Pretend you have 99 actions and only 2 Estates in your deck, with one Estate on top. So the other Estate has a 1 in 100 chance of being in each spot of your deck. 3 possible situations:
1) If the Estate is the very bottom card, then it makes no difference at all what order you play them in; you draw your whole deck. (1% chance)
2) If the Estate is the second from the top, then playing your Scrying Pool first means you draw nothing. Playing your Peddler first means you draw your deck. (1% chance)
3) If the Estate is in any other position at all (X from the top), then playing your Peddler first means drawing X cards, and playing your Scrying Pool first means drawing X+1 cards. (98% chance)
As you can see, 98% of the time it is better to play your Scrying Pool first. However, it is only better by 1 card. The 1% of the time where playing your Peddler first is better, it is 100 times better.
What you want to know is the distribution of the number of (useful) cards that you will draw is, depending on whether you play Scrying Pool first. It doesn't matter how much your choice to play Scrying Pool first "hurts you" in any one possible arrangement of cards in the deck. In the situation you presented, it seems like the distribution of the number of action cards that you draw will be the same whether you play Pool first or not.
Change it to 1 Estate on top plus 2 coppers in your deck and you'll have more coins for the turn, in the sense of
First Order stochastic dominance, by playing Pool first. Stef's simulation above presumably shows the same result (first order stochastic dominance in the number of coins generated by playing Scrying Pool first).
Despite all of these reasons to play Pool first, you may
regret it more if you play Pool first and the second Estate is the very next card.