer to the question 'is it the case that exactly one of the following statements are true?". So God 1 tells the truth and says "da".

D. God 1 is True, "da" means "no", and God 2 is False.

(1) is false

(2) is false

So neither statement is true. So "da" is the answer to the question "is it the case that exactly one of these statements is true?"

So "yes" is the answer to the question "Is 'da' the answer to the question 'is it the case that exactly one of the following statements are true?". So God 1 tells the truth and says "ja".

E. God 1 is False, "da" means "yes", and God 2 is Random

(1) is true

(2) is true

So both statements are true. So "ja" is the answer to the question "is it the case that exactly one of these statements is true?"

So "no" is the answer to the question "Is 'da' the answer to the question 'is it the case that exactly one of the following statements are true?". So God 1 lies and says "da".

F. God 1 is False, "da" means "yes", and God 2 is True

(1) is true

(2) is false

So exactly one statement is true. So "da" is the answer to the question "is it the case that exactly one of these statements is true?"

So "yes" is the answer to the question "Is 'da' the answer to the question 'is it the case that exactly one of the following statements are true?". So God 1 lies and says "ja".

G. God 1 is False, "da" means "no", and God 2 is Random

(1) is true

(2) is true

So both statements are true. So "da" is the answer to the question "is it the case that exactly one of these statements is true?"

So "yes" is the answer to the question "Is 'da' the answer to the question 'is it the case that exactly one of the following statements are true?". So God 1 lies and says "da".

H. God 1 is False, "da" means "no", and God 2 is True

(1) is true

(2) is false

So exactly one statement is true. So "ja" is the answer to the question "is it the case that exactly one of these statements is true?"

So "no" is the answer to the question "Is 'da' the answer to the question 'is it the case that exactly one of the following statements are true?". So God 1 lies and says "ja".

So in each case where God 2 is Random, God 1 answers "da" and in each case where God 2 is not Random, God 1 answers "ja".

QUESTION 2 (thanks to BostonFred, circa May 2006)

I now ask God 2 or 3: "Does 'da' mean 'yes'?"

If God 2 or 3 answers "da", then he is True. If God 2 or 3 answers "ja" then he is False.

QUESTION

I ask the same God: "Is 'da' the answer to the question "Is God 1 Random?'"

If I'm addressing True, "da" means that God 1 is Random and "ja" means that God 1 is False.

If I'm addressing False, "ja" means that God 1 is Random and "da" means that God 1 is True.

*** SPOILER ALERT! Click this link to display the potential spoiler text in this box. ***");document.close();

Does this mean all 3 questions are asked to the same god, because if that is true this is not solvable.

 

[Aabye]

Here's one that has been called the the hardest recreational logical puzzle ever created.

Three gods A , B , and C are called, in some order, True, False, and Random. True always speaks truly, False always speaks falsely, but whether Random speaks truly or falsely is a completely random matter. Your task is to determine the identities of A , B , and C by asking three yes-no questions; each question must be put to exactly one god. The gods understand English, but will answer in their own language, in which the words for yes and no are “da” and “ja”, in some order. You do not know which word means which.

I don't know the answer and haven't really ever tried to work it out, but it is apparently solvable. Perhaps with our combined powers...

Can we get to this one now?

http://people.ucsc.edu/~jburke/three_gods.pdf

Here's the solution I came up with. If someone wants to verify it, I'd be quite happy. I think it works.mytagid = Math.floor( Math.random() * 100 );document.write("

Since QUESTION 1 is the only confusing one, here's how it works:

The goal of the question is to determine whether God 2 is Random or not. If God 2 is Random, the QUESTIONS 2 and 3 are directed at God 3. If God 2 is not Random, QUESTIONS 2 and 3 are directed at God 2. It won't matter if God 1 is Random, since I'll be asking the next question to either God 2 or God 3, so I'll be sure to ask QUESTIONS 2 and 3 to a non-Random God.

QUESTION 1:

I ask God 1: "Is 'da' the answer to the question 'is it the case that exactly one of the following statements are true:

(1) You are False

(2) God 2 is Random?'"

So there are 8 cases to consider. Here they are, in order, w/ God 1's response and the reasoning behind it.

A. God 1 is True, "da" means "yes", and God 2 is Random.

(1) is false

(2) is true

So exactly one of the statements is true. So "da" is the answer to the question "is it the case that exactly one of these statements is true?"

So "yes" is the answer to the question "Is 'da' the answer to the question 'is it the case that exactly one of the following statements are true?". So God 1 tells the truth and says "da".

B. God 1 is True, "da" means "yes", and God 2 is False.

(1) is false

(2) is false

So neither statement is true. So "ja" is the answer to the question "is it the case that exactly one of these statements is true?"

So "no" is the anwer to the question "Is 'da' the answer to the question 'is it the case that exactly one of the following statements are true?". So God 1 tells the truth and says "ja".

C. God 1 is True, "da" means "no", and God 2 is Random.

(1) is false

(2) is true

So exactly one statement is true. So "ja" is the answer to the question "is it the case that exactly one of these statements is true?"

So "no" is the anwer to the question "Is 'da' the answer to the question 'is it the case that exactly one of the following statements are true?". So God 1 tells the truth and says "da".

D. God 1 is True, "da" means "no", and God 2 is False.

(1) is false

(2) is false

So neither statement is true. So "da" is the answer to the question "is it the case that exactly one of these statements is true?"

So "yes" is the answer to the question "Is 'da' the answer to the question 'is it the case that exactly one of the following statements are true?". So God 1 tells the truth and says "ja".

E. God 1 is False, "da" means "yes", and God 2 is Random

(1) is true

(2) is true

So both statements are true. So "ja" is the answer to the question "is it the case that exactly one of these statements is true?"

So "no" is the answer to the question "Is 'da' the answer to the question 'is it the case that exactly one of the following statements are true?". So God 1 lies and says "da".

F. God 1 is False, "da" means "yes", and God 2 is True

(1) is true

(2) is false

So exactly one statement is true. So "da" is the answer to the question "is it the case that exactly one of these statements is true?"

So "yes" is the answer to the question "Is 'da' the answer to the question 'is it the case that exactly one of the following statements are true?". So God 1 lies and says "ja".

G. God 1 is False, "da" means "no", and God 2 is Random

(1) is true

(2) is true

So both statements are true. So "da" is the answer to the question "is it the case that exactly one of these statements is true?"

So "yes" is the answer to the question "Is 'da' the answer to the question 'is it the case that exactly one of the following statements are true?". So God 1 lies and says "da".

H. God 1 is False, "da" means "no", and God 2 is True

(1) is true

(2) is false

So exactly one statement is true. So "ja" is the answer to the question "is it the case that exactly one of these statements is true?"

So "no" is the answer to the question "Is 'da' the answer to the question 'is it the case that exactly one of the following statements are true?". So God 1 lies and says "ja".

So in each case where God 2 is Random, God 1 answers "da" and in each case where God 2 is not Random, God 1 answers "ja".

QUESTION 2 (thanks to BostonFred, circa May 2006)

I now ask God 2 or 3: "Does 'da' mean 'yes'?"

If God 2 or 3 answers "da", then he is True. If God 2 or 3 answers "ja" then he is False.

QUESTION

I ask the same God: "Is 'da' the answer to the question "Is God 1 Random?'"

If I'm addressing True, "da" means that God 1 is Random and "ja" means that God 1 is False.

If I'm addressing False, "ja" means that God 1 is Random and "da" means that God 1 is True.

*** SPOILER ALERT! Click this link to display the potential spoiler text in this box. ***");document.close();

Does this mean all 3 questions are asked to the same god, because if that is true this is not solvable.

Correct, because that god could turn out to be Random and you can't get any reliable information about the other gods from his responses.I'm sure that it means that each question must be asked to exactly one of the gods, not that each question must be asked to the same god.

 

[ELM]

there are two jugs. one 5 gallon and one 3 gallon.

How do you get exactly four gallons of water in the 5 gallon jug?

It is easy, but there is more than one way to solve it. For many people doing it a different way is hard to arrive at because they get the first one so easily.

give it a whirl.

ps this was in Die Hard with a vengeance.

 

[videoguy505]

there are two jugs. one 5 gallon and one 3 gallon.

How do you get exactly four gallons of water in the 5 gallon jug?

It is easy, but there is more than one way to solve it. For many people doing it a different way is hard to arrive at because they get the first one so easily.

give it a whirl.

ps this was in Die Hard with a vengeance.

mytagid = Math.floor( Math.random() * 100 );document.write("Fastest way I can think of is fill the 5, pour it into the 3, leaving 2 inside the 5. Empty the 3, pour the remaining 2 in the 3. Fill the 5 again, then use it to fill (by pouring in 1 more gallon) the 3. There is now 4 in the 5.

I can't think of a way to do it by filling the 3 first, though.

*** SPOILER ALERT! Click this link to display the potential spoiler text in this box. ***

");document.close();

 

[Blind Tiger]

there are two jugs. one 5 gallon and one 3 gallon.

How do you get exactly four gallons of water in the 5 gallon jug?

It is easy, but there is more than one way to solve it. For many people doing it a different way is hard to arrive at because they get the first one so easily.

give it a whirl.

ps this was in Die Hard with a vengeance.

mytagid = Math.floor( Math.random() * 100 );document.write("Fastest way I can think of is fill the 5, pour it into the 3, leaving 2 inside the 5. Empty the 3, pour the remaining 2 in the 3. Fill the 5 again, then use it to fill (by pouring in 1 more gallon) the 3. There is now 4 in the 5.

I can't think of a way to do it by filling the 3 first, though.

*** SPOILER ALERT! Click this link to display the potential spoiler text in this box. ***

");document.close();

 

[Otis]

http://img124.imageshack.us/my.php?image=trigridzc2.gif

Explain this one.

Oh the humanity.

 

[videoguy505]

Not really a logic problem, but I always found it interesting. Prior to 1995, when the system started to fill up, area codes always had a 1 or 0 in the middle digit. This was done to differentiate areas, the ones with a middle 1 were mainly cities, the middle 0s were mainly for suburbs and large state areas.

Anyway, when the area code system was developed, the area codes were given out in a specific order.

Here are some of the original area codes:

201: New Jersey

202: DC

203: Connecticut

212: NYC

213: LA

214: Dallas

215: Philadelphia

301: Maryland

302: Delaware

303: Colorado

312: Chicago

313: Detroit

314: St. Louis

401: Rhode Island

412: Pittsburgh

801: Utah

802: Vermont

Can you figure out what the order is?

 

[The Man from Laramie]

Not really a logic problem, but I always found it interesting. Prior to 1995, when the system started to fill up, area codes always had a 1 or 0 in the middle digit. This was done to differentiate areas, the ones with a middle 1 were mainly cities, the middle 0s were mainly for suburbs and large state areas.Anyway, when the area code system was developed, the area codes were given out in a specific order.Here are some of the original area codes:201: New Jersey202: DC203: Connecticut212: NYC213: LA214: Dallas215: Philadelphia301: Maryland302: Delaware303: Colorado312: Chicago313: Detroit314: St. Louis401: Rhode Island412: Pittsburgh801: Utah802: VermontCan you figure out what the order is?

some part of it is density, no?

 

[spOOfy]

Not really a logic problem, but I always found it interesting. Prior to 1995, when the system started to fill up, area codes always had a 1 or 0 in the middle digit. This was done to differentiate areas, the ones with a middle 1 were mainly cities, the middle 0s were mainly for suburbs and large state areas.Anyway, when the area code system was developed, the area codes were given out in a specific order.Here are some of the original area codes:201: New Jersey202: DC203: Connecticut212: NYC213: LA214: Dallas215: Philadelphia301: Maryland302: Delaware303: Colorado312: Chicago313: Detroit314: St. Louis401: Rhode Island412: Pittsburgh801: Utah802: VermontCan you figure out what the order is?

some part of it is density, no?

I believe that is correct. In those days when area codes came to be, they had rotary phones (anyone here remember those?), so higher numbers were a P.I.T.A. So, they gave lower numbers to the more heavily-populated areas, I believe.

 

[The Man from Laramie]

some part of it is density, no?

I believe that is correct. In those days when area codes came to be, they had rotary phones (anyone here remember those?), so higher numbers were a P.I.T.A. So, they gave lower numbers to the more heavily-populated areas, I believe.

Always confused me why Boston got stuck with 617.

 

[videoguy505]

some part of it is density, no?

I believe that is correct. In those days when area codes came to be, they had rotary phones (anyone here remember those?), so higher numbers were a P.I.T.A. So, they gave lower numbers to the more heavily-populated areas, I believe.

Always confused me why Boston got stuck with 617.

Right, they are mainly in order of population by speed in which they can be dialed on a rotary phone.The best number, 212, went to NYC, the most populated city. LA and Chicago got the next two, 213 and 312.I believe Boston got demoted because in the NE corridor, there were too many similarly numbered cities already. They got royally screwed, though, with the long dial time of 617.

 

[The Man from Laramie]

2 1 22 1 33 1 22 1 43 1 34 1 22 1 53 1 44 1 35 1 22 1 63 1 54 1 45 1 36 1 22 1 73 1 64 1 55 1 46 1 37 1 22 1 83 1 74 1 65 1 56 1 47 1 38 1 22 1 93 1 84 1 75 1 66 1 57 1 48 1 39 1 22 0 13 1 94 1 85 1 76 1 67 1 58 1 49 1 32 0 23 0 14 1 95 1 86 1 77 1 68 1 59 1 42 0 33 0 24 0 15 1 96 1 87 1 78 1 69 1 52 0 43 0 34 0 25 0 16 1 97 1 88 1 79 1 62 0 53 0 44 0 35 0 26 0 17 1 98 1 89 1 72 0 63 0 54 0 45 0 36 0 27 0 18 1 99 1 82 0 73 0 64 0 55 0 46 0 37 0 28 0 19 1 92 0 83 0 74 0 65 0 56 0 47 0 38 0 29 0 12 0 93 0 84 0 75 0 66 0 57 0 48 0 39 0 23 0 94 0 85 0 76 0 67 0 58 0 49 0 34 0 95 0 86 0 77 0 68 0 59 0 45 0 96 0 87 0 78 0 69 0 56 0 97 0 88 0 79 0 67 0 98 0 89 0 78 0 99 0 89 0 9

I would guess that would be the ease of dial order.

 

[GTBilly]

Been reading so long that I do not remember the 30 cent question and if it had been answered. One of the coins is not a nickel:

The other one is a nickel and the one that is not a nickel is a quarter.

 

[fatguyinalittlecoat]

I am very bored, so anybody up for some conundrums? For those unfamiliar, here's how it works:1) I describe a situation that seems puzzling.2) You ask yes or no questions in an attempt to figure out what happened.3) I respond to the questions appropriately.4) We all have lots of fun.Don't participate if you're familiar with any of the conundrums. Here's the first one:

A man is lying drowned in a dead forest.

Ask away.

 

[WetDream]

What did the man look like?

 

[fatguyinalittlecoat]

What did the man look like?

Sorry, I can't answer that. Questions need to be in a "yes/no" format. P.S. But he was probably really sexy.

 

[Dragons]

Is he clothed?

Is there a container of water anywhere?

Is there an empty container anywhere?

Is there an ocean lake, river, stream, creek, puddle, or any other body of water nearby?