Three identical lightbulbs are connected to a battery and the lightbulbs are parallel. If the middle bulb burns out, what happens?

So the answer is: the light intensity of the other two bulbs remain the same (of course when I think about real life this makes perfect sense). However, I thought the intensity of the other two would increase. I thought so because if one bulb burns out, then total Resistance in the circuit decreases, and the total current for the circuit would then increase (I=V/R). This led me to believe that each of the remaining batteries in the circuit would get more current and have more light intensity. To my surprise, the explanation in TPR was (to paraphrase) that since there are only two bulbs left and not three, the current the battery provides decreases as their individual currents stay the same. To me it sounds like the battery can control how much voltage it supplies and I don't get that. Can anyone help me see where my thinking is incorrect or help explain it in a different way? I hate getting these concept type of questions wrong!

Thanks in advance

steve