Timeline for Two dice problem
Current License: CC BY-SA 3.0
10 events
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| Jul 17, 2013 at 13:09 | comment | added | whuber♦ |
@COOL You need to give m and j actual values, rather than symbolic ones, in order to have that Mathematica expression properly evaluated.
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| Jul 17, 2013 at 10:50 | comment | added | COOLSerdash |
@whuber I'm really sorry to bug you further with this, but the expression for which you provided the Mathematica code seems to be 1 whenever $x=1$, irrespective of the values of $j$ and $m$. What am I missing here? Another thing I can't reproduce is the numerical integration: When I use the following Mathematica code, the result is 0.0924: NIntegrate[Integrate[PDF[MultinormalDistribution[{3500/36,3500/36},{{(255500/1296),(255500/1296)*(-35/73)},{(255500/1296)*(-35/73),(255500/1296)}}],{x, y}],{x, 100.5,Infinity}],{y,100.5,Infinity}].
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| Jul 11, 2013 at 15:48 | history | edited | whuber♦ | CC BY-SA 3.0 |
Code formatting
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| Jul 10, 2013 at 14:42 | vote | accept | Ivana | ||
| Jul 10, 2013 at 13:28 | comment | added | whuber♦ | @COOL I added some equations for you. I hope there aren't too many typos in them :-). | |
| Jul 10, 2013 at 13:27 | history | edited | whuber♦ | CC BY-SA 3.0 |
added 978 characters in body
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| Jul 10, 2013 at 7:26 | comment | added | COOLSerdash | @whuber Thanks again for this wonderful answer. Forgive my ignorance, but I'm having trouble to understand the exact part. How exactly can we sum over $k$ and $j$? And what do we sum up, the polynomial at $x=1$? Does $p(x)^j$ mean that we take the whole polynomial to the $j$th power? | |
| Jul 9, 2013 at 23:12 | comment | added | Glen_b | Fantastic - a clear, detailed answer, covering all the bases - but also a very beautiful answer, wonderful tables and diagrams. A pleasure to read. | |
| Jul 9, 2013 at 22:58 | history | edited | whuber♦ | CC BY-SA 3.0 |
edited body
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| Jul 9, 2013 at 22:29 | history | answered | whuber♦ | CC BY-SA 3.0 |