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Everything posted by bonanova
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ESCAPE of MARTHA THE WATER PROBLEM -- LOCKED with LOCKERS
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The first arrangement can be read: the square root of 1 equals 11. The second arrangement [inside the spoiler] can be read: 11 divided by 1 equals 11 The puzzle asks you to move two matchsticks to make a valid equation. OK?
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Death Probability! Really Hard One!
bonanova replied to roolstar's question in New Logic/Math Puzzles
Agree. Alfred knows - and we know - either [1] Charlie is "the first" - in which case Possibility Two has been eliminated - i.e. determined to be false. or [2] Charlie is "the second" - in which case Possibility One has been eliminated - i.e. determined to be false.. Can you clarify why you say a possibility has not been eliminated? "The exact same three possibilities still exist ... " An option that has a zero fraction of the total probability [i.e. is false] is usually thought of as having been eliminated. The remaining possibilities - the options with non-zero probabilities - are: [1] Charlie and Alfred die - probability is 2/3. [2] Charlie and Bob die - probability is 1/3. If you want to hang onto your third possibility, then: [3] Albert and Bob die - probability is 0/3. [* - you can argue it's not relevant to Alfred, but it's certainly relevant to Bob.] Since Bob could have been named, but was not, Bob's chances of survival just doubled. Alfred's chances remained the same. Proof: Alfred's chances [of survival] stayed at 1/3. Charlie's chances went to zero. Bob's doubled, to 2/3. Since one and only one survives, these chances must sum to unity. This is why in Monty Hall you make the swap. -
Put things like that here: http://brainden.com/forum/viewforum.php?f=9
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Edited out a rant on some rogue user for spamming. Where did I put that eraser?
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If I had counted, my first reply would have been quicker. Nice puzzle.
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Death Probability! Really Hard One!
bonanova replied to roolstar's question in New Logic/Math Puzzles
Alfred knows: Charlie is either the first or the second. Alfred knows: One of those possibilities is in fact gone. Alfred knows: The liklihood of the other got doubled. -
The OP says ... Assume safely that the porbability [sic] of each gender is 1/2. All the "affecting factor" discussion does not pertain. This is a calculation of probability, not a study of the human reproductive process. The puzzle is, precisely, They have two kids, one of them is a girl, what is the probability that the other kid is also a girl. It has the same answer as: I flipped a fair coin twice. One of the outcomes was heads. What's the probability the other outcome was heads?
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Death Probability! Really Hard One!
bonanova replied to roolstar's question in New Logic/Math Puzzles
My comments in red. -
I believe your money is safe.
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Not a clue as to what this is supposed to mean. If A is oriented so that sides 653 are seen from some corner, orient B so that 152 are seen on the corresponding corner, 431 are seen on that corner of C, and 453 are seen on the corresponding corner of D. That's the notation he's asking for [but use the actual face numbers from your solution, not 152 421 and 563.] I'm pretty sure he's asking that the top faces need to be different colors; same for the bottom, front, back, left and right faces.
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Clarification needed: Usually the cubes are stacked, so that 4 faces are seen [apart from the ends] and the idea is that each of the four sets of 4 faces show all four colors. Are you posing the problem in such a way that all 6 faces of the four cubes are visible, and the six sets of 4 faces show all four colors? Sounds like you are, and that makes for an interesting puzzle. Will look at it more tonight. Good one.
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Death Probability! Really Hard One!
bonanova replied to roolstar's question in New Logic/Math Puzzles
Who's that standing over there in the corner, smiling at us? Why, it's ... Monty Hall! Let's see: Alfred is the door #1 that I chose, Bob is the door #2 that Monty opened to show a losing choice, and Charlie is the door #3 that I should trade my choice for. If only I had that choice! But I don't. Ooops, poor Al still has only a 1/3 chance of surviving. Charlie, on the other hand, now has a 2/3 chance of making it. Interesting ... because if it had been Charlie who had asked the question, and the guard had told Charlie that poor old Bob was gonna bite the big one, then Al would have the 2/3 chance. Right? -
For sure ... it was relatively quiet while you were "away". Merry Christmas* to all ... ! * or equivalent.
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If you can say that there was a tree in the forest, without an observer there to confirm it; If you can say that the tree fell, without an observer to confirm it; I will say with equal certainty that it made a sound, without an observer to confirm it. Corollary question: If a man speaks in the forest, and there is no woman there to hear him, is he still wrong?
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That was my original stance. I decided, in the spirit of the "gift-giving season" perhaps, that UR had earned some slack. The OP might have been stated more specifically about that point. [1] One way would state that you must select at least one toy to purchase. [2] Another way might state: You must pay for all your toys using coupons. Alternative [1] does not permit UR's answer, while [2] does. Universal quantifiers do not possess existential import. The statement "All idiots with IQ greater than 200 and three lefts hands are male" does not imply that at least one of them exists. If All is replaced by Some, the statement does imply at least one exists. I cut UR the slack because the OP does not express the idea of "some" toys or "some" coupons. Absent that language, I didn't explicitly rule out buying zero toys using zero coupons. That being the case, it seems more a matter of semantics than of logic.
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Wait, there's more ... [*]Tie the barometer to a string and lower it from the top of the building till it touches the ground. Measure the length of the string. [/*:m:cb09d] [*]Measure the length of the building's shadow, then measure the barometer's shadow. Use the ratio of shadow lengths to convert the length of the barometer into the height of the building. [/*:m:cb09d] [*]Every time somebody walks into or out of the building, stab them with the sharpened end of the barometer. Word of the 'Barometer Murderer' will eventually reach the building's owner, who will of course be forced to sell the building. The real estate advertisement should give the height of the building. [/*:m:cb09d] [*]Kill someone and leave the body and barometer on the roof of the building. The presence of the barometer will probably suggest death by scientific experiment. The TV news report probably will contain more information that might be relevant, such as the height of the building. [/*:m:cb09d] [*]With the barometer at ground level, draw line on building at top of barometer. Place bottom of barometer at the level of the line, draw second line on building at top. Repeat. Count the number N of marks made on the building. Multiply N by the height of the barometer. [/*:m:cb09d] [*]More simply, the building is N barometric units high.[/*:m:cb09d]
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Instant replay -- on further review ... I might not have closed the loophole on that point - should have said using at least one coupon. Zero may be a valid number of coupons to use.
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LTNS amigo....! Not a fair purchase.
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Interested to know roolstar's approach. Care to share, roolstar?
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You and your spouse invite four other couples to a party. Ten people in all. Prior to the party each person [other than you] knew a different number of people present. How many people did your spouse know? How many did you know? You may assume the following: If A knows B, then B knows A. Every male knows his partner. You might have invited people you don't know.
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Two days ago Jane celebrated her birthday. Today John, her twin brother, celebrates his. Explain.