The players were playing Chinese official mahjong.
The winning hand was supposed to be a Chicken hand, providing the required 8 points.
Another player saw a fan, e.g. 1-point fan such as Short straight, thus Chicken hand could no longer be counted and the value of the hand decreased.
Sherlock Miniature #2: Negative Fan
- Details
- Last Updated on Wednesday, 28 November 2012 18:00
- Written by Vitaly Novikov
Sherlock Holmes and Dr. Watson are drinking their morning coffee in the lounge. Watson is reading an article in “Mahjong News”:
‘Curious accident occurred at London’s Mahjong Championship. Some player declared mahjong and listed fan(s). The other player has added a fan to the winning hand and hence hand value decreased.’
“How may it have happened?”, Dr. Watson asked loudly. “And what is the number of such negative fan in the ‘Green Book’?”, questioned Mrs. Hudson.
“Please, calm down, my friends,” said Holmes. “It’s rather elementary! The matter is…”
Question: Please, explain what has happened.
Comments (2)
1Friday, 20 April 2012 12:46
Quentin
2Friday, 20 April 2012 16:15
Sylvain MALBEC
The winning player (A) counted a fan that needs to NOT have some other fan to be valid.
The other player (B) pointed out player A have a fan that invalidat A's fan.
A: "Chicken hand!"
B: "No, you have Self Draw/All Pung/whatever less-than-8-points fan. It's a false win."
According to the "freedom to choose the highest points" principle, the only way you have to decrease someoneelse hand value is to point out abusive fans.
The other player (B) pointed out player A have a fan that invalidat A's fan.
A: "Chicken hand!"
B: "No, you have Self Draw/All Pung/whatever less-than-8-points fan. It's a false win."
According to the "freedom to choose the highest points" principle, the only way you have to decrease someoneelse hand value is to point out abusive fans.
It is difficult to judge the difficulty of problems, but some did take me some time to solve.
I especially enjoyed the '32nd of December' and its fourth question.
Thanks to Vitaly for the problems, to Martin for hosting the "venue" and congrats to Sylvain and Scott for their success.
A: two kongs of the same suit (i.e. 8 one-suit tiles) and a kong of wind
So... winds are now suit tiles?
And they are in every suit?
Wow!
Looks like I've misunderstood the question and it was actually an easy one!
Was it because I was the only one to answer the question within the allotted time?
Just curious.
Thanks.
At first, it looks like each player had three pure melded kongs, two of them separated by two numbers (e.g. 1 and 4), and that their left-side neighbour is waiting for these two said kongs with a ryanmen (e.g. _23_).
But it turns out there are not enough tiles for that.
So, here's the trick:
Watson had: melded: 1111m 4444m 5555m, concealed: 23s EE.
Lestrade had: melded: 1111s 4444s 5555s, concealed: 23p SS.
Holmes had: melded: 1111p 4444p 5555p, concealed: 78m WW.
Mrs. Hudson had: melded: 6666m 9999m, concealed: RRRR(concealed kong) 23m NN, and erronously melded as flowers: 2223m.
It certainly "cut off all conceivable scenarios".