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0|0|0|||||Bertrand%27s Paradox|bole79||10:56:35|02/28/2008|I love the site and would like to point out a %22tiny%22 mistake in the %5Blink%3A http%3A%2F%2Fwww.cut-the-knot.org%2Fbertrand.shtml%7CBetrand%27s Paradox%5D article ... It%27s written above the last double applet%3A%0D%0A%0D%0A%5Bi%5D... for the second solution%2C the point is defined in its polar coordinates %28radius%2Bangle%29 which are selected independently ...%5B%2Fi%5D%0D%0A%0D%0AI think this could be written better. For if %5Bb%5Da random polar coordinate%5B%2Fb%5D is chosen%2C it is assumed that its radius and angle are chosen in such a way that points equally fills the inside of a circle %28square root for radius and linear for angle%29. But these are exactly the %22same%22 points as in the %22third%2C Cartesian Coordinates solution%22. I think it must be said that both%2C angle and radius are chosen linearly from their domains. This is not%2C I repeat%2C what is considered under %22choosing a random %28equally spaced%29 polar point%22.%0D%0A%0D%0A
2|1|0|||||RE%3A Bertrand%27s Paradox|alexb||11:55:26|02/28/2008|Thank you for the kind words.%0D%0A%3E For if a random polar coordinate is chosen%2C %0D%0A%3E it is assumed that its radius and angle are %0D%0A%3E chosen in such a way that points equally fills %0D%0A%3E the inside of a circle %0D%0A%0D%0AThere indeed was a mix-up on the page. Thank you for pointing this out. The applets and the explanations had to appear in the reverse order.%0D%0A%0D%0AFor the second solution%2C the point is chosen via its Cartesian coordinates%2C both chosen uniformly%2C with points outside the circle discarded. The probability is 1%2F4.%0D%0A%0D%0AFor the third solution%2C the point is chosen via its polar coordinates%2C both chosen uniformly%2C as suggested by E. T. Jaynes. This means that the point is in fact selected from a rectangle%0D%0A%0D%0A%26%2391%3B0%2C 1%26%2393%3B%26times%3B%26%2391%3B0%2C 2%26pi%3B%26%2393%3B %0D%0A%0D%0AThe probability is 1%2F2.%0D%0A%0D%0A