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On December 07 2010 20:57 aseq wrote:
Posted this on Page 4, no replies?
Two Envelopes Say you're given two envelopes, each one with a real amount of money in it. You have no clue how much is in each, but the only information that's given about it is that one contains double the amount of the other. You have to pick one, you can open it, look at the amount, and you can decide to then switch and pick the other.
Now explain this paradox: Just picking one (which would give you amount X) is not as good as picking one first, then switching to the other envelope, which gives you (0.5 * X * 0.5 + 0.5 * X * 2) / 2, which is 1.25 * X on average. But that means you could just divert from your initial choice and get more? Please explain!
You're wrong that X is your expected value for your first choice. If y is the amount in the bad envelope, you expect to see (y+2y)/2 dollars in either envelope.
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On December 06 2010 11:28 Trion wrote: Alright I bet no one will get this one.
Prove that there is no solution for x^n+y^n=z^n when n>2 and all numbers are natural.
One of the number theorists at my university has this; it's a T-shirt. The above pictures are the front and back. AWESOME.
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why would a man call a phone, when he has a perfectly good and working and accurate watche and a equally functional clock?
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On December 21 2010 10:35 Thoreezhea wrote: why would a man call a phone, when he has a perfectly good and working and accurate watche and a equally functional clock?
He wishes to order a Pizza?
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Figured I'd wake this thread up with a few deceptively easy ones:
1. 10 tigers come across a dead deer in the wild. They are unwilling to share the meal, so the meal would go to any one (the strongest) among them. However, if a tiger eats the deer, that tiger will suffer from food coma and is liable to get eaten by any one among the remaining tigers. That tiger, in turn, will suffer from food coma and is liable to get eaten by any one of the remaining tigers, and so on. Will any tiger eat the deer?
2. There's a game I don't know the name of, but it involves stones (or pebbles) placed in rows and columns like this:
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
There could be as many (whole) rows and columns as you'd like. The game has two players who take turns picking stones to remove. The stone they pick and *all* stones over and to the right of it will be removed, including any stones diagonally up and to the right of it. So if one picks the stone marked with a plus below, the stones marked with a dash will *also* be removed. The goal of the game is to avoid picking the last stone. Whoever has to pick that stone has lost.
. . . - - . . . - - . . . + - . . . . . . . . . . . . . . .
Who will win? The player who goes first, or the player who goes second? (Assume they play perfect).
Note that this is deceptively easy. You don't have to go through any trial-and-error or anything of the sorts, just come to a rather simple realization.
Good luck!
edit: Both solved by qrs (just browse through his posts on this page if you want the answer).
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On January 25 2011 16:47 Mayfly wrote:Figured I'd wake this thread up with a few deceptively easy ones: 1. 10 tigers come across a dead deer in the wild. They are unwilling to share the meal, so the meal would go to any one (the strongest) among them. However, if a tiger eats the deer, that tiger will suffer from food coma and is liable to get eaten by any one among the remaining tigers. That tiger, in turn, will suffer from food coma and is liable to get eaten by any one of the remaining tigers, and so on. Will any tiger eat the deer? 2. There's a game I don't know the name of, but it involves stones (or pebbles) placed in rows and columns like this: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . There could be as many (whole) rows and columns as you'd like. The game has two players who take turns picking stones to remove. The stone they pick and *all* stones over and to the right of it will be removed, including any stones diagonally up and to the right of it. So if one picks the stone marked with a plus below, the stones marked with a dash will *also* be removed. The goal of the game is to avoid picking the last stone. Whoever has to pick that stone has lost. . . . - - . . . - - . . . + - . . . . . . . . . . . . . . . Who will win? The player who goes first, or the player who goes second? (Assume they play perfect). Note that this is deceptively easy. You don't have to go through any trial-and-error or anything of the sorts, just come to a rather simple realization. Good luck!
1. Yes. Assuming all the tigers are rational, if I were one of the tigers, I would eat a deer. I could sleep in peace knowing if any of the tigers chose to eat me, he too would die. Because of this, no tiger would dare eat me.
2. The player who goes first. On your first move you pick this pebble: . - - - - . - - - - . - - - - . x - - - . . . . . Then regardless of what the other player does, just mirror it, and you can't lose.
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Both answers wrong. You're on the right track, though. However, if you were to pick the stone you said, you would definitely lose to a perfect player. You're thinking too much about which stone to pick, it's more about who will win.
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What move set could a perfect player do that would allow him to win from that spot.
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+ . . . . . . . . .
edit: That's why I made it asymmetrical. If it was symmetrical (i.e. a perfect square) it would've been a brilliant move on your part. But as I said, it could be any amount of rows and columns, and I'm looking for a general proof that one of the two players can force a win.
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Oh, ok. I didn't realize it could be any number of rows or columns.
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For the tigers: + Show Spoiler +If there is one waking tiger left + food, of course he eats the food. If there are two waking tigers left + food, neither tiger dares eat the food because he will be eaten after he falls asleep. Therefore if there are three tigers left + food, one tiger can safely eat the food. Therefore if there are four tigers left, no tiger can safely turn himself into food. etc.: if there are an even number of tigers left, no one eats. 10 is an even number. For the stones: + Show Spoiler +If it's true that the game is either always won by the first player or always won by the second player, then it must be always won by the first player. Proof: suppose it were always won by the second player. Then on his first move, the first player could just slice off a row or column to reach a different version of the game where he is the second player, and thus the winner. This leads to a contradiction, so the game must be always won by the first player.
However, this doesn't prove that the game is always won by the first player--maybe some versions of the game are won by the first player and some by the second.
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Correct, dear sir! (edit: for the tigers).
The stones still need some work but you're very much on the right track. (Your proof is actually quite valid, but can be made even more general).
edit 2: Some hints: The game is won by player number + Show Spoiler +. You need to already know that this game (and every game that has no ties) has a means to win it by perfect play, since it has no ties. It's not that proof I'm looking for here.
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LOL damnit i love this, but can someone explain to me something? the one about the monks..ive read the answers and stuff and apparently most of u say if no1 kills themselves within the 1st night it means 2 or more have red eyes...if no1 kills themselves within the 2nd night that means 3 or more have red eyes...etc, can u guys explain this? i dont understand >_< thanks
TL;DR: why, if no monk commits suicide on night 1, means that 2+ monks have red eyes?
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Some riddles: Horses and Tracks + Show Spoiler +You have 25 horses, and each one runs at a unique, constant speed. You have exactly one, straight race track, and it has five lanes, each lane only being able to fit one horse. You have no stopwatch, so the only way to determine a horse's speed is to race it directly against another horse, and see which one is faster. You can put five horses on the track, have them all go, and then record the order of the five. That counts as exactly one race. Your job is to find the three fastest horses in as few races as possible Answer: + Show Spoiler +You can find the three fastest horses in 7 races. First, split the horses into five groups, 1, 2, 3, 4, and 5. Race each of the groups, and then give the horses themselves identifiers, so now you have horse 1A, 1B, 1C, 1D, 1E, 2A, 2B, and so on. In each group, the fastest one is A and the slowest is E. Clearly, all horses that are D or E have no chance of being in the top three fastest, so you can ignore them. Now, race 1A, 2A, 3A, 4A, and 5A. For simplicity's sake, let's say that they finished in numerical order. That means that 4A and 5A are definitely not the top three fastest, and therefore, neither are 4B, 4C, 5B, or 5C. Ignore all of these. You also know that 1A is the fastest of all 25 horses. 1A beat 2A, who in turn beat 2B. 2C is therefore not in the top 3. 2A also beat 3A, so 3B and 3C are not in the top 3. This leaves you with five horses: 1B, 1C, 2A, 2B, and 3A. Race these five, and take the top two finishers. That is seven races
The Cruel King + Show Spoiler +There is a kingdom with population 51. There are 50 people, and the king. The king decides one day to line the fifty people up, and put either a red or blue hat on each person. The people are all facing the same way, so that the 50th person can see the hats of the 49 in front of him, but not his own. The first person, of course, can not see any hats. The king orders them to go down the line, from the 50th to the 1st, and guess the color of their own hat. If they get it wrong, the king will shoot them on the spot. If they get it right, they are allowed to live. Before they are lined up, they are allowed to devise a strategy to save as many lives as possible. What strategy do they use, and how many lives can they save? The strategy has to guarantee a certain number of people saved, so even though you could possibly save all fifty people if they all guess red, the worst case scenario is that they all die. Answer: + Show Spoiler +They can guarantee that 49 people are saved, and the other person has a 50/50 chance of living. Their strategy is for the 50th person (i.e. the first to guess) to look in front of him, and count the number of red hats. If it is even, he will guess red. If it is odd, he will guess blue. His guess has absolutely nothing to do with the color of his own hat. Let's say he guesses red. Then the 49th person knows that between him and the 48 people in front of him, there are an even number of red hats. He now counts the number of red hats in front of him. If it is odd, then the last red hat must be on his head. If it is even, then he is wearing a blue hat. He guesses correctly, and it just continues down the line. 50th guy is kind of screwed, and everyone in front of him lives.
The Cruel King, part 2 + Show Spoiler +The Cruel King is pissed that 49 (possibly 50?) people survived his first little activity. He devises a new one. He takes a group of seven people, and also seven index cards. On these index cards, he writes an integer between 1 and 7, inclusive, and he may repeat numbers. These seven people will be put into a room, and the card will be placed on their forehead, such that each person can see the other six cards, but not his own. Once in the room, they are not allowed to discuss anything. Then, all at once, they must each shout out a number. If at least one person guesses his own number, they are all saved. If nobody guesses their own number correctly, they are all killed. They are again allowed to devise a strategy prior to this challenge. What strategy do they use? Answer: + Show Spoiler +Each of the people will go into the room assuming that the grand total is a different least residue mod 7. There are exactly 7 least residues mod 7 (0 through 6), and of course, the total sum must be one of these. Then each person adds up the other six numbers, and determines what their own number must be given their respective assumptions. The person who has the right guess mod 7 will guess his own number right.
Oh, and this: http://www.cut-the-knot.org/Probability/LightBulbs.shtml
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On January 25 2011 17:18 Mayfly wrote:Correct, dear sir! (edit: for the tigers). The stones still need some work but you're very much on the right track. (Your proof is actually quite valid, but can be made even more general). edit 2: Some hints: The game is won by player number + Show Spoiler +. You need to already know that this game (and every game that has no ties) has a means to win it by perfect play, since it has no ties. It's not that proof I'm looking for here. Ah, I see. Hint (for those that want one): + Show Spoiler +The effect on the board of one player choosing a stone and the next player choosing a stone that is neither above nor to the right of the first is the same as if the first player had selected that second stone.
Last night, I didn't see how I could use that to prove that one player could force a win, but this morning I found the missing piece.
The winning strategy for player (named in spoiler above) falls into two parts: 1. + Show Spoiler +Select the top right stone 2. If this is part of a winning strategy, then this player has a win. If it is not, then + Show Spoiler +after player 1 selects the top right stone, player 2 must have a win (since there are no ties and it is not a winning strategy, it is a losing strategy). Then Player 2's next move is a winning move. The winning strategy for player 1 is simply to play that move first. This move will always be playable, as per the insight in the spoilered "hint".
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That was the solution I was looking for. Good job.
Exactly the same thing but another way to put it: + Show Spoiler +Imagine you're player 1 and that you're choosing the top-right-most stone. It is either a) the first step to a forced win, or b) a losing strategy. If the former then we have no problem. If the latter, then look at player 2's options. Can he make a move that you couldn't have made? Since you took the top-right-most stone, ANY stone he will pick would've also removed that stone had it been there. If indeed it is a losing strategy to pick the top-right-most stone, then you're to blame since you could have made the winning move player 2 will respond with. Hence, player 1 has a win here.
The winning starting move if you're curious is + Show Spoiler +to remove all but one stone in either the row or the column if the NxM board is asymmetric. If it is symmetric, just pick the stone top-right of the bottom-left-most stone.
I still don't know what this game is called, if anyone knows please tell me.
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On January 26 2011 01:03 Mayfly wrote:That was the solution I was looking for. Good job. Exactly the same thing but another way to put it: + Show Spoiler +Imagine you're player 1 and that you're choosing the top-right-most stone. It is either a) the first step to a forced win, or b) a losing strategy. If the former then we have no problem. If the latter, then look at player 2's options. Can he make a move that you couldn't have made? Since you took the top-right-most stone, ANY stone he will pick would've also removed that stone had it been there. If indeed it is a losing strategy to pick the top-right-most stone, then you're to blame since you could have made the winning move player 2 will respond with. Hence, player 1 has a win here. The winning starting move if you're curious is + Show Spoiler +to remove all but one stone in either the row or the column if the NxM board is asymmetric. If it is symmetric, just pick the stone top-right of the bottom-left-most stone. I still don't know what this game is called, if anyone knows please tell me.
Player 1 loses if there is only 1 stone . This is the only case where the induction of taking the 2nd player's move fails.
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The answer for the Forcefield Detainment one is so stupid, why doesn't the guard just say that he'll shoot the first one to leave?
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I don't know what it's called, but it's probably my favorite riddle. (to anyone that knows how this goes, I apologize for butchering how it's told >.> )
The Pill Problem + Show Spoiler +A man has been diagnosed with a terminal illness that can only be treated with two specific pills, we'll call them pill A and pill B. He must take one of each simultaneously at the same time of every day or else he will die. Since he cannot miss a day, he is only given a specific amount of pills which will last him exactly until he is given more pills; there are no leftovers or extras ever. These two pills are exactly the same in size, shape, color, taste, smell and weight; in other words, they are completely similar in appearance and are otherwise indistinguishable to all the senses. To alleviate this problem, the two bottles that contain pills A and B are sufficiently labeled to avoid confusion.
He takes the pills every morning at the same time. One morning, he wakes up late and has to rush to the bathroom so that he can take his pills and not die. In his haste, he mistakenly grabs one of pill A and two of pill B. He does not notice this until he is about to take the pills. He doesn't remember where each sat in his hand when he grabbed the pills. Taking all three right now would only prolong the inevitable; he would only die at a later date. He has to determine which is pill A and which is pill B right now, take them, then return the pill B to the bottle, or else he dies. What does he do?
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On January 26 2011 06:37 thai_quan_doh wrote:I don't know what it's called, but it's probably my favorite riddle. (to anyone that knows how this goes, I apologize for butchering how it's told >.> ) The Pill Problem+ Show Spoiler +A man has been diagnosed with a terminal illness that can only be treated with two specific pills, we'll call them pill A and pill B. He must take one of each simultaneously at the same time of every day or else he will die. Since he cannot miss a day, he is only given a specific amount of pills which will last him exactly until he is given more pills; there are no leftovers or extras ever. These two pills are exactly the same in size, shape, color, taste, smell and weight; in other words, they are completely similar in appearance and are otherwise indistinguishable to all the senses. To alleviate this problem, the two bottles that contain pills A and B are sufficiently labeled to avoid confusion.
He takes the pills every morning at the same time. One morning, he wakes up late and has to rush to the bathroom so that he can take his pills and not die. In his haste, he mistakenly grabs one of pill A and two of pill B. He does not notice this until he is about to take the pills. He doesn't remember where each sat in his hand when he grabbed the pills. Taking all three right now would only prolong the inevitable; he would only die at a later date. He has to determine which is pill A and which is pill B right now, take them, then return the pill B to the bottle, or else he dies. What does he do? Hmm, if he chooses at random he has a 2/3 chance of guessing right, but that's not an answer. If he has enough time, he can + Show Spoiler +crush the pills to powder and mix them homogeneously so he knows exactly what he has (3 pills worth of 1/3 A 2/3 B) but your statement of the problem seems to exclude that. I can't think of any way to distinguish "indistinguishable" pills.
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