unreality
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oops sorry lol. The Princess was something I was thinking about for a future problem... and I put it in there without thinking ;D wow. hehe. Nice catch, bonanova.
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yeah! I made another one! lol You've played in three games of zarball! You've played in five games of zarball! But are you game for the Royal Zarball Tournament? This is how the tournament works: First Round A vs B = I C vs D = J E vs F = K G vs H = L Second Round I vs J = M K vs L = N Final Round M vs N = WINNER!!!!! ******** Here are the competitors and your chances of beating each of them: You, the Prisoner Prince - 1/1 chance as usual. You can always beat the Prince, every time Queen- 1/2 chance as usual. You can beat the Queen half of the time King- 1/4 chance as usual. You have a quarter chance of beating the King Duke- 1/3 Earl- 2/3 Jester- 0. The Jester will beat you every time- I mean, all he does in his spare time is juggle! But don't worry- the Jester's only weakness is the Peasant, who beats him every time Peasant- 1/2, except when playing the Jester. The Peasant beats the Jester every time Except for the special cases of the Jester, Peasant and Princess, the chances stay the same for beating you when they play each other (however they must be made relative to each other). For example, if you have a 1/3 chance of beating the Duke and a 2/3 chance of beating the Earl, it follows that the Duke is twice as good as the Earl, so if they played against one another, the Earl would have a 2/3 chance of winning, and the Duke would have a 1/3 chance of winning. However this isn't true. Okay, so down to the actual question: You failed to escape your imprisonment the first two chances... this is your last chance, says the King. He is allowing you to arrange the starting bracket of the Royal Zarball Tournament. Remember the key in this post- the 8 starting positions are the letters A through H. How should you arrange the eight competitors on the bracket to give you the maximum chance of winning the Royal Zarball Tournament and go free? What is that chance? Is there more than one solution?
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Storm and Bonanova got it MysteryKidakah, try writing out the five letters and looking at the probabilities for all three. I'll give you a hint: the winning probability for the right answer is 1/4, or 0.25, for all three combinations btw bonanova is wrong about jkyle's followup question's answers- nobody can be played twice in a row. Making a 1/1 chance impossible for any of the three combos. Here's a follow-up question:
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True story or did you/someone make it up? Either way, pretty sweet ;D
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Me too, but nobody has gotten the correct solution yet. (And twoaday was in three different universes it seemed lol, cuz he gave three dif answers, none of them working). And to whoever changed the topic title, thanks! (though I meant "Five Games of Zarball" instead of "Four Games of Zarball", Just "Five" is fine too lol ) The answer:
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You chose to face KQK in your last zarball competition for freedom. Unfortunately, you lost the second game against the Queen! It's okay though, the King loves playing zarball, you'll get another chance- and you did, when the Prince came to the castle for a visit. A good thing (for you) is that you can beat the Prince every single time you play him, no matter what. Yeah, he's that bad. You have a 1/2 chance to beat his mother, the Queen, and only a 1/4 chance to beat the King. The King comes to your cell with the challenge: you will play 5 games of zarball against the royal family (which is the King, Queen and Prince). You can play them in any order you so choose, though none of the family members can play more than twice total, and none of them can play twice in a row either. If you win the first, fourth and fifth game, you will go free. If you win both the second and third game, you will go free. If you win both the third and fourth game, you will go free. (You could get two of those combinations, or even all three, and still go free. You just have to succeed at one of those combinations) Maybe it's not the best choice probability-wise, but you don't want to put a higher chance in any of the three combinations, to put more eggs in one basket, so to speak. You want your chances to be equal for all three. So how should you have the King arrange the games if you want the highest equal chance for all three combinations for going free? (ie, an example would be to have a 1/16 chance to win games 1,4 and 5, a 1/16 chance to win games 2 and 3, and a 1/16 chance to win games 3 and 4... though I can tell you right now 1/16 isn't the highest possible)
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I'm posting a sequel now
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Everyone got it right ;D and bonanova nailed the Probability Proof of it. Good job!
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Here's an inspirational quote to live your life by: "Don't live your life by inspirational quotes."
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lol I think you're misunderstanding me... I would be the last one to believe in a fortune cookie. I believe completely in free will ;D it was just a paradox for paradox's sake (but apparently there is some arguement over whether it's a paradox or not, though it doesnt matter cuz its pointless)
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The problem can be solved by logic/intuition/whatever, which is what I did (and PolishNorbi), though you can also solve it with pure probability ;D
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You are a prisoner to a land ruled by a King and Queen, both of whom are pretty nice, luckily for you. The King enjoys riddles, math challenges, and physical games as well. Often he comes and chats with you in your cell. You have only been there two weeks when he says he has an idea. He will let you go free if you can win two games in a row of zarball. Zarball is a 1-on-1 game of agility, strength, speed and accuracy, played throughout the land. You think you're pretty good at it. You know that the King is one of the best, though, and the Queen is also pretty good, though nowhere near as good as the King. The King says: "If you can win two games in a row out of three games, you shall go free." He scratched his goti and looked at you with an interested face. "The three games will be alternating of me and my wife. You can play me first, the Queen second, and me last. Or you can play the Queen first and last, and me second, in the middle. KQK or QKQ. Either way is fine." What do you do?
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There are ten Super-Self-Cooling light bulbs in a room. To prevent loss of unnecessary heat energy, they are specifically designed to instantly cool themselves down when turned off. In fact they try to cool themselves when turned on as well, to make sure their circuit doesn't break, and the bulbs are made of a resilient glass that can't shatter. You are in a room with a closed door leading into another room where these ten bulbs are. You can control the ten light bulbs from the room you are currently in. You have to be able to tell which bulb corresponds with which dial. Yes, dial. Fortunately for you, each Super-Self-Cooling lightbulb isn't a simple on-off switch. It's a dial, with settings from 0 to 4 to level the brightness. Yep, 5 levels of brightness... since there are ten bulbs, it should be easy, right? Wrong. Not that it would be easy anyway, with 5 settings for 10 bulbs, but there's more: The circuitry is screwed up. One of the dials is useless, and another dial controls its own light bulb as well as the light bulb that used to correspond with the useless dial- so one dial controls nothing and another dial controls two light bulbs at once. There are two cases like that! Two useless dials, two doubled-up dials. There is no way to figure out beforehand which dials are useless or not, all the rewiring problems are above the other room. Since your job is to figure out how each of the 8 working dials with 5 setting levels affect the 10 bulbs, what do you do?
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The meaning of life is the continued existence of life.
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I've heard of these before, there are lots of these, I remember a circle with leprechauns on it for some reason. Right now I dragged it to my desktop and i'm lookin at all 63 frames to tell ya what happened
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(wow the game is slow on this forum ) The one unquestionable fact happened suddenly because something desired to act intelligently. He... (It)... fainted
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possibilities: A=alive D=dead H=honestant S=swindelcant A-HH A-HS A-SH A-SS D-HH D-HS D-SH D-SS * take AHH, if they are both honestants, then both are lying. Cross AHH out * take AHS, the first guy is telling the truth but then the second guy WOULDNT be lying. Cross AHS out * take ASH, the first guy's statement works if the second guy is the honestant. But the second guy's first statement is a contradiction. Cross ASH out * take a**, the first guy's statement only works if your companion dies, so this is already a contradiction. Cross a** out we've eliminated all the alive possibilities your companion is dead am I right? just in case, I will look at the rest: DHH- the second guy is lying, contradiction. Cross DHH out DHS- this fits as far as I can tell DSH- contradiction, cross it out DSS- guard 2 is telling the truth, contradiction leaving DHS your companion is dead guard 1: Honestant guard 2: swindelcant