Why can a coupled wavefunction of 2 nucleons seemingly arbitrarily give two different expressions? When looking at the Isospin representation of a proton-neutron system (with the notation $|I,I_3\rangle$), you can go from an uncoupled to a coupled representation like this:
$$\textstyle|\frac{1}{2},+\frac{1}{2}\rangle|\frac{1}{2},-\frac{1}{2}\rangle=\sqrt{\frac{1}{2}}(|1,0\rangle+|0,0\rangle)$$
However, you can also find the following antisymmetric wavefunction when switching the first two kets around (or switching proton and neutron):
$$\textstyle|\frac{1}{2},-\frac{1}{2}\rangle|\frac{1}{2},+\frac{1}{2}\rangle=\sqrt{\frac{1}{2}}(|1,0\rangle-|0,0\rangle)$$
If you now want to calculate the probability of finding the proton-neutron system in a $|1,0\rangle$ state, you have to take the square of the inproduct of the above expressions with $\langle1,0|$, which gives $\frac{1}{2}$ for both expressions. 
Do you now have to add these 2 probabilities together, since they should be equivalent, though one is symmetric and the other is asymmetric.
However, adding the probabilities means the probability to find the proton-neutron state in the $|1,0\rangle$ state is 1, but you could reason similarly for $|0,0\rangle$ and so find a total probability of 2 which doesn't seem right.
It seems weird that the wavefunction is different if you assume a proton-neutron vs. a neutron-proton system.
As a background for this question, I need to calculate the relative probabilities for 2 reactions, one of which is a pion and deuteron reacting to form a proton and neutron, so I'm writing left and right side in the coupled representation and taking the inproduct.
 A: It seems there is a bit of confusion here.  A state like
$$
\textstyle\vert \frac{1}{2},\frac{1}{2}\rangle_1\vert \frac{1}{2},-\frac{1}{2}\rangle_2 \tag{1}
$$
does not transform into itself under permutation of the particle index so is neither symmetric nor antisymmetric.  As written, (1) describes distinguishable nucleons: the first nucleon is certainly the proton and the second is certainly the neutron.
If your nucleons are indistinguishable, you need to work with properly symmetrized states
$$
\vert\psi_\pm\rangle =\frac{1}{\sqrt{2}}\Bigl(\textstyle\vert \frac{1}{2},\frac{1}{2}\rangle_1\vert \frac{1}{2},-\frac{1}{2}\rangle_2\pm 
\vert \frac{1}{2},-\frac{1}{2}\rangle_1\vert \frac{1}{2},\frac{1}{2}\rangle_2\Bigr)
$$ 
which are basically the $\vert 1,0\rangle$ and $\vert 0,0\rangle$ states you already have.
To compute the probability of finding the proton-neutron system (as a system of distinguishable particles) in $\vert 1,0\rangle$, you have 
$$
\textstyle
\vert \langle 1,0\left[\vert \frac{1}{2},\frac{1}{2}\rangle_1\vert \frac{1}{2},-\frac{1}{2}\rangle\right]\vert^2 =\frac{1}{2}\, .
$$
This is the probability of having the first nucleon a proton, and the second nucleon a neutron, in the isospin state $\vert 1,0\rangle$. 
At this point you can ask about the probability of having the first nucleon a neutron, and the second a proton, and this must be $1/2$ since the probabilities must sum to 1.  You can also ask about the probability of having the first nucleon a proton, and the second nucleon a neutron, in the isospin state $\vert 0,0\rangle$, which again must be $1/2$ for the same reason that the probabilities sum to $1$.
