quantum mechanics - Energy measurement from position eigenstate

Given that the eigenstates of the position operator can be written as $\delta(x-x')$, and suppose we measure a particle in an infinite potential with walls at $x=0$ and $x=L$. I measure the particle to be in the position $x=L/2$, so the particle is in the eigenstate $|x \rangle = \delta(x-L/2)$. Suppose now that I want to measure the energy of the particle. The eigenstates of the energy operator are given by:

$$|\psi_n\rangle = \sqrt{\frac{2}{L}}\sin \left( \frac{n\pi x}{L} \right)$$

In order to measure energy I understand that I have to expand the original eigenstate in terms of the new energy eigenstates:

$$|x\rangle = \sum|\psi_n\rangle\langle\psi_n|x\rangle$$ where the probability of collapse into the eigenstate is given by:

$$P_n = |\langle\psi_n|x\rangle|^2$$

But now I sort of run into an issue. Sure then, I can say that:

$$\langle\psi_n|x\rangle = \int \sqrt{\frac{2}{L}}\sin \left( \frac{n\pi x}{L} \right)\delta(x-L/2)dx$$ and since

$$\int \delta(x-x')f(x)dx = f(x')$$ I can say

$$\langle\psi_n|x\rangle = \sqrt{\frac{2}{L}}\sin \left( \frac{n\pi }{2} \right)$$ and, $$P_n=|\langle\psi_n|x\rangle|^2 = \frac{2}{L}\sin^2 \left( \frac{n\pi}{2} \right)$$

I know that this means that all odd values of n are equally probably and all even values are not possible, but probability is supposed to be dimensionless, so what's happened here? What rookie error have I made?