Random Magic Paths - is it truly random?

[Ivan Pedroso]

@Alexti

Ahhhh yeah - that was sloppy.

(7/8)^20*8 is NOT the probality of getting exactly one path missing. You are absolutely right about it not garanteeing that the other 7 paths are present.

Back when I have had time for a better look (sorry 'bout posting too hastilly )

[alexti]

Ivan Pedroso said:
@Alexti

Ahhhh yeah - that was sloppy.

(7/8)^20*8 is NOT the probality of getting exactly one path missing. You are absolutely right about it not garanteeing that the other 7 paths are present.

Back when I have had time for a better look (sorry 'bout posting too hastilly )

Curiously enough, the correct formula (I'm 99% sure ) looks very much like yours, only the sign alternates (see my earlier post, coefficients are the same, only the sign is different).

[Ivan Pedroso]

alexti said:

Curiously enough, the correct formula (I'm 99% sure ) looks very much like yours, only the sign alternates (see my earlier post, coefficients are the same, only the sign is different).

Yeah - your initial formula is correct. But I must admit that it could do with some explanations. 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.)

The alternating signs are introduced when probabilities of non-independent events are added.

Same as when non-disjoint sets are unified:

U(A,B) = A + B - I(A,B)
U: unified
I: intersection (written as an upside-down "U")

With 3 sets A,B, and C you get:
U(A,B,C) = A + B + C - I(A,B) - I(A,C) - I(B,C) + I(A,B,C)

With 8:
U(all eight) = A + B + ... + H - I(all with two) + I(all with three) - I(all with four) and so on with plus and minus alternating between the groups of intersections.

One of the ones in the group called I(all with four) could be: I(A,B,C,D) or I(A,B,D,F) or ... well anyone with four letters

- - - - we'll use the above stuff now - - - -

"A" above could mean no FIRE pick in 20 sages in a row. "B" no AIR and so on.

Then U(A,B,C,D,E,F,G,H) is all the sets that can be constructed with 20 sages where any one path is missing.

P(getting one of the sets in U(A,B,C,D,E,F,G,H)) is then the probability of getting a row of 20 sages with any one path missing. But as Alexti said, if you add together P(A)+P(B)+ ...+P(H) you will NOT get P(getting one of the sets in U(A,B,C,D,E,F,G,H)).
Because P(A) is the the probability of getting 20 sages without seeing any FIRE paths. But a series of 20 with all sorcery paths will then be a part of A, B, C, and D and would thus get counted 4 times instead of only once. The alternating signs ensures that these "extra countings" gets added and deducted correctly, in order to only count the relevant contributions once. The above describes how P(getting one of the sets in U(A,B,C,D,E,F,G,H)) should be calculated.

P(A) = (7/8)^20 (so are P(B) and P(C) and so forth)
so the first part (the one with A + B + C...) is thus:
8*(7/8)^20

P( I(A,B) ) = (6/8)^20 (and so are P( I(A,C) ) and bla bla)
the re are 28 ways to make these parings, so the second part is:
- 28*(6/8)^20

The third part is ( 8!/(5!3!)=8*7*6/(3*2)=56 ways to combine three letters from the eight available):
+ 56*(5/8)^20

And so on and so on... resulting in:
((7/8)^20 * 8c7) - ((6/8)^20 * 8c6) + ((5/8^20 * 8c5) - . . . + ((1/8^20 * 8c1) = 0.4694

As stated by Alexti and misunderstood by me, but now hopefully clear to all

[BigDaddy]

That equation looks very familiar, but I've misplace my book... (not my advance statistics books with all their distributions and tests, just the basic one).

Anyway, for anyone who stumble across this with some arcane pattern and thinks it might not be random, remember that radomness is elusive even to statisticians. Understanding randomness could be likened to understanding the actual magnitude of infinite. We can use it and test it, just like infinite, but our perception of it will never be quite accurate. So, consider a random number off of an infinitely long number line. . .

[bone_daddy]

So, does this have to do with true randomness not being possible within a system containing a limited number of variables?

[BigDaddy]

A computer can never make really random numbers, a computer scientist has ensured me that that is the case as I believed it was. Instead it uses the clock to produce seemingly random numbers. The numbers themselve seem very much random, in fact, you could call them "virtually random".

True randomness is a rather complicated term, but according to my text randomness is part of anything who's probability of occurence or omission is <100%. So, I'd say that the probability of failure of a device, or the probability of throwing heads on a coin are still random, regardless of their distribution. Distributions just describe some specific random systems. So, I'd have to say no. It's just easier to feel you've grasped random with a coin flip.

Considering a random point on an infinitely long numberline is just a good way to illustrate a humans inability to truly understand certain concepts. Cosider a point on a numberline, if you take an infinitely small movement from that number to any other position on that numberline, there are an infinite number of uniques points between the first and the second, until the movement = 0.

Similarly, we make do with random models and distribution that resemble the things we can test. So, consider this, the probability of flipping heads or tails on a coin 1000 times is 9.3326E-302. IF the coin is flipped an infinite number of times, the probability of this occuring is 1.

In this example we have a demonstration of true randomness, with only two choices (heads or tails).

[Saber Cherry]

Via's CPU's produce random numbers, using special hardware. A Turing Machine cannot make random numbers, but modern computers don't have all the limitations of Turing machines (as shown by Via).

[bone_daddy]

Ok, so although the occurence of any given number within the set of variables will be random, there will be an emergent pattern (the distribution) which is not. Is my comprehension of what you are saying correct?

Please forgive my ignorance concerning stats and probs, but I've never taken any math courses beyond basic algebra and trig. Never really needed them. I have a tendency to only learn things as I need them, unless it's something I find interesting. :-/

[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

---

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.)