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Old February 20th, 2005, 12:38 AM

alexti alexti is offline
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Default Re: Random Magic Paths - is it truly random?

Quote:
Ivan Pedroso said:
Yeah - your initial formula is correct. But I must admit that it could do with some explanations.

Thanks for the explanation why my formula is obvious Indeed, that's where it comes from. It reminds me of my teacher in the university who was used to say "and obviously *something* follows". After it would take me couple of hours to figure out why it was obvious

Quote:
Ivan Pedroso said:
I looked at it (briefly, I'll admit) and could not see what it all meant, or what the idea behind the factors and numbers were. And when I saw that p(i)=0 for i=8 I just disregarded the whole thing and wrote up my (wrong) ideas. (now I get it, and that p(8)=0 is perfectly fine. It's the probability of getting 20 sages in a row with eight(!) paths missing - zero of cause.)

In general case, formula is
Code:

p=sum[i=1..m]((-1)^(i+1)*C(i,m)*p(i)), where p(i)= ((M-i)/M)^N


where M is a number of picks and m <= M is number of picks in a subset, so the formula gives the probability of missing one (or more) of the paths out of subset of m paths.
I've just substitued 8 for m and M for our case, which produced curious p(8)=0
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