Probability calculus on random bitstrings If I randomly choose 2 bitstrings on length ($n$) with $n$ to be an even number, what is the probability, parametrized on $n$, that at least $\frac{n}{2}$ of the bits are equal?
In my mind, since the random choice of every single bit is considered independent, I presume that the total probability is $\frac{1}{2}$, but I'm sincerely confused.
Edit:
Let me give a little example:
If $\text{bitstring}_1(4) = 1001$ and $\text{bitstring}_2(4) = 1010$ the statement is true because they have in common at least $\frac{n}{2}$ bits.

If we assume that all bitstrings are equally likely then the problem is quite simple.  Each matched pair of bits in the bitstrings will match with fixed probability $\tfrac{1}{2}$ and the number of matches has distribution:
$$M \sim \text{Bin}(n, \tfrac{1}{2}).$$
Consequently, the probability of at least $n/2$ matches (for even $n$) is:
$$\begin{align}
\mathbb{P}(\text{At least } \tfrac{n}{2} \text{ matches})
&= \sum_{m=n/2}^n \text{Bin}(m|n, \tfrac{1}{2}) \\[6pt]
&= \frac{1}{2^n} \sum_{m=n/2}^n {n \choose m} \\[6pt]
&= \frac{1}{2} \Bigg[1 + \frac{1}{2^n} {n \choose n/2} \Bigg]. \\[6pt]
\end{align}$$
