`At 1/19/04, Hal Finney wrote:`

However, here is an alternate formulation of my argument which seems to be roughly equivalent and which avoids this objection: create a random program tape by flipping a coin for each bit. Now the probability that you created the first program above is 1/2^100, and for the second, 1/2^120, so the first program is 2^20 times more probable than the second.

That's an interesting idea, but I don't know what to make of it. All it does is create a conflict of intuition which I don't know how to resolve. On the one hand, the following argument seems to make sense: consider an infinite sequence of random bits. The probability that the sequence begins with "1" is .5. The probability that it begins with "01" is .25. Therefore, in the uncountably infinite set of all possible infinite bit-strings, those that begin with "1" are twice as common as those that begin with "01". However, this is in direct conflict with the intuition which says that, since there are uncountably many infinite bit-strings that begin with "1", and uncountably many that begin with "01", the two types of strings are equally as common. How can we resolve this conflict?

That's an interesting idea, but I don't know what to make of it. All it does is create a conflict of intuition which I don't know how to resolve. On the one hand, the following argument seems to make sense: consider an infinite sequence of random bits. The probability that the sequence begins with "1" is .5. The probability that it begins with "01" is .25. Therefore, in the uncountably infinite set of all possible infinite bit-strings, those that begin with "1" are twice as common as those that begin with "01". However, this is in direct conflict with the intuition which says that, since there are uncountably many infinite bit-strings that begin with "1", and uncountably many that begin with "01", the two types of strings are equally as common. How can we resolve this conflict?

`-- Kory`