[noaharc]

Yep, I assumed they were normalized. I don't see the canceling out, though.

If you do poorly on the first test, the x-coordinate is low/close to y-axis. You're then expected to do better on the second test, so the y-coordinate is high. This will flatten the left half of the line.

As you said, if you do well on the first test, the x-coordinate is high, and the y-coordinate is low. This will flatten the right half of the line.

Right?

[mbrubeck]

I'll assume up front that thee scores on the two tests are independent, since that's the scenario in which regression to the mean applies. (It also applies if they are correlated but have some independent "noise" component, but that complicates things.)

Your mistake is here:

"If you do poorly on the first test... you're then expected to do better on the second test, so the y-coordinate is high."

Actually, under my assumption, a student who does poorly on the first test is no more or less likely than anyone else to do well on the second test. Their y-coordinate will not be "high" in absolute terms; on average, it will be the same as the mean for the whole class. The regression exists because the group has a low starting point, not because it has a high ending point. As a group the high scorers will regress to the mean, not past it. (In the case where scores are partially correlated, the group will regress toward the mean.)

For example, suppose scores are independently, uniformly distributed on both tests. Then your scatterplot will have dots distributed uniformly over its entire area - obviously this does not change if you switch axes. And yet there is regression to the mean. Divide the graph into quadrants. If you look at the right half of the plot (high scorers on first exam), you'll see that there are as many in the upper quadrant as the lower (their mean on the second exam is the class mean). Same for the left side of the plot. This isn't order-dependent; you'll find that the high-scorers on B also regress to the mean when you look at their A scores.