Random Magic Paths - is it truly random?

[alexti]

BigDaddy said:
Based on his knowledge of statistics he made an estimate, thats what the ~ is for.

The exact value is given by
((7/8)^20 * 8c7) + ((6/8)^20 * 8c6) + ((5/8^20 * 8c5) + . . . + ((1/8^20 * 8c1) = .647163

What I said is that the formula is wrong and so the result was off by about 18%.

BigDaddy said:
If he didn't figure it out exactly he's a pretty good guesser.

And he did say it properly:

Probability of not getting 1 or more paths after recruiting 20 random paths.

That was said right, only the result was wrong it's approximately 47%, not 65%.

If you've missed my point, consider the case of getting random out of 3 magic paths and doing just 3 rolls. Ivan's formula gives (2/3)^3*3c2 + (1/3)^3*3c1 = 3*8/27 + 3/27 = 27/27 = 1 - probability of missing 1 or more paths, which is clearly wrong because you have positive probability of getting 3 different paths in 3 rolls.

Note. Number of combinations of N out of M is notorious for having many different notations (like McN, C(N,M) (M,N) etc). So we're all talking about the same coefficients