tag:blogger.com,1999:blog-264226589944705290.post4877576396117838973..comments2023-11-05T03:45:25.001-08:00Comments on God Plays Dice: The asymptotics of "27 lines on a cubic"Michael Lugohttp://www.blogger.com/profile/15671307315028242949noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-264226589944705290.post-81371218644448006352008-11-27T07:57:00.000-08:002008-11-27T07:57:00.000-08:00This is a very nice argument ;), but I think some ...This is a very nice argument ;), but I think some technical work is required to turn in into a "formal" proof.<BR/><BR/>Here is something that may be troublesome... You say: "Furthermore, since the sum is a sum of a bunch of small contributions, it's approximately normal."<BR/><BR/>The variables you use are indeed independent, but not identically distributed. How does the asymptotic normality follow? Moreover, asymptotic normality gives you information about the distribution, and not the individual probabilities, where you need a local limit law.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-38476385121926243922008-11-27T05:35:00.000-08:002008-11-27T05:35:00.000-08:00Efrique,you're right, it should be n/3; I mistyped...Efrique,<BR/><BR/>you're right, it should be n/3; I mistyped. (What follows is correcet based on it being n/3, though.)Michael Lugohttps://www.blogger.com/profile/15671307315028242949noreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-85876782768704205952008-11-26T21:02:00.000-08:002008-11-26T21:02:00.000-08:00Could you explain how you get to "2n(n-1)/[3(2n-1)...Could you explain how you get to "2n(n-1)/[3(2n-1)]" and then from there to n^2/3?<BR/><BR/>It may just be that I'm overtired, but that doesn't look right to me.<BR/><BR/>I could believe that I've lost a factor of two doing the summation, but that last step - surely that becomes n/3, right?Efriquehttps://www.blogger.com/profile/08526031804261484547noreply@blogger.com