Second Generation Ride Chains

Math.

A whoooooole lotta math.

Alright, to put this gigantic ass block of calculations into words. The probability of you riding out the entire chain is 39.3%. Ignoring 1 Grade 2, you have a 71.6% chance to ride the Grade 2 if you run 2 of it, a 77.4% chance to ride it if you run 3, and an 82.2% chance of riding it at 4. Because the difference in probabilities between 3 and 4 are so little, it's more than justifiable to not run 4 if you can find something more worthwhile to take that unit's place. I can only recommend to use 2 if you aren't totally reliant on your Vanguard being at 11K defense. Also, with this sort of deck, the grade ratios required for a consistent deck are (assuming you are running 4 of the Grade 1 of the chain) at least 12 Grade 1s,  at least 7 Grade 2s, and 8 Grade 3s, leaving 6 open slots (hopefully Grade 1s and 2s) open for player preference to mess around with.

If there are any questions at all about Second Generation Ride Chains, please ask so here or in the request box, and I'll try to answer logically or with math as soon as I can.

(the rest of this article is just mathematical proofs of my work, you can entirely ignore them unless you too are a mathematician and want catch any mistakes I might make)

~To note, all these probabilities assume you are going first with 4 Grade 1s and Grade 3s of the chain in the deck with a mulligan of 3, and 1 damage per opponent's turn.

The Probability to Ride From Grade 0 to Grade 3:

Grade 1:
Probability to not get the Grade 1:
(45/49)*(44/48)*(43/47)*(42/46)*(41/45)*(43/47)*(42/46)*(41/45)*(40/44)
(41*41*43*43)/(3*23*46*47*47)
3108169/7011366

Probability to get the Grade 1:
1-3108169/7011366=3903197/7011366=(29*134593)/(3*23*46*47*47)

55.7% chance of getting the Grade 1 by turn 1.

Grade 2:
Grade 1 searches for it, 100%

Grade 3:
Probability you don't get it by turn 1:
(45/49)*(44/48)*(43/47)*(42/46)*(41/45)*(43/47)*(42/46)*(41/45)*(40/44)
(41*41*43*43)/(2*3*23*23*47*47)

Probability you don't get it by turn 2:
(39/43)*(38/42)*(37/41)+(4/43)*(39/42)*(38/41)
(13*19*37)/(7*41*43)+(2*13*38)/(7*41*43)
9139+988/(7*41*43)=(13*19*41)/(7*41*43)
(13*19)/(7*43)

Probability you don't get it by turn 3:
(39/43)*(36/40)*(35/39)+(39/43)*(4/40)*(36/39)+(4/43)*(37/40)*(36/39)+(4/43)*(3/40)*(37/39)

(35*36*39)/(39*40*43)+(4*36*39)/(39*40*43)+(4*36*37)/(39*40*43)+(3*4*37)/(39*40*43)

(7*9)/(2*43)+(18)/(5*43)+(6*37)/(5*13*43)+(37)/(2*5*13*43)

[(7*5*9*13)+(18*2*13)+(2*6*37)+(37)]/(2*5*13*43)=5044/5590

(2*97)/(5*43)

Probability you draw your first G1
(29*134593)/(2*3*23*23*47*47)

Probability you draw your second G1
1-[(44/47)*(43/46)*(42/45)*(41/44)*[(40/43)*(39/42)+(3/43)*(40/42)]]
1-[[(41*42*43*44)/(44*45*46*47)]*[(2*2*2*5/43)]]
1-(2*2*41*14)/(3*23*47)
947/3243
(947)/(3*23*47)

Probability you draw 1 of 4 extra Grade 3s
1-(13*19*41*41*43)/(2*3*7*23*23*47*47)
1-17853901/49079562
31225661/49079562
31225661/(2*3*7*23*23*47*47)

Probability of Grade 1 successfully searching
[(29*134593)/(2*3*23*23*47*47)]*[(947)/(3*23*47)]*[31225661/(2*3*7*23*23*47*47)]
[115420271302291499/1115964206574387156]

[(41*41)/(3*23*23*47*47)]*[(13*19)/(7)]*[(97)/(5)]*[1000543935272095657/1115964206574387156]

Probability of failing to get Grade 3:
40296986036054539081231903/136927586165871105087264180

Probability of getting a Grade 3:
96630600129816566006032277/136927586165871105087264180

(3903197/7011366)*(96630600129816566006032277/136927586165871105087264180)=

The Probability of Riding the Entire Chain is:
377168268534899630985047165489569/960049422105459026591271104669880, or 39.3%

The probability of riding the Grade 2:

Probability of not riding 1 Grade 2:
(41*41*43)/(47)*(1/7)*(1/47)*(1/23)*(1/47)*(1/23)*(44)*(43)
Probability of riding 1 Grade 2:
247697133/384456569=64.4%

Probability of not riding 2 Grade 2s:
(41*41*43)/(23*47)*(1/7)*(1/47)*(1/23)*(43)*(1/47)*(11/2*23)*(43/3)*(1/2*2)*(41)
11*41*41*41*43*43*43/2*2*2*3*7*23*23*23*47*47*47
60276721417/212220026088
Probability of riding 1 of 2 Grade 2s:
151943304671/212220026088=71.6%

Probability of not riding 3 Grade 2s:
(41*41*43)/(23*47*47)*(1/47)*(43/23)*(41/3)*(2*11/47)*(43/23)*(1/3)
2*11*41*41*41*43*43*43/3*3*23*23*23*47*47*47*47*
120553442834/534339708543
Probability of riding 1 of 3 Grade 2s:
413786265709/534339708543=77.4%

Probability of not riding 4 Grade 2s:
(41*41*43*43)/(2*3*23*23*47*47)*(13*41*41*43)/(2*23*23*47*47)
2920681137751/16386417728652
Probability of riding 1 of 4 Grade 2s:
13465736590901/16386417728652=82.2%

To set up a consistent deck ratio, each possibility to ride a grade should surpass 92.8%

Grade 1s required to break 92.8%:
12?
(37/7)*(1/47)*(17/23)*(7/47)*(17/23)*(11/3)*(2)
235246/3505683
12.

Grade 2s required to break 92.8% (3 Grade 2 form)
3+4?
(crap, I accidentally deleted the numbers)
7

Grade 3s required
8