This is a good oneSuppose you're on a game show, and you're given the choice of 3 doors: Behind one door is a car; behind the other 2, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?
Enjoying stabbsy's balls. Think I've arrived at a solution for the extension bit. Took a couple of tricky (for me at least) insights and full pen and paper job to get there. Also assumes you're allowed to number/keep track of balls between weighings? Solution is pretty long and convoluted so maybe missed an elegant simple solution?
I’m still struggling with the blue eyed islanders. I think it’s one of these proof by induction things...
Instead, if the host joins the remaining 2 boxes together and asks you, the choice becomes obvious. Because you know the host will not throw the prize away, this is identical to joining the remaining 2 boxes together....If you think of it in this way with more and more boxes, it becomes more obvious.
Quote from: Banana finger on August 07, 2021, 03:14:59 pmInstead, if the host joins the remaining 2 boxes together and asks you, the choice becomes obvious. Because you know the host will not throw the prize away, this is identical to joining the remaining 2 boxes together....If you think of it in this way with more and more boxes, it becomes more obvious.That’s a really elegant solution! I’ve always just worked through the numbers, but even though you know it’s right it still feels unintuitive. Great way of illustrating what’s going on.
On an otherwise deserted and isolated island, 200 perfect logicians are stranded. The islanders are perfectly logical in every decision they make, and they will not do anything unless they are absolutely certain of the outcome. However, they cannot communicate with each other. They are forbidden from speaking with one another, or signing, or writing messages in the sand, else they be shot dead by the captain of a mysterious ship that visits the island each night.Of the 200 islanders, 100 have blue eyes, and 100 have brown eyes. However, no individual knows what color their own eyes are. There are no reflective surfaces on the island for the inhabitants to see a reflection of their own eyes. They can each see the 199 other islanders and their eye colors, but any given individual does not know if his or her own eyes are brown, blue, or perhaps another color entirely. And remember, they cannot communicate with each other in any way under penalty of death.Each night, when the captain of the ship comes, the islanders have a chance to leave the barren and desolate spit of land they have been marooned on. If an islander tells the captain the color of his or her own eyes, they may board the ship and leave. If they get it wrong, they will be shot dead.Now, there is one more person on the island: the guru, who the islanders know to always tell the truth. The guru has green eyes. One day, she stands up before all 200 islanders and says:I see a person with blue eyes.Who leaves the island? And when do they leave?
It's been around quite a bit so you may have seen it already, but I enjoyed the blue eyed islanders puzzleQuoteOn an otherwise deserted and isolated island, 200 perfect logicians are stranded. The islanders are perfectly logical in every decision they make, and they will not do anything unless they are absolutely certain of the outcome. However, they cannot communicate with each other. They are forbidden from speaking with one another, or signing, or writing messages in the sand, else they be shot dead by the captain of a mysterious ship that visits the island each night.Of the 200 islanders, 100 have blue eyes, and 100 have brown eyes. However, no individual knows what color their own eyes are. There are no reflective surfaces on the island for the inhabitants to see a reflection of their own eyes. They can each see the 199 other islanders and their eye colors, but any given individual does not know if his or her own eyes are brown, blue, or perhaps another color entirely. And remember, they cannot communicate with each other in any way under penalty of death.Each night, when the captain of the ship comes, the islanders have a chance to leave the barren and desolate spit of land they have been marooned on. If an islander tells the captain the color of his or her own eyes, they may board the ship and leave. If they get it wrong, they will be shot dead.Now, there is one more person on the island: the guru, who the islanders know to always tell the truth. The guru has green eyes. One day, she stands up before all 200 islanders and says:I see a person with blue eyes.Who leaves the island? And when do they leave?
...The gods understand English, but will answer all questions 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.
Quote from: Steve R on August 07, 2021, 02:49:58 pmEnjoying stabbsy's balls. Think I've arrived at a solution for the extension bit. Took a couple of tricky (for me at least) insights and full pen and paper job to get there. Also assumes you're allowed to number/keep track of balls between weighings? Solution is pretty long and convoluted so maybe missed an elegant simple solution?Don’t think there’s an elegant solution, I used pen and paper. I think you have to go through all the options and yes, you can keep track of balls between weighings.
Three very logical friends, Amy, Bob and Carol, sat in a circle wearing hats. Each hat had a number on it so all three could see the others' numbers but not their own. They were told that the numbers were three different positive single digits.They each made a statement in turn and were told to raise their hands when they knew their own number. Amy said: "Carol's number is greater than Bob's." No one raised their hand.Bob then said: "The sum of Amy's and Carol's numbers is even." On hearing this, Carol raised her hand. Even then, Amy and Bob did not raise theirs. But after a pause, and once it had become clear that Amy wasn't going to raise her hand, Bob raised his and then Amy raised hers.What were Amy's, Bob's and Carol's numbers?
Quote from: Steve R on August 07, 2021, 11:55:08 pmThree very logical friends, Amy, Bob and Carol...NSFW : Amy 6, Bob 4, Carol 8?
Three very logical friends, Amy, Bob and Carol...
Ok so this one is fucking 9AThree gods A, B, and C are called, in no particular 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 all questions 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.
Close but one number is wrong
Quote from: Banana finger on August 07, 2021, 06:41:57 pmOk so this one is fucking 9AThree gods A, B, and C are called, in no particular 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 all questions 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.Here is an answer (not my own, but posting because once you see it the puzzle is significantly downgraded): NSFW : Let E be a function that takes you from a question q to the question [If I asked you 'q' in your current mental state would you say 'ja'?]If you ask any god E(q), a response of 'ja' indicates that the answer to q is affirmative. If they give a response of 'da' this indicates that the answer to q is negative. Given this, the puzzle becomes incredibly simple. You just ask the first god E(are you random)? If they answer 'ja then you know they are random, if they answer 'da' then you know they are not random. If you know the first is random you then ask the second god E(are you true?). If you get a 'ja' you know the second is true, and therefore the final one is false. If you get a 'da' you know the second is false, and the final one is true. If the first god answers 'da' to the first question you know they are not random. So, you ask if they are true. If 'ja' then they are true, if 'da' they are false. You can then just proceed as before and do the others by process of elimination. Ask second if they are random. If 'ja' then they are random and final one is whatever is left. If 'da' then the final one is random, and middle one is whatever is left. Here is why a response of 'ja' to E(q) always entails an affirmative answer: Suppose the question is 'are you random?', now consider the possible responses we get from each god when asked E(q): God = True: Ja = yes. The answer to q is 'da' (no), so true will answer q with 'da' (no). So true will answer E(q) 'da' (no). Ja = no. The answer to q is 'ja', so true will answer q with 'ja' (no). So, true will answer E(q) 'da' (yes)God = False: Ja = yes. The answer to q is 'da' (no), so false will answer q with 'ja' (yes). So, false will answer E(q) with 'da' (no).Da = yes. The answer to q is 'ja' (no), so false will answer q with 'da' (yes). So, false will answer E(q) with 'da' (yes). God = Random. Random is telling the truth (T(random))Ja = yesThe answer to q is 'ja' (yes), so T(random) will answer q with 'ja' (yes). So, T(random) will answer E(q) with 'ja' (yes). Ja = noThe answer to q is 'da' (yes), so, T(random) will answer q with 'da' (yes). So, T(random) will answer E(q) with 'ja' (no). Random is lying (F(random))Ja = yesThe answer to q is 'ja' (yes), so F(random) will answer q with 'da' (no). So, F(random) will answer E(q) with 'ja' (yes). Ja = noThe answer to q is 'da' (yes), so F(random) will answer q with 'ja' (no). So, F(random) will answer E(q) with 'ja' (no).