How many points does Mrs. Hudson have? (Easy ‘Sherlock problem’!)
- Details
- Created on Monday, 25 April 2011 05:05
- Last Updated on Wednesday, 28 November 2012 18:00
- Written by Vitaly Novikov
Last mystery ‘Mrs. Hudson's Problem’ turned out to be a real tough one to solve. Today we offer you a much easier mini-mystery of ‘paramahjong’. Please, try to solve it but ... don't push hard and have fun!
(The solution to Mrs. Hudson's Problem is published here. Please check also Adrie van Geffen's answers!)
Remember: you can find all of Vitaly's contributions here. (In the Extra menu, please click 'Sherlock Holmes Mahjong Mysteries'.)
So, Sherlock Holmes, Dr. Watson, inspector Lestrade and Mrs. Hudson are playing mahjong at Baker Street 221 B.
At the table Mrs. Hudson has:
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Last taken tile (from the wall -- added later, V.N.) is
Question: How many points does Mrs. Hudson have (MC Rules)?

All tiles in MJ-set are don't have backs! And, when this deal are played, mrs. Hudson prepare dinner, so she was little scatterbrain. When start-hands are sorting, she accidentally rotate 3 tiles of "1 dots" faces to gentelments! But they used tile-holder sticks, so gentelments can't see this mistake - tiles almost fully hided by sticks! So, mrs. Hudson thinking it was a concealed pung of white dragons, because white dragon in this set looking as clear tile, no any figures.
Thru all game mrs. Hudson trying to collect "all types". It was a nice bonus, when she collected concealed mixed straigth. So, when she taked last tile from wall and it was WW, she count next fans^
1) mixed straigth - 8
2) all types - 6
3) fully conc. hand - 4
4) single wait - 1
5) dragon pung
6) last tile - 8
- Exuse me, mrs. Hudson! - saying Holmes. - It's very strange, but i have white dragons pung in my hand too. So, pls, remove points for all types and dragon pung. When mistake are discovered mrs. Hudson remove 8 pts and add 3 pts for tile-hog and terminals-pung.
but if she happily calls "Hu!" and presents her up-to-now concealed tiles it would be -60 by the rules, as she should pay a penalty of 20 to every other player for calling Hu with a dead hand (consisting of 15 tiles) - anyway being a beginner in a friendly game, I guess, Holmes will just blame himself, that he didn't notice the moment, when Mrs. Hudson got a "long hand" and declare the game void... ,-)




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".