$2^{12} = 4096$ possible outcomes and we want to count how many of these enable the team to win. To do this, we let the 12 players form 6 pairs so that the members of each pair are diametrically opposite. Note that the positions of four players form a rectangle if and only if they form two pairs.

Call a pair 'bad' if the two players put up flags of the same colour. Note that if there are three or more bad pairs, then at least two pairs will both have chosen the same colour, so the team loses. Hence for the team to win, there are three possibilities:

* There is no bad pair — that means for each pair the two players will put up flags of different colours (2 possibilities for each pair), so there are altogether $2^6 = 64$ possibilities for this case.

* There is one bad pair — there are 6 choices for the bad pair, and regardless of whether a pair is good or bad, there are 2 possible combinations of flag colours for each pair. Hence there are $6 \times 2^6 = 384$ possibilities for this case.

* There are two bad pairs — there are $\binom{6}{2} = 15$ choices for the bad pairs, and one bad pair must put up red flags and the other bad pair must put up blue flags (otherwise the team loses), giving 2 possibilities. There are 2 possible combinations of flag colours for each of the other 4 pairs, and so there $15 \times 2 \times 2^4 = 480$ possibilities for this case.

It follows that the answer is $\frac{64 + 384 + 480}{4096} = \frac{29}{128}$.