Originally Posted by

**JeWiSh** For the first part, you just calculate in order the odds of what HAS to happen for player A to win on his second draw.

So, first he has to draw a red ball, the odds of which are 3/8. Then, in one situation Player B draws a white, which is a 5/7 chance. Lastly A has to choose another Red, which would be a 2/6 chance. $\displaystyle \frac{3}{8} * \frac{5}{7} * \frac{2}{6} = \frac{30}{336}$

However, B could have chosen a red ball on his draw, the odds of which were 2/7. Then, for B to get the last red ball would have been a 1/6 chance.$\displaystyle \frac{3}{8} * \frac{2}{7} * \frac{1}{6} = \frac{6}{336}$ So the total chance is $\displaystyle \frac{36}{336} = \frac{3}{28}$