bonanova
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Everything posted by bonanova
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Paradox Not truth, nor certainty. These I forswore In my novitiate, as young men called To holy orders must abjure the world. "If ..., then ...," this only I assert; And my successes are but pretty chains Linking twin doubts, for it is vain to ask If what I postulate be justified Or what I prove possess the stamp of fact. Yet bridges stand, and men no longer crawl In two dimensions. And such triumphs stem In no small measure from the power this game, Played with the thrice attenuated shades Of things, has over their originals. How frail the wand, but how profound the spell! Clarence R. Wylie, Jr., 1948
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Yup. Any number N of randomly chosen points will tend to divide the interval into N+1 equal partitions. What - they don't all tend to fall at the middle? Individually, yes. Interesting.
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Sure it is. All the points are randomly chosen. I ran a program that just chose two points and verified this. Agree with unbounded result of 1/2.
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Numbers are balls being weighed. Each outcome = lighter, equal, heavier = points to the next weighing. After three weighings the light and heavy balls are shown.
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The average length of the sides of the 3 million triangles of the previous post [random vertices] was .664.... That is just slightly less than 1/3 of the diameter, not of the average chord length. If that helps our thinking.
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Repeat post. Locked.
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Yes I thought they had it -- but then realized they were re-using symmetric points. Rather than say more, I just excluded symmetry to see if the other points would be found...
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Determine the probability p(convex) that 4 random points placed on the unit disk form a convex quadrilateral. Here is an algorithm for averaging the exact result for a million random triangles.
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Shorter still I think.
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Hats on a death row!! One of my favorites puzzles!
bonanova replied to roolstar's question in New Logic/Math Puzzles
How about: Every prisoner, each in his own transparent soundproof shell, has a keyboard with a black button and a red button . The keyboard unlocks when it's his turn, and locks again after one of the buttons is pushed. When the jailer decides to do so, he presses a button that displays "prisoner 13 has chosen BLACK", e.g., on a large screen that all can see. The turn passes to the next prisoner. etc. No extra words, no verbal intonations, no prisoner-chosen delays, etc.... -
Hats on a death row!! One of my favorites puzzles!
bonanova replied to roolstar's question in New Logic/Math Puzzles
You've pretty well answered your question already. If the OP is equivalent to You may answer "BLACK" or you may answer "RED". If you say anything else, all will die. Then saying "BLACK or RED" is a fatal idea. If it's equivalent to You may answer "BLACK or RED". If you say anything else, all will die. then your idea is the only permissible approach, and all will escape by saying it. It's fairly clear the first meaning is the intended one. -
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Dividing by 256 gives their arrival probabilities of .004 .031 .109 .219 .273 .219 .109 .031 .004 [to 3 places] Squaring these give the meeting probabilities as .000 .001 .012 .048 .075 .048 .012 .001 .000 [3 places] Adding these gives the overall probability of meeting. as .196 [3 places] Y-san, Prime and Garrek99 all got this. Prime got it Y-san gave the correct explanation apparently with a calculation error. Garrek99 made the assumption the arrival probabilities were the same, leading to the result for a 5-path square and equal speeds. I ran across this problem a while back, published with wrong answers. I did the math a couple days ago and thought I'd share the fun.
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It can be shorter than that.