So I finally got my V5 3 days ago 
Turned out to be better than I expected 
Anyway, I came across a couple of sources (including cutex.info, bigcubes.com, Erik's video, etc.) claiming different numbers of cases for the last two tredges: one said 2, another said 7, another said 12, another said 15... Point is, I was completely confused.
So this was what I did: Defining the last two midges as solved, we have 4! = 24 cases. And after many many many hours of work, I've categorized them into 8 groups.
A
B
*
*
C
*
*
D
E
F
G
H
All cases within the same group can be transformed into the asterisked case either by doing a U2, or by flipping one of the tredges. You never have to do both, i.e. U2 and flip one of the tredges. And to flip one of the tredges, you need an average of exactly 2 moves, you can see how I arrived at this number later on.
So in other words, excluding the solved case, there are only 7 algorithms needed for the last two tredges, and there isn't even a need to mirror any of them. I suck at mirroring on the fly (be it L/R or F/B), so this is definitely good for me.
To flip one of the tredges:
0 moves: 1 case
1 move: 3 cases
2 moves: 13 cases
3 moves: 5 cases
That gives us an average of (0+3+26+15)/22 = 2 moves.
Now, AvG versus bigcubes.com. With the bigcubes.com tredge-pairing method, the probability of ending up with each case is obvious, you just have to count the number of cases in each group:
A: 1/3 (easiest, intuitive case)
B: 1/6
C: 1/6
D: 1/12
E: 1/12
F: 1/12
G: 1/24
H: 1/24
With AvG, it's a little different: Cases C, D, E, F and G can only occur if they were already like that right from the start, i.e. once the centers were done. So the combined probability of cases C, D, E, F and G occurring is 1/(24*23) + (1+4+2+2)/(24*23*22*21) = 0.00185, that's 0.2%. (Someone double check this for me please!)
So actually, the only cases that ever appear are A and B (and of course H). Solving A is intuitive. So you only need to learn one alg to solve B if you're doing AvG tredges.
If you get cases C, D, E, F or G, consider yourself unlucky and do the alg for either A or B to reduce the case to A or B. There really isn't any point in learning algs for these cases because they occur so rarely that you don't really get to practice them at all and you'll probably take longer to recall them than to do 2 algs.
Besides, for cases like D and G, you probably would've recognized them while pairing tredges and fixed them while it was still easy.
P/S: Keep in mind that this entire post is made by someone who just got his V5 3 days ago, and hasn't even decided which tredge-pairing method to use. It might be full of errors so don't just take what I said for granted, double check the cases and probabilities and whatever on your own and please feel free to point out my mistakes. I don't claim to know much about each tredge-pairing method, but whatever I've just presented represents my independent study on the last 2 tredges.
Turned out to be better than I expected Anyway, I came across a couple of sources (including cutex.info, bigcubes.com, Erik's video, etc.) claiming different numbers of cases for the last two tredges: one said 2, another said 7, another said 12, another said 15... Point is, I was completely confused.
So this was what I did: Defining the last two midges as solved, we have 4! = 24 cases. And after many many many hours of work, I've categorized them into 8 groups.
A
B
C
D
E
F
G
H
All cases within the same group can be transformed into the asterisked case either by doing a U2, or by flipping one of the tredges. You never have to do both, i.e. U2 and flip one of the tredges. And to flip one of the tredges, you need an average of exactly 2 moves, you can see how I arrived at this number later on.
So in other words, excluding the solved case, there are only 7 algorithms needed for the last two tredges, and there isn't even a need to mirror any of them. I suck at mirroring on the fly (be it L/R or F/B), so this is definitely good for me.
To flip one of the tredges:
0 moves: 1 case
1 move: 3 cases
2 moves: 13 cases
3 moves: 5 cases
That gives us an average of (0+3+26+15)/22 = 2 moves.
Now, AvG versus bigcubes.com. With the bigcubes.com tredge-pairing method, the probability of ending up with each case is obvious, you just have to count the number of cases in each group:
A: 1/3 (easiest, intuitive case
)B: 1/6
C: 1/6
D: 1/12
E: 1/12
F: 1/12
G: 1/24
H: 1/24
With AvG, it's a little different: Cases C, D, E, F and G can only occur if they were already like that right from the start, i.e. once the centers were done. So the combined probability of cases C, D, E, F and G occurring is 1/(24*23) + (1+4+2+2)/(24*23*22*21) = 0.00185, that's 0.2%. (Someone double check this for me please!)
So actually, the only cases that ever appear are A and B (and of course H). Solving A is intuitive. So you only need to learn one alg to solve B if you're doing AvG tredges.
If you get cases C, D, E, F or G, consider yourself unlucky and do the alg for either A or B to reduce the case to A or B. There really isn't any point in learning algs for these cases because they occur so rarely that you don't really get to practice them at all and you'll probably take longer to recall them than to do 2 algs.
Besides, for cases like D and G, you probably would've recognized them while pairing tredges and fixed them while it was still easy.
P/S: Keep in mind that this entire post is made by someone who just got his V5 3 days ago, and hasn't even decided which tredge-pairing method to use. It might be full of errors so don't just take what I said for granted, double check the cases and probabilities and whatever on your own and please feel free to point out my mistakes. I don't claim to know much about each tredge-pairing method, but whatever I've just presented represents my independent study on the last 2 tredges.
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