The GP Duke build on blog

I had to take another look at this clan and specifically the Duke build to help a team mate. I agree Duke is not the best GP build, but it's what he has at the moment, and there's no time to change it.
Anyway, got a few questions:
1- Why is the Duke build 14/11/8 when the recommended ratio for a rank 3 chain is 15/10/8?
2- Why the presence of damage unflippers and the absence of Sleygal Daggers on a build that is extremely light on counterblast uses until final turn? (I am assuming the opponent isn't bad and will guard the Viviannes while there are open RG zones)
3- Wouldn't a few Garmores be "necessary" as backup vanguards? The probability of getting a Duke by turn 3 is 59% on average (going first 57%, going second 61%) not counting the effect of the grade 0 search and not counting mulligan. So I guess about 25% of the games we will be playing a Gigantech as a Grade 3 vanguard, which is awful.
4- Nimue needs a counterblast, so it can't use it's effect on turn 1, only on turn 2 at the earliest. Isn't making a column with a Nimue backed by a booster (another grade 1 or a grade 0) on turn 2 terrible from a scalability point of view? The ride chain can get rid of 1 RG in that column on turn 3, but that requires the full ride chain to work and that will only work on a third of the games. Doesn't sound a good bet just to make the opponent guard a weak RG attack on turn 2 that he would probably guard anyway.

1- That ratio mostly pertains to the riding aspect of a Ride Chain deck, not its overall performance. Among other things, the difference is small anyway so it's not something to worry about.
2- While I'm not sure about Tripp, Alice mentions that it was mostly optional between Nimue and Sleygal Dagger, Nimue just being the unit she picked out of the two. Again, not something to worry too much over.
3- Well, actually the probability is more like 528036725260529/688229544603384, or 76.7% when you incorporate a mulligan 3 and the Grade 1. While this doesn't include the probability of riding a Grade 1 and 2, it's a really high probability of riding Duke right away. Along with this, being a Limit Break, you don't necessarily need to use Duke right away in the first place. Sans the Ride Chain skill, there's very little Duke could be doing for much of the midgame anyway other than be a possible 11K body. So long as one can be drawn by the late game, it's not much of a loss anyway. Again, not something to worry about.
4- Depends on how you look at it. With Gareth, that's still a 15K column right there, and that also assumes you gun for Nimue to be the attacker in the first place, where it can just as easily serve as a 7K booster for Viviane and Tripp. There's a lot of wishy washy scenarios that revolve around explaining Nimue which I don't really want to dive into after grabbing that probability of Duke, but for the most part Nimue doesn't face those sort of problems. And again, it's mostly optional anyway. If you opt to use Sleygal Dagger, just use it, and don't worry about Nimue.

TehNACHO wrote:

3- Well, actually the probability is more like 528036725260529/688229544603384, or 76.7% when you incorporate a mulligan 3 and the Grade 1.

Would it be a terrible boredom to explain how you get to this number? I couldn't yet find a good way to make probability calculations involving ride chains and I'd really like to know.

Multiply the probability of not drawing it regularly by the probability of not getting it through the Ride Chain skill.

TehNACHO wrote:

Multiply the probability of not drawing it regularly by the probability of not getting it through the Ride Chain skill.

Are you factoring in that if you don't have the grade 2 in your starting hand, when using the grade 0 skill there might be the grade 2 and the grade 3 in those 7, and you will add the grade 2? That's the part that gives me a headache in rank 3 chains calculations.

(45*44*43*42*41*43*42*41*40)/(49*48*47*46*45*47*46*45*44)*[(39/43)*(38/42)*(37/41)*[(36/40)*(35/39)+(4/40)*(36/39)]+(4/43)*(39/42)*(38/41)*[(37/40)*(36/39)+(3/40)*(37/39)]]

(45*44*43*42*41*43*42*41*40)/(49*48*47*46*45*47*46*45*44) is the opening hand under the assumption of a mulligan 3.

[(39/43)*(38/42)*(37/41)+(4/43)*(39/42)*(38/41)] Is turn 2. (39/43) and (4/43) is there to assume you took one damage between the turns, and separates out. This is rejoined afterwards by the plus.

[(36/40)*(35/39)+(4/40)*(36/39)] and [(37/40)*(36/39)+(3/40)*(37/39)] are turn 3. Once again, assumes one damage and thus is split up upon 36/40 to 4/40 and 37/40 to 36/39.

Next, you find the probability that you don't pull anything from the Grade 1. You'll need to find the probability you successfully pull the Grade 1:

1-(45*44*43*42*41*43*42*41*40)/(49*48*47*46*45*47*46*45*44)

And then the probability that you are able to get the Grade 3 from the search
1-(39*38*37*36*35*34*33)/(43*42*41*40*39*38*37)

And finally, multiply the two and subtract from one to find the probability that you don't get a Grade 3 from the search.

1-[1-(45*44*43*42*41*43*42*41*40)/(49*48*47*46*45*47*46*45*44)]*[1-(39*38*37*36*35*34*33)/(43*42*41*40*39*38*37)]

Multiply this and the probability you don't draw a Grade 3 normally and subtract from one:

1-(45*44*43*42*41*43*42*41*40)/(49*48*47*46*45*47*46*45*44)*[(39/43)*(38/42)*(37/41)*[(36/40)*(35/39)+(4/40)*(36/39)]+(4/43)*(39/42)*(38/41)*[(37/40)*(36/39)+(3/40)*(37/39)]]*[1-[1-(45*44*43*42*41*43*42*41*40)/(49*48*47*46*45*47*46*45*44)]*[1-(39*38*37*36*35*34*33)/(43*42*41*40*39*38*37)]]

And so is the probability you can get.

LightLightning wrote:

Are you factoring in that if you don't have the grade 2 in your starting hand, when using the grade 0 skill there might be the grade 2 and the grade 3 in those 7, and you will add the grade 2? That's the part that gives me a headache in rank 3 chains calculations.

Again, the probability I gave was only for Spectral Duke. It wasn't factoring Grade 1s and 2s.

TehNACHO wrote:

[(39/43)*(38/42)*(37/41)+(4/43)*(39/42)*(38/41)] Is turn 2. (39/43) and (4/43) is there to assume you took one damage between the turns, and separates out. This is rejoined afterwards by the plus.

[(36/40)*(35/39)+(4/40)*(36/39)] and [(37/40)*(36/39)+(3/40)*(37/39)] are turn 3. Once again, assumes one damage and thus is split up upon 36/40 to 4/40 and 37/40 to 36/39.

You can always ignore damage when calculating the probability of draws. It's basically the same argument we had a couple months ago about soul charging effects and probability to draw triggers. Drawing cards number 1, 2, 3 or 2, 3, 4 or 17, 24, 38 from a randomized deck has exactly the same probability of finding any card.

[(39/43)*(38/42)*(37/41)+(4/43)*(39/42)*(38/41)]
is the same as [(39/43)*(38/42)]

[(36/40)*(35/39)+(4/40)*(36/39)]
is the same as [(36/40)]

[(37/40)*(36/39)+(3/40)*(37/39)]
is the same as [(37/40)]

TehNACHO wrote:

Multiply this and the probability you don't draw a Grade 3 normally and subtract from one:

1-(45*44*43*42*41*43*42*41*40)/(49*48*47*46*45*47*46*45*44)*[(39/43)*(38/42)*(37/41)*[(36/40)*(35/39)+(4/40)*(36/39)]+(4/43)*(39/42)*(38/41)*[(37/40)*(36/39)+(3/40)*(37/39)]]*[1-[1-(45*44*43*42*41*43*42*41*40)/(49*48*47*46*45*47*46*45*44)]*[1-(39*38*37*36*35*34*33)/(43*42*41*40*39*38*37)]]

I'm not 100% sure on this part but I don't think you can multiply the probability of getting it through draws with the probability of getting it through the chain, because they are not "independent" (not sure if this is the correct term) events. Basically what I mean is the draws for the grade 3 and for the grade 1 needed for the chain are using the same "event".
Doing math for ride chains really gives me a headache...

LightLightning wrote:

You can always ignore damage when calculating the probability of draws. It's basically the same argument we had a couple months ago about soul charging effects and probability to draw triggers. Drawing cards number 1, 2, 3 or 2, 3, 4 or 17, 24, 38 from a randomized deck has exactly the same probability of finding any card.

Not exactly, as these are two different scenarios. This scenario pertains to grabbing just one, and thus we're working from the chance to fail at each event and then subtracting from one to find the success. In the other one, it showed averages of how many triggers you would expect, which worked differently in this because it took every scenario and averaged them out. The math for both are highly different, and that oversimplification would just lead to inaccuracies for these numbers.

LightLightning wrote:

I'm not 100% sure on this part but I don't think you can multiply the probability of getting it through draws with the probability of getting it through the chain, because they are not "independent" (not sure if this is the correct term) events. Basically what I mean is the draws for the grade 3 and for the grade 1 needed for the chain are using the same "event".
Doing math for ride chains really gives me a headache...

Close, but I'm not doing that. I'm multiplying the failures of both and subtracting from one to find a success in either and/or both events. Also, failing to get anything from the chain and doesn't affect your ability to draw it and failing to draw it doesn't effect the probabilities around getting it from the chain, so multiplying the failures together still works.

TehNACHO wrote:

The math for both are highly different, and that oversimplification would just lead to inaccuracies for these numbers.

Where are you seeing inaccuracies? You think it's just a coincidence that the 3 "simplification" examples I gave you had the exact same results as your calculations?

TehNACHO wrote:

Close, but I'm not doing that. I'm multiplying the failures of both and subtracting from one to find a success in either and/or both events. Also, failing to get anything from the chain and doesn't affect your ability to draw it and failing to draw it doesn't effect the probabilities around getting it from the chain, so multiplying the failures together still works.

Let me simplify that with an example:
4 cards (A, B, C, D), 1 draw

What is the probability of:
1) drawing A
OR
2) drawing B?
(empirically we know it's 0,5 of course)

If we did your calculations from what I understood:

Probability of failing 1 = 0,75
Probability of failing 2 = 0,75

Probability of success = 1-([probability of failing 1]*[probability of failing 2]) = 1 - 0,5625 = 0,4375

Please correct me if I didn't understand your calculations as I'm not sure I did.

LightLightning wrote:

Where are you seeing inaccuracies? You think it's just a coincidence that the 3 "simplification" examples I gave you had the exact same results as your calculations?

Could you draw out an equation. You may be right, but simply giving me these simplified fractions but not telling me where they go or how they apply to each other doesn't help me at all understand where you're trying to point at.

LightLightning wrote:

Let me simplify that with an example:
4 cards (A, B, C, D), 1 draw

What is the probability of:
1) drawing A
OR
2) drawing B?
(empirically we know it's 0,5 of course)

If we did your calculations from what I understood:

Probability of failing 1 = 0,75
Probability of failing 2 = 0,75

Probability of success = 1-([probability of failing 1]*[probability of failing 2]) = 1 - 0,5625 = 0,4375

Please correct me if I didn't understand your calculations as I'm not sure I did.

I don't think you did '^ ^

TehNACHO wrote:

LightLightning wrote:

Where are you seeing inaccuracies? You think it's just a coincidence that the 3 "simplification" examples I gave you had the exact same results as your calculations?

Could you draw out an equation. You may be right, but simply giving me these simplified fractions but not telling me where they go or how they apply to each other doesn't help me at all understand where you're trying to point at.

Those simplified fractions are simply the draw probability (or in this case the "faillure to draw" probability) had you not damage checked anything at all.

Just forget the damage checks / soul charges / anything else that goes between draws, you don't need to take those into account as they don't matter at all in a randomized deck. Your calculations will be much more simple and the results will be exactly the same. This "property" is only valid for random events like damage checks and soul charges though, if it were the case of searching your deck for a specific card and sending it to the drop zone you would have to take that into account.

Here's the calculation for that ABCD scenario:


The difference is pick one uses OR not XOR. In OR you add the probabilities (and you'll know you're probably not wrong to use OR so long as the P doesn't meet or exceed 1). In XOR, you do the following:

(A+B) - A*B

(0.25 + 0.25) - (0.25 * 0.25) = 0.4375

Notice that's the same answer as multiplying the failures and subtracting from one? That's because XOR is the same methodology Nacho was using in dependent probability, where it belongs. However the pick 1 of 4; 2 successes scenario is not dependent actually. It's only dependent once you pick 2 or more since the first pick isn't changed by anything. After that, you use XOR.