TL;DR: a non-uniform key with a certain entropy requires less (but not much less) guessing attempts on average than a uniform key with the same entropy, and then only if adversaries know and make good use of the distribution of the keys.
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The method leading to the least expected number of attempts is to make attempts from most to least probable key. For $N=4$ and `0` having probability $p\le 1/2$ and `1` probability $1-p$, that is
~~~
1111
0111 1011 1101 1110
0011 0101 0110 1001 1010 1100
0001 0010 0100 1000
0000
~~~

The expected number of attempts is
$$\begin{array}{lll}
&1\,(1-p)^4\\
\quad+&(2+3+4+5)\,p(1-p)^3\\
\quad+&(6+7+8+9+10+11)\,p^2(1-p)^2\\
\quad+&(12+13+14+15)\,p^3(1-p)\\
\quad+&16\,p^4\\
=&1+10p+15p^2-10p^3\end{array}$$

Plotting that expected number of attempts against the entropy-based approximation $(2^\mathsf{entropy}+1)/2$ with $\mathsf{entropy}=-N(p\,\log_2(p)+(1-p)\log_2(1-p))$:
[![plot][1]][1]

Thus $(2^\mathsf{entropy}+1)/2$ is a good approximation of the expected number of attempts for $N=4$. Working on the general case.


 [1]: https://i.sstatic.net/HogX2yOy.png