This made me wonder, "which draft deck is better, 1, 2, or 3 pack rats?" I am assuming that in RTR draft the maximum number of rats you can have is 3, though analysis that includes a general solution for a certain number of pack rats would be super sweet because technically it's possible, even if everyone is drafting correctly (I.E. taking pack rat over everything else all the time), to have 6 pack rats (you open 3 packs with at least 1 of 2 possible pack rats, the person to your right/left/right opens packs with 1 pack rat, 1 foil pack rat).
- We have a 40 card deck made of N pack rats and 40-N swamps.
- We start the game on the play, meaning that we do not draw an extra card on our first turn.
- At the beginning of play we will draw 7 cards. If it is not guaranteed that we will have 1 pack rat and 3 lands in play by turn 3 (playing 1 land per turn), we mulligan, meaning we draw 6 cards.
- We repeat this process until we have mulliganed to 2 cards, at which point we keep the hand and play the game regardless of what that hand is.
So, for example, in a deck that has 2 pack rats, a hand of "Swamp, Swamp, Swamp" would be a mulligan. A hand of "swamp, swamp" may be a winning hand, but only if our next two draw steps are "swamp, pack rat" in any order. A hand of "pack rat, pack rat" loses. A hand of "pack rat, swamp" loses if we draw a second pack rat. A hand of "pack rat, pack rat, swamp" wins.
My question to you is, what build of N pack rats and 40-N n swamps, 0 < N < 7, N is a positive integer, gives us the highest chance of "winning"?
(I honestly have no earthly idea how to approach this because I'm really bad at statistics)