Reachmahjong.com

As we all know when we have concealed hand with 3 koutsu, and shanpon-machi, we have to get tsumo for suuankou. But what about iipeikou, when we have wait (with concealed hand) and someone discards that . Will it be iipeikou or not? Until recently I have thought that yes, because it is concealed hand, I can even have riichi, so no matter how you look at it, it fits the requirements. But after I have though about suuankou, analogically iipeikou should not be counted if not made by tsumo. But if that’s so, then iipeikou would be even more hard to get, and still it is worth not much.
So anyone knows for sure if iipeikou should be counted in (for example) that situation:
ron on

Yes, this still counts as iipeikou. The reason why suuankou doesn\'t count on a ron is because while the hand is closed, one of the four mentsu isn\'t, which is the important part of the hand. This is also why suuankou tanki counts even on a ron, since the four mentsu are still closed.

I have never thought about that mentsu can be open while \"hand\" in general is closed, but it looks like it is it. So when I have, for example riichi with shanpon machi, and won on discard, I will count (fu) this koutsu as min-koutsu not an-koutsu, yes? And in iipeikou case it is min-shuntsu, but this yaku needs closed hand not closed shuntsu, and in shuntsu we don\'t count fu so it doesn\'t matter?

Yes.
When you ron, the last set (or pair) is open, while the hand is still concealed (if no other set is open).

For iipeikou, the hand needs to be concealed. So it doesn\'t matter if the last set is open or not.

For sanankou (and suankou), the 3 (4) koutsu need to be concealed, not the hand. Here, the way you complete the last koutsu is determinant.
If you ron the last koutsu, it will be open.
To have an an-koutsu you need to tsumo it, or win on an other set/pair.