I'm guessing he used the same method Chriskang used eons ago (i.e. dwelving into the depths of the game's code, getting the algorithm, calculating and then testing the results).
that's what i did, but i could have misunderstood the algorithm. I've never programmed in flash and i dont even know how to start, i also dont know how a flash program works :p
Still i think what i've done is accurate.
I'd love to hear a little more about the methodology. jmizzle says above that only zanz knows the current algorithm, and afaik you aren't zanz. 
You mention using four cards in the spins but some previous sources point to five. And you didn't mention the number of repicks to find a non-pillar card.
I'm not zanz indeed
, and that's why didn't post the algorithm, maybe he will keep that secret. But i think its interesting to see what the real chances are to win when fighting FG, especially when you're a newb to the game and have a hard try to win vs FG for no rewards :p
The general methodology is not so complex (due to the fact that afaik there is no 10% chance to turn the first card into a shard). Here is what i did:
The probability to win a card from a FG in 1 spin is equal to the sum of the probabilities to win a specific card in 1 spin. P[win a card in 1 spin] = SUM(P[win card n° i in 1 spin]) i going from 1 to inf (whe check all cards)
the slot machine has 3 reels, each reel has the same 4 cards on it (and no, its not 5) . To calculate the probability to win a specific card C, you must know how many times that card is on the reel (otherwise said, how many card C are among the 4 cards).
if n° C is 0 : probability to win the card is = 0 (you cant win the card if it is not present on the reel)
if n° C is 1 : probability to win the card is = (1/4)*(1/4)*(1/4) = 1.5625 %
if n° C is 2 : probability to win the card is = (2/4)*(2/4)*(2/4) = 12.5 %
if n° C is 3 : probability to win the card is = (3/4)*(3/4)*(3/4) = 42.18 %
if n° C is 4 : probability to win the card is = 1
Ill call those probabilities P(w0),P(w1),P(w2),P(w3),P(w4)
On the tables i've listed the probability a specific card C will be choosen for the reel, lets call it P(c). This mean that each of the 4 cards has a probability of P(c) to be the card c, and (1-p(c)) to be another card. P(c) depends on the algorithm.P(c) is very low for towers,unless you deck is solely made of towers :p
To have 4 cards C in the reel, the algorithm has to choose 4 times the card C and the probability is given by P(c)*P(c)*P(c)*P(c).
To have 3 cards C in the reel, the algorithm has to choose 3 times the card C and 1 time another card and the probability is given by 4*P(c)*P(c)*P(c)*(1-P(c)). In this case there are 4 possibilities, one where the non C card is choosen first, one where it is choosen 2nd, 3rd, ...
To have 2 cards C in the reel, the probability is given by 6*P(c)*P(c)*(1-P(c))*(1-P(c)). There are 6 possibilities (6 is the number of ways to choose 2 objects among 4).
To have 1 cards C in the reel, the probability is given by 4*P(c)*(1-P(c))*(1-P(c))*(1-P(c)).
I'll call those probabilities P(c1) P(c2) P(c3) P(c4)
So the probability to win card C is given by:
P(w0)*P(c0) + P(w1)*P(c1) +P(w2)*P(c2) +P(w3)*P(c3) +P(w4)*P(c4)
= 0 + (1/4)^3 * 4 * P(c) * (1-P(c))^3 + (2/4)^3 * 6 * P(c)^2 * (1-P(c))^2 + (3/4)^3 * 4 * P(c)^3 * (1-P(c)) + P(c)^4
I´m always careful with this kind of "epic post". There are many people in this forum, who talk about probabilities and make beginner mistakes like not using the right distributions , independence mistakes, adding probabilities, although you are not allowed to do this.
Did you use the algorithm of the Slot Wheel Mystery (http://elementscommunity.org/forum/index.php/topic,590.0.html), but with 4, not 5 cards ?
I dont say i did any mistakes, but i've verified the calculus methodology with a simulation(the two being base on my understanding of the algorithm). And the hard part was to find the slot machine mechanism, this is basic probabilities, but i'm aware its very easy to do mistakes.
And the algorithm mentionned in the first page is outdated.
