Just a thought. Given that the slot wheel picks 5 cards at random, making 50 attempts to find non-pillar cards before giving up and including pillars, it seems like the best 'deal' for a player rare-farming would be a deck constructed of:
Rare 1x6
Rare 2x6
Pillarsx18
This way, the computer has a 2 in 5 chance of finding one of the rares on each of those 50 attempts, giving an infintesimally small chance of getting pillars in the mix, but the player still has a 52% chance of winning a card by winning the game.
This, of course, applies only to those grand masters who have 6 of a given rare to stack into a deck. 
The only question remaining is: what rares to put in? I think the best for the newbs would be Pulv and Eternity, but that's just me...
Essence, that is what I've been thinking too. that is why I tried to do close to that (Shard x 6, Pulverizer x 6, Eternity x 2, pillar x 17). My intention though was to do Eternity x 6, Pulverizer x 6, tower x 18 if I had 6 Eternities/Pulverizer. Unfortunately other than Shard, the only rare I have 6 of is probably Owl's Eye, Miracle, Fahrenheit, which people don't seem too interested in.
However, with this setup I think you will still be getting a decent number of pillars, however it is still going to give you a good chance to win rares. My understand is it goes like this:
Find 1 card at random from deck. find another card at random from deck, etc... (repeat 5 times, so you have 5 cards from deck, some of which may be the same thing (ex it could pick 3 eternities, 1 pulverizer, 1 pillar).
If ANY of 5 cards picked above are a pillar, redo the whole selection If after 50 tries you can never get a set of 5 that does not include a pillar, then just go with whatever you got (so if on the 51st try the result was 2 eternities, 2 pulverizer, 1 pillar, then that is what the wheel would have). Depending on if a card is 'removed' once it is picked in the first part, the chances of getting a group of 5 without a pillar on each individual try would be either:
(if card is eliminated from possibilties once selected)
12/30 * 11/29 * 10/28 * 9/27 * 8/26
= 95040/17100720
=~ 1/200
(if card is not eliminated from possibilities once selected)
(12/30)^5
= 248832/24300000
=~ 1/100
So in the first case since you try this 50 times, about 23% times you will end up with a spin that has no pillars (1 - (199/200)^30). If it is the 2nd then 40% the time (1 - (99/100)^30) you will end up with a spin with no pillars. Alternatively if you had a deck of 6 of three rares, and 12 pillars:
18/30 * 17/29 * 16/28 * 15/27 * 14/26
= 1028160/17100720
=~ 1/16
or could be:
(18/30)^5
= 1889568/24300000
=~ 1/13
So if you have 18 pillars it seems like most spins will include some pillars in it. However, when they didn't (somewhere between 1/4 and 1/2 of the time), they would have only 2 cards (or possibly even 1 on super rare occassions) in it so you would have a very good chance of winning.
With 12 pillars and 3 rares, somewhere between 96% (1-(15/16)^50) and 98% (1-(12/13)^50) of the time your spins should not include any pillars.
To go the next step and say which of these is more optimal is a bit trickier. Would probably need to create a simulation for that... perhaps I shall do that.
