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Originally Posted by W
You're dead wrong, and could prove it to yourself by simply doing it over with a different random seed.
Yes, the different bitstrings are dependant on eachother, so if you have 1001, you know you need atleast 4 bits more to find 0001, and if it's there you've already found 0010 0100 and 1000 as well. But for each of the different bitstrings, the probable position is entirely random and with equal distribution.
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i have a number ways of 'proving' to myself that they are equal, on average. my problem is i have what appear to be better proofs that they aren't, and i see no way of refuting them.
i've got one better now:
1111
vs.
1110
in any random string, we reach the first time we have three ones in a row (the prerequisite for a 1111 or a 1110).
00010111
k?
from here, how far is it to the last digit of 1110?
we can calculate.
1*1/2 + 2*1/4 + 3*1/8 + 4*1/16..... = ~1.64
a 0 is coming along pretty damn soon.
from the exact same point, how far on average to the last digit of 1111?
1*1/2 + (average distance from scratch to the last digit of 1111+1)*1/2
because there's 1 chance of the next digit being a 1 and 1 chance of a 0, and having to start over.
even if god intervened here, and everytime after that frustrating 0 you immediately got 1111,
that's 1*1/2 + (4+1)*1/2 = 3. which is still greater than 1.64.
in reality (average distance from scratch to last digit of 1111) is closer to 18 than 4, so 1111 takes ~ (9-1.64) ~ 7 longer than 1110.
counterintuitive =/= wrong