Count the functions that have an entry with a specific structure Consider the function $f: \{0,1\}^n\rightarrow \{0,1\}^{2n}$
I want to count all the functions that have one entry where its first n bits are equal with the last n bits.
Assume we target entry number 10 and g is one of those functions that have the aforementioned property: $m=\{0,1\}^n;g(10)=m||m$
The total number of functions assuming $2^n$ rows where each row has $2n$ bits is $2^{{2^n}2n}$
I would say, if we want a specific entry of the function to have that structure we can first count the rest of the rows of the function which have: $(2^{n}-1)*2n$ bits and then since we want the first n bits to be equal with the last to the entry that we didn't take into account, we add another n bits, so total for each function: $(2^{n}-1)*2n+n$.
Total functions: $2^{(2^{n}-1)*2n+n}$
So the probability: $\frac{2^{(2^{n}-1)*2n+n}}{2^{2^n*2n}}=\frac{1}{2^n}$
Do i have it right? Is there an easier way to think about it? The result is to trivial to require that amount of effort.
Edit (added some more details):
One example (for $n=2$):
$F: \{0,1\}^2 \rightarrow \{0,1\}^{4}$
and I want to count all functions that have a specific entry, e.g 00, with a specific structure(first n bits equal with the last n bits). $F(00)=\{0000, 1010, 0101, 1111\}$ in this case

Entry 00| 0000
Entry 01| 0100
Entry 10| 0001
Entry 11| 0100

For example here $F(00)=0000$ and since $00=00$, this function should be counted


Entry 00| 0001
Entry 01| 0000
Entry 10| 0000
Entry 11| 0000

In this function since $F(00)=0001$ and $00\neq 01$, it should not be counted
If you want the function to have the property for one input, say $x$, which is fixed ahead of time, you just focus on that output vector
$$
f(x)=f(x_1,\ldots,x_n)=(z_1,\ldots,z_{2n})
$$
and the condition $z_i=z_{i+n}$ holds with probability

$$P(given ~row ~is ~good)=\frac{2^{n}}{2^{2n}}=2^{-n},$$
since in this case you don't care what is happening in the other rows. 

If you want this property to hold only for your selected input, this happens with probability

$$P(given~ row ~is ~good) (1-P(given ~row ~is ~good))^{2^n-1}=2^{-n}[1-2^{-n}]^{2^{n}-1}$$ 
 which can be estimated for large $n$ from above and below by using
$$
1-k\varepsilon+\frac{k(k-1)}{2}&lt;[1-\varepsilon]^k&lt;1-k\varepsilon
$$
where $\varepsilon$ is small and positive.