The play goes:
A draws first.
B draws second.
A draws again and wins. But A can only win with two red balls, so A drew a red ball on the first draw. B could not have drawn a red on his first draw.
Now the probabilities. With 3 out of 8 balls being red, the first draw probability of red=3/8.
B doesn't draw a red ball so the probability of this is 5/7 because there are only 7 balls left and 5 are white.
A draws a red ball on his second turn. Only 6 balls remain and there are 4 white balls and 2 red. So the probability is 2/6=1/3.
The combined probabilities are 3/8*5/7*1/3=5/56.