Not to belabor the point (too late), but I think this illustration might help shed some light.
I have a simulation with 30,057 random card draws. Of those 30,057, there are 9,110 results of the bottom card being an Estate, yielding 30.3%. This is to be expected. So far, so good.
A) Of those 30,057, there are 2,570 results where the first hand has 0 Estates. Of those 2,570, there 1,536 results of the bottom card being an Estate, yielding 59.8%. This coincides with an earlier claim that the probability is 60%.
B) Likewise, there are 12,464 results where the first hand has 1 Estate. Of those 12,464, there are 5,102 results of the bottom card being an Estate, yielding 40.9%. Again, this coincides with the claim that the probability is 40%.
C) There are 12,484 results where the first hand has 2 Estates. Of that, there are 2,472 results of the bottom card being an Estate, yielding 19.8%.
D) Naturally, 0% of the 2,539 results where the first hand has 3 Estates had an Estate on the bottom.
This is where the condition is important. If you add all these up (probability of A times the sample size of A and so forth), then you get the final number of 30%. It is absolutely true that the last card (indeed any one card) being an Estate is 30% when you don't know anything about the deck. Once you know the first hand, then you know that you are either in case A, B, C, or D. The probability that the final card of a fresh shuffle being 30% is still there, but you are now no longer looking at a fresh shuffle. You are now looking at one of four cases. Knowledge of the first hand gives you better information.