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With both switches open, both transistors are in cut-off (not conducting), the lamps are out, and resistor end of the switches are at +E voltage through the left lamp and R1, in series, and through the right lamp and R2 in series. If the right switch is the first one closed, the current supplied from +E, through the left lamp and R2 supplies base current to turn on T2. T2's collector-emitter voltage drops to about 0.2V and supplies a current path through the right lamp, lighting it. Because the base current for T1 would be provided through R1 from T2's collector, which is now only 0.2V, T1 cannot be turned on...it's locked out. E has to be disconnected to reset the game.
Look at the datasheet values for high collector currents: https://www.fairchildsemi.com/ds/PN/PN2222A.pdf
At 500mA the hfe is not 100. For guarantied collector-emitter saturation I usually use hfe/10. Base current would be Ilamp/(hfe/10). R1 and R2 would equal the supply voltage minus the base-emitter voltage, divided by the base current. ~ (E-Vbe)/((Ilamp/(hfe/10))
R3 and R4 are there just to hold the bases at ground when the switches are open...so...anything a lot higher than R1 or R2..... I'd use R1=R3x100 up to R3x10000
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