Yeah, that was one of my solutions but it only happens 2.7% of the time. Of course, unless my math is wrong.

I think your math is wrong.

I compute probability of drawing your 2 opening purchases and 3 coppers on turn 3 is P(first purchase in top 5 cards)*P(second purchase in top 5 cards | first purchase in top 5 cards)*P(first 3 other cards are copper) = 5/12*4/11*(7/10*6/9*5/8) = 35/792 = 4.4%.

Or by counting (using nCr to denote the binomial coefficient "n choose r"):

number of successful turn 3 hands = 5C2*2*7C3 (choose locations of the 2 opening purchases in the top 5 cards, they can be in either order, choose the locations of the 3 estates in the remaining cards)

number of possible deck arrangements = 12C7*5C2*2 (choose the locations of the 7 copper out of 12, choose the locations of the 2 non-estates of the remaining 5, the 2 can be in either order)

So P(success) = 7C3/12C7 = 35/792 = 4.4%.

This 4.4% number actually shows up in the "basic opening probabilities" blog entry (

http://dominionstrategy.com/2011/03/09/basic-opening-probabilities/) as the probability of drawing cccss on turn 3, and Jack Rudd got the same thing, so I'm pretty sure it's right. Of course, this is still not as good as your 4.8% solution, assuming your math is right for that solution.