i dont think your dimostration is exact. At least in exposition: remember my deck is 16 towers, 6 graboids and 6 shireker, so your cases must be numbered with towers 1-16, graboids 1-6 and shriekers 1-6, doing all possible combinations. You have even to take count of all exchnge possibilities between cards in different turns, and all possibilities of unplayed cards. Then compare that number to all possible combinations (302 for turn 1) and you'll got it. Seems a huge work just to get an idea of it calculating turn 1...
And there are even more cases: just shriek+ seven towers then graboid then shriek and more others. Supposing you start you leave the most effective cases :those starting with 1 card more!
Seems you argued to get the right, not to discover truth
My considerations:
At first glance it think that hands with lot of towers and graboids come often. You almost everytime have 2 shriekers on turn 3, but sometimes, as i've said, 3. And sometimes is not so uncommon, even quite common, practise demonstrate it.
I dont think to have exagerated: Just said i was searching for something faster than mono fire and with a ridicolous percentage of bad draws (by now, for me, 0 losses with just graboids and towers upgrade)
One day, when i'll be bored as you, i'll get you exact percentage dates
Btw, i took away antlion and sword and got 2 unupgraded (=cheaper) shields, useful if you've not so much creatures upgraded
EDIT: btw, what BS means?
EDIT: i've edited this reply just twice, to improve exposition and try not to sound aggressive (because i'm not, but my basic english sounds not so fair sometimes)
Just to help muddy the waters a little more
, let's look at the probabilities. First I'm using a slightly modified version to help give easier percentages (18 pillars [60%], 6 graboids [20%], 6 shriekers [20%]). In your opening 8 card hand, you should have 4-5 [4.8] towers and 3-4 [3.2] creatures (1-2 each of graboids and shriekers). On turn 2, you should have 5-6 (5.4) towers and 3-4 (3.6) creatures. By turn 3, you'll have 10 cards with 6 towers and 4 creatures (2 graboids and 2 shriekers). This is the most probable 3-turn-draw based on the percentages of your different cards.
So let's look at the most probable situation.
Turn 1:
5 Towers, and 1) 2 Graboid and 1 Shrieker ; or 2) 2 Shrieker and 1 Graboid
Play 5 towers and Graboid
2 damage, 7 quantum
Turn 2:
Draw 1a) Shrieker / 2a) Graboid ; or 1b, 2b) Tower very possibly
Play 1a) Graboid or possibly 1b) Graboid and Tower ; or 2a) Graboid or possibly 2b) Tower and nothing
1a) 14 damage, 9 quantum ; 1b) 14 damage, 11 quantum ; 2a) 14 damage, 9 quantum ; 2b) 12 damage, 14 quantum
Turn 3:
Draw 1a, 2a) Tower ; or 1b, 2b) Shrieker/Graboid
Play 1a) Tower and Shrieker or possibly 1b) Shrieker; or 2a) Tower and Shrieker or 2b) Shrieker and Graboid
1a) 44 damage, 8 quantum ; 1b) 44 damage, 9 quantum ; 2a) 44 damage, 8 quantum ; 2b) 34 damage, 9 quantum
Turn 4
Draw Tower
Play Tower and Shrieker
1a, 2a) 84 damage, 8 quantum ; 1b) 84 damage, 9 quantum 2b) 74 damage, 9 quantum
Turn 5
Draw Shrieker/Graboid
Play Shrieker/Graboid
1a, 1b, 2a) 134 or 126 damage ; 2b) 124 or 116 damage
Sorry it's a little confusing. Just trying to show the major probabilities.
Probabilities favor having 3 shriekers by turn 3 by a little, but honestly it doesn't matter. Either way, you'll have 4 shriekers by turn 4 and finish them by the end of turn 5. This is on average based on the percentages of each card you have. Hope that helps a little.