The deflection of a simply-supported beam under a concentrated load at midspan, regardless of the cross-section, is equal to:
$$\delta = \dfrac{PL^3}{48EI}$$
So the only thing that changes in your cases is the inertia, $I$. If Type 3 is the same as Type 1, only inverted, then their deflection should be the same.
If the sum of the top and bottom thicknesses is equal in all the cases (for instance, Type 1 is $1 + 3$, Type 2 is $2 + 2$, and Type 3 is $3 + 1$), then the inertia of Type 2 will be greater than that of Type 1 and 3. To prove this, we need to dive into the equations.
Using the following dimensions:

we can calculate all the properties. Type 2 is merely the case where $t_1=t_2$. The centers of gravity of these cross-sections are:
$$\begin{align}
\overline{y} &= \frac{h_1b_1\frac{h_1}{2} - h_2b_2\left(t_1 + \frac{h_2}{2}\right)}{h_1b_1-h_2b_2} \\
&= \frac{h_1b_1\frac{h_1}{2} - h_2b_2\left(t_1 + \frac{1}{2}\left(h_1-t_1-t_2\right)\right)}{h_1b_1-h_2b_2} \\
&= \frac{h_1b_1\frac{h_1}{2} - h_2b_2\frac{h_1}{2} - h_2b_2\left(\frac{t_1-t_2}{2}\right)}{h_1b_1-h_2b_2} \\
&= \frac{\frac{h_1}{2}\left(h_1b_1 - h_2b_2\right)}{h_1b_1-h_2b_2} - \frac{h_2b_2\left(\frac{t_1-t_2}{2}\right)}{h_1b_1-h_2b_2} \\
&= \frac{h_1}{2} - \frac{h_2b_2\left(t_1-t_2\right)}{2\left(h_1b_1-h_2b_2\right)}
\end{align}$$
For Type 2, this simplifies to $\frac{h_1}{2}$, as expected, since $t_1=t_2$, zeroing the second term.
Now, the inertia of these sections is equal to:
$$I = \frac{b_1h_1^3}{12} + b_1h_1\left(\frac{h_1}{2}-\overline{y}\right)^2 - \frac{b_2h_2^3}{12} - b_2h_2\left(t_1 + \frac{h_2}{2}-\overline{y}\right)^2$$
For Type 2, this simplifies to
$$I_2 = \frac{b_1h_1^3}{12} - \frac{b_2h_2^3}{12}$$
So, to see that Type 2 will always have a larger inertia, we need to find the difference between $I_{1,3}$ and $I_2$:
$$\begin{align}
\Delta I &= I_{1,3} - I_2 \\
&= b_1h_1\left(\frac{h_1}{2}-\overline{y}\right)^2 - b_2h_2\left(t_1 + \frac{h_2}{2}-\overline{y}\right)^2 \\
&= b_1h_1\left(\frac{h_1}{2}-\overline{y}\right)^2 - b_2h_2\left(t_1 + \frac{1}{2}\left(h_1-t_1-t_2\right)-\overline{y}\right)^2 \\
&= b_1h_1\left(\frac{h_1}{2}-\overline{y}\right)^2 - b_2h_2\left(\frac{1}{2}\left(h_1+t_1-t_2\right)-\overline{y}\right)^2 \\
&= b_1h_1\left(\frac{h_2b_2\left(t_1-t_2\right)}{2\left(h_1b_1-h_2b_2\right)}\right)^2 - b_2h_2\left(\frac{t_1-t_2}{2} + \frac{h_2b_2\left(t_1-t_2\right)}{2\left(h_1b_1-h_2b_2\right)}\right)^2 \\
&= \left(\frac{t_1-t_2}{2}\right)^2\left(b_1h_1\left(\frac{h_2b_2}{h_1b_1-h_2b_2}\right)^2 - b_2h_2\left(1 + \frac{h_2b_2}{h_1b_1-h_2b_2}\right)^2\right) \\
&= \left(\frac{t_1-t_2}{2}\right)^2\left(b_1h_1\left(\frac{h_2b_2}{h_1b_1-h_2b_2}\right)^2 - b_2h_2\left(\frac{h_1b_1}{h_1b_1-h_2b_2}\right)^2\right) \\
&= \left(\frac{t_1-t_2}{2\left(h_1b_1-h_2b_2\right)}\right)^2\left(b_1h_1\left(h_2b_2\right)^2 - b_2h_2\left(h_1b_1\right)^2\right) \\
\Delta I &= \underbrace{\left(\frac{t_1-t_2}{2\left(h_1b_1-h_2b_2\right)}\right)^2}_{\geq0}\underbrace{b_1h_1b_2h_2}_{>0}\underbrace{\left(h_2b_2- h_1b_1\right)}_{<0} \\
\end{align}$$
Since $\Delta I$ is therefore the product of three values, two of which are positive (or zero) and one of which is negative (since $h_1$ and $b_1$ are greater than $h_2$ and $b_2$, respectively), $\Delta I \leq 0$. This therefore means that $I_{1,3}$ (the inertia of Types 1 and 3) will always be lesser than $I_2$ (the inertia of Type 2). Therefore, the deflections under Type 2 will be smaller than those of Types 1 and 3.
All this above assumes that the material is isotropic and has the same behavior under both tension and compression. Something like reinforced concrete, where the concrete under tension is useless since it'll just crack, Type 3 is probably the best, since it puts the most concrete in the compression zone and leaves the steel to resist the tension.
This answer was initially incorrect, stating that Types 1 and 3 would have a larger inertia. I assumed the parallel axis component would not be so significant, and we all know what happens when one assumes. After seeing @kamran's answer, I double-checked my work and realized I was sorely mistaken. I apologize for the inconvenience.