1. (Compulsory 3) A bag contains 6 balls of the same shape and size, including 2 black balls, 2 white balls, and 2 red balls. It is stipulated that drawing a black ball scores 0 points, drawing a white ball scores 1 point, and drawing a red ball scores 2 points; when drawing these balls, no one can see the color of the balls. First, player A draws 3 balls and does not return them to the original bag, then player B draws the remaining 3 balls. The player with the higher total score wins.
(1) Find the probability that A and B end in a tie;
(2) If you can choose the order of drawing the balls, would you choose to draw first or second, and why?
Four, 1. Let black balls be numbered 1, 2; white balls be numbered 3, 4; and red balls be numbered 5, 6. Then all possible outcomes of player A drawing the balls are as follows:
$$
\begin{array}{l}
123,124,125,126,134,135,136, \\
145,146,156,234,235,236,245, \\
246,256,345,346,356,456 .
\end{array}
$$
(1) In the event of a tie, both player A and player B should score 3 points each. Therefore, player A should draw one black, one white, and one red ball, which can happen in 8 ways. Hence, the probability of a tie is $P_{1}=\frac{8}{20}=\frac{2}{5}$.
(2) For player A to win, the score can only be 4 or 5 points, meaning the 3 balls drawn can only be 2 red and 1 white, 1 red and 2 white, or 2 red and 1 black, which can happen in 6 ways. Thus, the probability of player A (the first drawer) winning is $P_{2}=\frac{6}{20}=\frac{3}{10}$.
Therefore, the probability of player B (the second drawer) winning is
$$
P_{3}=1-P_{1}-P_{2}=\frac{3}{10}=P_{2} \text {. }
$$

Hence, the probability of winning is the same for the first and second drawers.