Okay, I I've got an equation. I think it's correct but it looks a little unintuitive so I'm going to go through my thought process and if I made a mistake someone can point it out.

Anyway the chances of any one card pulled from the top of the deck being the card you want is y/40 where "y" is the number of copies of that cards. However, the chances of a single copy of that card being in your opening hand it greater than 7y/40 because if, for example, you draw it sequentially the chance of the first card being it is indeed y/40, but if it "whiff" the chance of the second card being it is y/39 because your deck now has one card less that isn't the card you want. As a result the chances of the desired being in your opening hand are actually

P = (y/40 + y/39 + y/38 + y/37 + y/36 + y/35 + y/34)

This is equivalent to 7y/37, so if you have one copy of the card in your 40 card limited deck the probability of drawing it in your opening hand is approximately 19%. To factor in drawing additional cards over the course of several turns you can simply extend the equation to

P = 7y/37 + SUM:(y/(34-t))

Where "t" is the turn number on the draw and SUM: is a summation. Because as far as I'm aware of reddit lacks a good equation formatter and I'm just improvising I'll run this equation through an example to demonstrate in case it isn't clear.

Say you have "y" copies of the card in your deck and you want to find out the chances of drawing at least one of them by turn 3, so t=3.

P = (7y)/(37) + (y/(34-1) + y/(34-2) + y/(34-3))

If y = 2 dragons as in OP's case the probability of drawing a dragon by turn 3 is 56.6%. Of course this equation can be used in constructed or even EDH (though in that case "y" would always =1), you would just need to change the deck size.

The more generalized equation is

P = 7y/(d-3) + SUM:(y/((d-6)-t))

Where y = number of copies of the desired card, t = number of turns passed, on the draw, and d = initial deck size.