Saturday, April 21, 2018

Saki 189: Succession

Someone uploaded the translated Saki chapters to MangaDex. Thank you~ I've been wanting to do that for weeks but always kept leaving it for later. Too many chapters. The other series too.

Saki ch189 (Mega) v2
Saki ch189 (Mediafire) v2

It's like the cover from a future Shinohayu chapter
many years from now or something, not that I know though.

v2: OK I've made a change to the joke in page 7. After Sawaya asks if she can drop the honorific, instead of calling Kyouko by her name or surname, she suddenly makes up a name for her as a joke. On top of that, it's not even a girl's name, it's "Saburou" which also means "third son". It's so obvious now but it took me some time to figure out Sawaya makes up that name out of the blue. I kept thinking "Who the hell is Saburou?" 😅


  1. Thank you!

    Himeko may have surpassed the key, but Ryuuka surpassed Himeko.

    Now that we see how precise Ryuuka's "Limitless Heaven" really is, I'm starting to understand why her Tokipathy ability gives her perfect win predictions.

  2. That's a weird way to tell us about the last hand Ritz. Also that chapter felt a bit short... But well, I didn't expect a chapter today (hadn't realize it had almready been 2 weeks since the last) so whatever. Thanks for the hard work

  3. Himeko having a 3man and 9sou wait seems unrealistic given a 12 han hand since chintsu's out and 9sou was her high out. Thanks for the chapter.

    1. Well let's see Richii, Tsumo, San Ankou, 2 Fanpais, 3 doras... Not good enough, but with 1 kan we could add 3 more doras and it would work.
      It would be a bit weird though, it may have been a mistake

    2. This comment has been removed by the author.

    3. She declared riichi though which would make that a suuankou. The likely thing would be it an error for a 3/9m out, possibly trading ittsu, a dora or iipeikou for the low out. This is all speculation though.

    4. Oh right. Forgot about that. If that's the case then forget about what I said above.
      Thanks for pointing that out.

    5. The only possibility I could think of Himeko getting 12han right now with riichi is Himeko made a kan, with her betting on an ippatsu for the 13th han, which Ryuuka kills it off.

      Riichi,Tsumo,San Ankou, Dora 8.

      My brain is fried now.

    6. She definitely has riichi.
      It seems she doesn’t have double riichi.
      She definitely has menzen tsumo because she has riichi and wins by self-draw.
      She doesn’t have houtei raoyui or chankan because she wins by self-draw.
      While unlikely, she can have haitei raoyue or rinshan kaihou (but not both).
      She can have tan’yao only if she wins on 3 man.
      She could have ippatsu only if Ryuuka had not called pon and had let her win on 9 sou. It’s not clear if she could have ippatsu at all, since she is said to have “declared riichi one turn in advance”.
      She can have haitei raoyue only if she wins on 3 man because we saw 9 sou was not the last tile.

      She can’t both win on 9 sou and have rinshan. 9 sou would be her live-wall draw, and upon drawing it she would either win immediately without rinshan, or if she declared kan, she wouldn’t get another 9 sou from the dead wall because she’d already have them all in her hand.

      She definitely has two pairs of numbers, one in man and the other in sou (but not necessarily 3 man and 9 sou). Whichever suit she wins on, the pair of that suit will become a koutsu.

      She can’t have shousangen because her pair is numbers.
      She can’t have honroutou because she has 3 man or an adjacent tile.
      She can’t have hon’itsu or chin’itsu because she has both man and sou.
      She can’t have toitoi because combined with menzen tsumo it would be suuankou.
      She can’t have pinfu, chiitoitsu or ryampeikou because she has a koutsu upon winning.

      She can’t have three or more yakuhai because she has a koutsu of numbers upon winning and combined with menzen tsumo this makes suuankou.

      She can’t have chanta or junchan: if she did have chanta or junchan, any 2 man tiles in her hand would have to be part of complete 123 shuntsu (and the only other man tiles could be 1 man forming some koutsu and a pair) in case she won on 9 sou, but she would have to have a 12 taatsu in case she won on 3 man; both of these conditions can’t hold at the same time.

      She can have sankantsu only if she has rinshan. If she already had three kan before her last draw, she would have to have the shampon wait 99s33m, which is toitoi and suuankou.

      Yaku that remain:

      1 fan:

      * iipeikou
      * up to two yakuhai
      * tan’yao (yasume)
      * haitei raoyue (yasume)
      * rinshan kaihou (yasume)

      2 fan:

      * sanshoku doujun
      * ittsuu
      * san’ankou
      * sankantsu
      * sanshoku doukou

      Each of the 2-fan yaku requires either three shuntsu or three koutsu.
      In addition, iipeikou is incompatible with any of them.
      Sanshoku doujun and ittsuu are also incompatible with each other.

      If she has three shuntsu, she has no yakuhai because the fourth set is her koutsu of numbers, and she has no rinshan kaihou because she has no other koutsu that she could kan.

      * If she has sanshoku doujun, she can have at most tan’yao and haitei raoyue for 3 man or neither for 9 sou. That’s 6 and 4 fan respectively.
      * If she has ittsuu (which is incompatible with tan’yao), she can have at most haitei raoyue for 3 man or nothing extra for 9 sou. That’s 5 and 4 fan respectively.
      * If she has iipeikou, she can have either one yakuhai or tan’yao plus either haitei or rinshan for 3 man or only one yakuhai for 9 sou. That’s 5 and 4 fan respectively.
      * If she has san’ankou without sanshoku doukou, she can have two yakuhai, sankantsu and rinshan for 3 man or just two yakuhai for 9 sou. That’s 9 and 6 fan respectively.
      * If she has sanshoku doukou, she can’t have yakuhai because she needs a shuntsu to avoid toitoi/suuankou, and she can’t have tan’yao because the potential koutsu in sou will instead be treated as a pair, so she can only have sankantsu and rinshan for 3 man or san’ankou for 9 sou. That’s 9 and 6 fan respectively. In addition, she can have sanshoku doukou only with one of the two winning tiles but not with both, because she must have a shuntsu+pair shape to avoid toitoi/suuankou but the pair becomes a koutsu only for one of the winning tiles.
      * If she has none of the above, she has at most two yakuhai and either haiten or rinshan for 3 man or only two yakuhai for 9 sou. That’s 5 and 4 fan respectively.

    7. Note that in every case, her score with 9 sou can’t be higher than with 3 man except if she has ippatsu (and has none of haitei, sankantsu and rinshan).

      On top of this, she can have some dora. I doubt that anyone but her would declare kan in this hand, so there’s probably only one dora indicator unless she has kantsu herself. That’s at most 4 dora. Plus there are four (?) red fives, for a total of 8 dora. It would be more interesting to find a hand that doesn’t rely on dora so much, but I don’t think it’s even possible to have 8 dora like this in a hand without kantsu. Any hand must contain 3n+2 sou tiles chaining up to 9 sou, 3n+2 man tiles chaining to 3 man and 3n pin tiles, and since we want to have 5 of each suit and two of 5 pin, we need five sou tiles, five sou tiles and a koutsu of 5 pin. The five tiles of sou and man must be composed of a pair and a set, and it’s impossible to have four identical tiles in this shape. But there’s a chance to have 8 dora if we include the winning tile. The sou tiles must contain 5 and 9, so they must be 55599. We need a shuntsu in man to avoid toitoi/suuankou, and one option is 33345. Now if Himeko draws 3 man, she’ll have all four red fives and four 3 man, so if 3 man is a dora, she’ll have 8 dora. Unfortunately, this means winning on 9 sou gives her a cheaper hand, but we’re told otherwise. So if she is to have at least 8 dora, someone must have declared kan.

      If she is to have 7 dora without any kan declarations, she needs 5 fan from yaku. This is only possible if she has san’ankou and a single yakuhai. To have 7 dora she needs at least one red 5 pin, so two of her sets are 555p and a yakuhai, leaving only 7 tiles for sou and man. Either sou or man will have to be a lone pair, so she doesn’t have the red five of that suit. In the other suit, she must have shuntsu+pair, so she can’t have four identical tiles either. So she has only 6 dora at most, and a 7-dora hand is impossible.

      If she is to have 6 dora without any kan declarations, she needs 6 fan from yaku. This is only possible if she has san’ankou and two yakuhai. Again, either sou or man has to be a lone pair and the other suit can have at most three identical tiles, so she can have only three regular dora and only one red five. So a 6-dora hand is also impossible.

      Since she can’t have more than 6 fan from yaku when she wins with 9 sou, I conclude that someone must necessarily have declared kan and Himeko must have either kan dora or kantsu of regular dora in her hand.

      How’s this for a scenario: 9 sou is not her winning tile. And the 3 man is not the tile she drew from the wall in her turn. Instead, whatever tile she drew let her call kan, and 9 sou would also have let her call kan. The kan replacement tile turned out to be 3 man. Either using 9 sou would have given her ippatsu or 9 sou itself is a dora but the tile she actually drew isn’t. She doesn’t need any kan dora, and even if 9 sou is a dora, she doesn’t need a kantsu of dora when she wins on 3 man. If 9 sou is a dora, her hand could be EE 999(9)s 999(9)p 12999m: riichi, tsumo, rinshan, chanta, san’ankou, sanshoku doukou, dora 3 or 4. If 9 sou is not a dora, her hand could be WhWh RRR(R) GGG 999(9)s 12m: riichi, tsumo, rinshan, chanta, san’ankou, yakuhai 2, shousangen, just one dora 1m, 2m or 3m and optionally ippatsu.

    8. -She had 9sou, which eliminates tanyao already.
      -Rinshan is definitely out if she is relying on ippatsu for the 13th tile since both can't exist at the same time.
      -The other players will definitely won't call kan since they knew it's impossible to win Himeko in her key round. It's like giving a freebie to her. So the kan will have to be from her.
      -Also Himeko isn't Saki. She couldn't have known the replacement tile is a 3m.

      After analyzing, this is the conclusion I can come out with is Himeko is betting on either an ippatsu or a haitei for the 13th han, since:
      -Mairu said she tried until the very end
      -Himeko mentioned she is scared of losing the last draw
      -Tomokiyo mentioned she riichi one turn in advance

      Which means there are 8 tiles left when she riichi, as she was afraid of riichi-ing at the last go round(from bad experience), which Ryuuka killed off both her chance of an ippatsu and a haitei as she called and afraid Ryuuka might call again and she will lose her last draw.

      So my conclusion will still be almost the same, with one kan:
      Riichi, Tsumo, San Ankou, Dora 8, with 2 chance of either an ippatsu or a haitei for the 13th han.
      Ritz please be nice and reveal her hand please. I rest my case.

    9. Oh, right. She can’t have both rinshan and ippatsu. OK, then my very last scenario doesn’t work (and if she has rinshan and would also have had rinshan after drawing 9 sou, then 9 sou must be a dora).

      I did address the rest.
      • She could have tan’yao, but only when winning on 3 man; and since we know that the 9 sou hand is more valuable, either she doesn’t have tan’yao or it’s offset by something else in the 9 sou hand (but it can’t be, as I showed earlier).
      • I also don’t think other players will call kan, although it’s possible if it lets them break her ippatsu or haitei, for example.
      • And yes, the only way she could know the rinshan tile is if she actually won with rinshan. This is why in my last paragraph I suggest that the 3 man may in fact be the rinshan tile.

      If she is betting only on haitei for the 13th han, then she wouldn’t get a higher hand with the 9 sou than with the 3 man, because the 9 sou definitely isn’t the haitei tile; only the 3 man may be. But you’re right: she could be betting on getting one fan from either ippatsu (9 sou) or, if ippatsu fails, haitei (a tile we never saw because it was never reached). And Ryuuka’s pon kills both, and Himeko wins without either. I didn’t consider this.

      It does sound a bit like haitei to me too based on the same lines you quoted, so that’s promising.

      In any case, either 9 sou is a dora or she would get ippatsu with it. (In fact, if the 3 man is itself the haitei tile, then it would have to be both at the same time.) There is no hand shape that would give her more yaku fan for winning with the 9 sou other than ippatsu. And as I showed earlier, someone, whether Himeko herself or someone else for whatever reason, must have declared kan at least once by the end of the hand; otherwise she doesn’t have enough fan.

      If she has kan or ura dora*, one possible hand is SSS WWW 78999s 33m with dora one of the winds and either 9 sou (then she isn’t betting on ippatsu) or the other wind (then she is): riichi, tsumo, san’ankou, yakuhai 2, dora 6 with chance of either dora 7 or ippatsu (or haitei).

      (* We know there’s some ura dora that she didn’t get. But since kan have been declared, there are multiple ura dora indicators and it might just be possible that she got some ura dora but not all. But I wouldn’t bet on that.)

      If you want her to have 8 dora, 55578999s 33555m with dora 5s and 5m and red fives in sou and man: riichi, tsumo, san’ankou, dora 8 with chance of ippatsu (or haitei). Alternatively, she has two kantsu of dora: NNNN 66678s 2222p 33m with dora N and 2p; or one kantsu of dora and some extra kan dora and red fives: 99s 2222555p 33345m with dora 2p and 4m and red fives 5p, 5p, 5m. The final option is that there have been at least two kan declarations and hence there are at least three dora indicators, which allows more freedom.

      If the regular dora is a five, she could also have a kantsu of regular dora and red fives but no kan or ura dora: 55599s 5555p 33345m with dora 5p and all four red fives.

      If she has neither kan or ura dora nor a kantsu of dora, I think her only option is rinshan as suggested in the last paragraph of my previous comment, and then 9 sou is a regular dora and she doesn’t have and wouldn’t have ippatsu or haitei. But since it sounds like she may in fact have been hoping for ippatsu and haitei, this probably isn’t it.

    10. You should probably re-read the condition of tanyao. Also remember she doesn't know which tile is which, so it is a bet, only her chance of winning is there with her key power. Lazy to read the rest lol.

    11. Um, maybe you should read the rest then. I don’t understand what you think I’m getting wrong.

      66678s 2222p 338888m is waiting for 69s 3m and has tan’yao when completed with 3m.

    12. As a quick TL;DR, I did end up agreeing with you that she’s probably hoping to get either ippatsu or haitei. Dora 6 or 7 and one or two yakuhai looks more likely to me than dora 8, but it doesn’t matter much.

  4. Why do Kyouka and Sawaya both agree that Sawaya wouldn't have discarded a North if they had known Himeko only had a sanbaiman, instead of a yakuman

    1. To be precise is if they had known that Ryuuka suspected that Himeko only had sanbaiman.

      In that case, Ryuuka discards a North because she knows that she can not be overtaken by Himeko's sanbaiman. If Kyouko and Sawaya had known that, they would never have risked the possibility that South 4 ends suddenly due to the four equal wind discards rule. After seeing Kyouko discard a North, Sawaya would have discarded something else.

      In the other case, if everyone had thought Himeko had yakuman, Sawaya can discard a North because she assumes correctly that Ryuuka would never do something so stupid as to discard another North, because that just assures Himeko's final victory.

    2. Also Sawaya realized after she discarded the North that there is the possibility of the Yakuman is instead just a Sanbaiman, which explains her oops moment.

    3. I think Sawaya discards the North a bit carelessly and the oops moment is when she realizes that the round might end abruptly if the other two also discard North.

      If Sawaya had known that Ryuuka suspected that Himeko had sanbaiman, she would have put more thought into her discard. Because then she would know Ryuuka and Himeko have a common interest in going to the bonus round but Sawaya has no reason to help them.

    4. I actually said that based on what Sawaya said to Kyouko about not discarding that Pei if she knew about the Sanbaiman. It's as you said that she knew Ryuuka wouldn't have discarded the 4th Pei if it will be a Yakuman, but she probably realize that what if it wasn't a yakuman and just a Sanbaiman, then Himeko and Ryuuka would have the chance to go for a suufon, which she immediately regret her choice of discard.

    5. Oh yes, I see your point now.

    6. Sorry for the confusion then. Probably should explain clearly earlier.

  5. I hope I will also see this girls playing again some time, eeh. :)
    Thx for jokes explanation and interesting mahjong discussion :D

  6. Thanks for the translation!!