Probability concept that distinguishes likelihood of sequences 0110101011101... and 000000000000...? Say we have a coin and want to decide if it is fair or not. We flip it many times. Consider two cases.


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*Say the result is a sequence like 0110101011101...

*The result is 000000000000...


In the first case the assumption that the coin is fair sounds reasonable, while in the the second the coin is obviously completely biased. However, for a truly fair coin the probabilities of the two sequences are the identical. What is then the concept that distinguishes the first one as more likely? How does one quantify that?

Consider also the third case


*The result is 010101010101010...


This actually does not look like a coin at all, but it seems to pass naive tests which boil down to comparing number of zeros to number of ones. Is there a sense in which it is less random then (1)? Or I have to invent a new rule if somebody gives me a more cleverly crafted sequence? 

The sequences themselves do have the same probability. However, what makes the first seem more likely to you is the fact that it has a more even number of ones and zeros. Instead of testing the likelihood of the exact sequences, you can test the likelihood of the number of ones and the number of zeros. 
For a fair coin, this is binomial with $p = \frac{1}{2}$, meaning that the probability that the number of ones in a sequence of length $n$ is $k$ is 
$P_{\frac{1}{2}}(k) = {n\choose{k}}\frac{1}{2}^n$
The only thing that differentiates the sequences is essentially ${n\choose{k}}$, which represents the number of sequences of length n with k ones. This would indeed give a higher probability to your first sequence. 
Another way of looking at it, is treating $p$ as unknown. For a given $p$, such that $P(x_i = 1) = p = 1 - P(x_i = 0) \forall i$, the probability of a sequence $x_1, ... x_n$ is 
$P(x_1, ... x_n | p) = p^{ |\{i : x_i = 1\}|}\cdot (1 - p)^{|\{i : x_i = 0\}|}$ 
Now, which $p$ maximizes the likelihood for each sequence? How far is it from the true $p = \frac{1}{2}$ for a fair coin? 
