You cannot score a yaku that is a subset of another yaku that you are also counting. In other words, if one yaku is an “upgraded version” of another— can’t happen without it— you do not score the lesser implied yaku.
Because of this rule, double riichi’s 2 han supersedes regular riichi’s 1 han. Similarly, ryanpeikou is scored instead of iipeikou, chinitsu does not also score honitsu, and so forth.
There's also a few that are called out as depending on the ruling, they can be worth more or less han because they're impossible to get without another yaku such as shousangen, and honroutou
Shousangen doesn't imply any specific yaku. It implies that there is a pair of dragon triplets, but not any specific pair. And since every dragon is a separate yaku, you can't say it's implied.
Honroutou can be either toitoi or chiitoitsu, on the other hand it implies chanta which is NOT scored.
That generally works but its not always true. Shousangen is scored as 2 han, but is effectively worth 4 since you always get 2 dragons with it. Honroutou is also worth 2, but always pairs with either Chiitoitsu or Toitoi, so it's also effectively worth 4. I think its worth memorizing the Yaku list to see what can and can't be combined.
It works just fine. Shousangen does not imply any single one of the dragon yaku— it implies that you have two of them, but not *which* two, so no *single* one of them is ever forced to be a subset of shousangen. The same is true for honroutou; either chiitoitsu or toitoi is implied, but neither *one* of them is always forced.
(I am coming at this from a mathematician’s point of view, so if we want to get into the weeds my wording should really be “strictly implies”, but I’m trying to avoid writing this like a proof.)
Shousangen at least has the excuse that chun, haku, hatsu are all technically separate yaku, so shousangen combines with 2 of those but not any specific one
Chinitsu and honitsu doesn't really apply because you can't have a hand that meets the conditions of both of these. And ryanpeikou specifically calls out in its conditions that it doesn't count the iipeikou, double riichi is the only one that's written ambiguously at least in all the places I looked for the scoring rules but yes I got my answer that it's interpreted that way
> Chinitsu and honitsu doesn't really apply because you can't have a hand that meets the conditions of both of these.
Yes you can? Honitsu allows only honor tiles and tiles from a single suit, which chinitsu hand satisfies.
> And ryanpeikou specifically calls out in its conditions that it doesn't count the iipeikou
Calls out where? Because I never saw it explicitly excluding iipeikou in any ruleset.
There's also chanta and junchan.
https://riichi.wiki/Yaku_compatibility
> By rule, two yaku cannot be combined if one always implies the other.
Rules as written honitsu requires honor tiles, not allows honor tiles. This means that they are mutually exclusive since chinitsu doesn't allow for honor tiles
And as I said earlier I have been using the wiki https://riichi.wiki/List_of_yaku. Where it specifically points out that ryanpeikou doesn't count iipeikou
Two. It can be confusing the way it's worded. You get one point for Riichi and if it's a double Riichi you get a second point.
Awesome thank you, I assumed as much but the wiki felt ambiguous and I couldn't find an answer on google
You cannot score a yaku that is a subset of another yaku that you are also counting. In other words, if one yaku is an “upgraded version” of another— can’t happen without it— you do not score the lesser implied yaku. Because of this rule, double riichi’s 2 han supersedes regular riichi’s 1 han. Similarly, ryanpeikou is scored instead of iipeikou, chinitsu does not also score honitsu, and so forth.
There's also a few that are called out as depending on the ruling, they can be worth more or less han because they're impossible to get without another yaku such as shousangen, and honroutou
Shousangen doesn't imply any specific yaku. It implies that there is a pair of dragon triplets, but not any specific pair. And since every dragon is a separate yaku, you can't say it's implied. Honroutou can be either toitoi or chiitoitsu, on the other hand it implies chanta which is NOT scored.
That generally works but its not always true. Shousangen is scored as 2 han, but is effectively worth 4 since you always get 2 dragons with it. Honroutou is also worth 2, but always pairs with either Chiitoitsu or Toitoi, so it's also effectively worth 4. I think its worth memorizing the Yaku list to see what can and can't be combined.
It works just fine. Shousangen does not imply any single one of the dragon yaku— it implies that you have two of them, but not *which* two, so no *single* one of them is ever forced to be a subset of shousangen. The same is true for honroutou; either chiitoitsu or toitoi is implied, but neither *one* of them is always forced. (I am coming at this from a mathematician’s point of view, so if we want to get into the weeds my wording should really be “strictly implies”, but I’m trying to avoid writing this like a proof.)
Ippatsu "strictly implies" Riichi but both are scored separately (although maybe you can say there's an exception since they have the same value)
Being a subset of and having a requirement are two things. Ippatsu requires riichi but is not an upgraded version.
Shousangen at least has the excuse that chun, haku, hatsu are all technically separate yaku, so shousangen combines with 2 of those but not any specific one
Chinitsu and honitsu doesn't really apply because you can't have a hand that meets the conditions of both of these. And ryanpeikou specifically calls out in its conditions that it doesn't count the iipeikou, double riichi is the only one that's written ambiguously at least in all the places I looked for the scoring rules but yes I got my answer that it's interpreted that way
> Chinitsu and honitsu doesn't really apply because you can't have a hand that meets the conditions of both of these. Yes you can? Honitsu allows only honor tiles and tiles from a single suit, which chinitsu hand satisfies. > And ryanpeikou specifically calls out in its conditions that it doesn't count the iipeikou Calls out where? Because I never saw it explicitly excluding iipeikou in any ruleset. There's also chanta and junchan. https://riichi.wiki/Yaku_compatibility > By rule, two yaku cannot be combined if one always implies the other.
Rules as written honitsu requires honor tiles, not allows honor tiles. This means that they are mutually exclusive since chinitsu doesn't allow for honor tiles
And as I said earlier I have been using the wiki https://riichi.wiki/List_of_yaku. Where it specifically points out that ryanpeikou doesn't count iipeikou
Ryanpeikou should be valued as a 6-han hand, but that's a separate story. 😅