OK, here’s the deal.

I will put myself into the ground as a hypothetical teammate and go through the logic.

First, and most important, I assume that there is a solution. This is key, because it reduces the options of the show.

Second, I assume that everyone on the team is facing the same wall, but could be on either side. This is not stated in the problem, but seems to be assumed. If you don’t assume this, and instead assume there are multiple walls and teammates are facing any direction willy-nilly, it is unsolvable.

Third, I assume that everyone on my team is smarter than I am and can work through this logic as quickly or more quickly than I can.

Now, I know that if all four of the team members are on the same side, then team member #4 could see all three heads in front of him and therefore within a couple of seconds, he would shout out the answer. So if I don’t hear anything for a couple of seconds, I assume all four of us aren’t on the same side of the same wall.

Further I assume that since the test is presumed to be solvable, that there are not two team members on each side of the wall, because that would be unsolvable.

This means I am forced to conclude that there is one team member on one side of the wall, and three on the other.

Now, if I look at the wall and see only wall, I know that I am either the single team member on the 1 side, or else I am the first team member on the 3 side. Either way I know that I can’t determine the X on my own head. So I remain silent.

If I look at the wall and see two heads, with the same color X on them, then I know my X color and will shout it out. If I see two heads with alternating colors, then I can’t determine my color and so must remain silent.

Which leaves head # 2 on the 3 side.

If I look at the wall and see one head with a black X on it, I know that I am on the 3 side. If I don’t hear head #3 behind me yell out “White” within a very short time, I know that he must not be seeing two black Xs, meaning…

Hey! My X must be white!