For a real battery, why is two identical bulbs in parrallel slightly more dim than just one in series Image for reference:
$$\begin{align}
&r = 0.8\,\Omega\\
&\mathscr{E} = 1.5\,\mathrm V\\
&A = B = 6\,\Omega.
\end{align}$$

The current when the switch is open, the only other resistor is the A bulb, so total resistance is 6.8 ohms, but when it's closed the total resistance should be 3.8 ohms (2 6 Ohm resistors in parallel = 3 ohm $R_{\mathrm{eq}}$), but that seems to not be the case.
Why would the current be less when it is closed? The way the question is asked it seems like it should be less, but the math makes it seem like more.
 A: You are correct that the total current will be more, however half of the current will be going through the section of wire connected to light bulb A, and half of the current will be going through light-bulb B. Therefore, even though you may have more current going through the entire circuit, the current going through A and B are individually smaller.
A: 
Why would the current be less when it is closed? The way the question
  is asked it seems like it should be less, but the math makes it seem
  like more.

First, I assume you mean that the current through bulb A is less when the switch is closed than when the switch is open.  Clearly, the current through the battery is more when the switch is closed so why should there be less current through bulb A?
Here's a hint:  if $r = 0$, closing the switch does not change the current through bulb A and doubles the current through the battery.
Try finding the voltage across bulb A when the switch is open and then when the switch is closed as a function of the internal resistance $r$.  You should be able to do this using voltage division and the information you've already given.
