Probabilities: In what condition 2 players have same chance to win the game? 
There is a shooting competition between A and B. A is the first to shoot. The player who hits
  the target sooner is the winner. If player A has shooting accuracy of
  30% what should be player B's shooting accuracy in order that 2
  players have same chance to win the game?

I don't understand why the answer isn't 30%. Can somebody explain this a little?
 A: Player A goes first, so the chances are not symmetrical.
If player A has 100% accuracy, and goes first, then of course player B has a 0% chance of winning regardless of accuracy.
If player A has 0% accuracy, then player B has 100% chance of winning so long as he has greater than 0% accuracy.*
Now in this case player A has 30% accuracy.  You can see that if player B has 100% accuracy, he has a 70% chance of winning.  And if player B has 0% accuracy, he has a 0% chance of winning.  By varying B's accuracy, you can attain any chance of winning for B between 0% and 70% inclusive.
So please note that it's only possible to solve this question because player A's accuracy is less than (or equal to) 50%.  If player A had 51% accuracy, there would be no possible way that player B could have an equal chance of winning.

*I'm going to use "he" as these are hypothetical people and it makes the writing flow better for me.
A: Let $p_A=0.3$ and $p_B$ be the probabilities that $A$, resp. $B$, hit the target in one shot, and let $P_A$ and $P_B$ be the probabilities that $A$, resp. $B$, win the game. Then
$$P_A=p_A+(1-p_A)(1-p_B)P_A,\qquad  P_B=(1-p_A)\bigl(p_B+(1-p_B)P_B\bigr)\ .\tag{1}$$
Now we want $P_A=P_B={1\over2}$. Plugging this into $(1)$ and solving for $p_B$ gives
$$p_B={p_A\over 1-p_A}\ .$$
Since we are given that $p_A=0.3$ the resulting requirement is $$p_B={3\over7}\ .$$
