Random Magic Paths - is it truly random?

[alexti]

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

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:

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

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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

[Ivan Pedroso]

alexti said:
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

Hehehe. I can relate to that , I've heard and read it many times too. What about: "The proof of [bla bla] is left to the reader as an exercise" - that occured alot in my text books.

(It's a typical situation in math, until you get the point you are more or less in the dark, but when you get it, it really feels obvious afterwards. All teachers really could do with remembering their fumblings in the dark in their early days.)