Open the first door first, assuming that the first four keys are not only opened by the last one, that is five times.

The second door has been used 1, and there are four keys left, but none of the first three, that is, four times.

And then loop. . . The last 5+4+3+2+ 1= 15 (times)

In order to win, A will produce six cards after taking the penultimate card. Let A take four cards first, then B, and then count the remaining cards as multiples of six. Finally, after B takes it, there will only be 5~ 1 card.