Intuition behind "the probability that the first person chooses the red ball before the second person is not 1/2, they take turns choose balls" 
There are 6 balls, 2 of them are red. A and B choose them one by one, prove that the probability that A takes a Red first, before B, is not 1/2

What is the intuitive explanation behind it? Of course one can directly find the probability, but why is such the case?
 A: If you want to compare the probabilities of two events, a good way is often to match up outcomes that lead to one event with outcomes that lead to the other event.
Here, we can consider the following matching: given an order in which balls are drawn, swap the balls drawn by A with the balls drawn by B. For example, if the order is "blue, blue, red, blue, red, blue", then A draws "blue, red, red" and B draws "blue, blue, blue". Swapping those, we get "blue, blue, blue, red, blue, red" where A draws "blue, blue, blue" and B draws "blue, red, red".
How does this affect who wins (if drawing the first red ball is "winning")?


*

*In most cases, this pairs up an "A wins" order with a "B wins" order. If A's sequence contains a red ball earlier than B's sequence, then after swapping the two sequences, the reverse will hold.

*However, in cases where A first draws a red ball on the $k^{\text{th}}$ turn, and so does B, this pairs up an "A wins" order with another "A wins" order (since A's $k^{\text{th}}$ turn comes before B's $k^{\text{th}}$ turn). 


(In fact, the second kind of cases always pair an order of ball drawings with itself. For example, "blue, blue, red, red, blue, blue" is paired with itself. But this does not matter for the argument. In cases with more than 2 red balls, the second kind of cases can pair two different "A wins" orders.)
We've matched up "A wins" orders with "B wins" orders, one for one, and had a few "A wins" orders left over. Therefore A is more likely to win.

This argument also works for any number of blue and red balls, where calculating the exact probabilities might be more annoying. If there is exactly one red ball, then A and B have an equal chance of winning, because there are no cases of the second type. Otherwise, A has a greater chance of winning.
A: Assume that all six balls are selected, with person $A$ choosing first in each of the three rounds.
As you observed correctly in the comments, if there were only one red ball among the six balls, then person $A$ and person $B$ would be equally likely to choose the red ball.  That is because the only red ball is equally likely to be in any of the six positions in the sequence of selections and each person has three chances to select a ball.
However, if there are two red balls among the six balls, the first red ball that is selected must be selected during one of the first five selections.  Since $A$ selects first in each round, person $A$ has three of those five selections, while person $B$ has just two, giving person $A$ more chances than person $B$ to select the first red ball.  Moreover, if neither red ball is picked during the first two rounds, person $A$ is guaranteed to pick a red ball before person $B$ since both balls left are red and person $A$ chooses first.
