Q/A 2: Can you please point me out which part of the mystery state that Mrs. Hudson didn't had a non-winning hand at some point?
Q/A 4: Ho! I had misunderstood the question! I thought ALL four winning hands must had an 88-points fan.
‘Four Wins of Mrs. Hudson’ - the solution
- Details
- Created on Thursday, 03 May 2012 10:14
- Last Updated on Wednesday, 28 November 2012 18:00
- Written by Vitaly Novikov
When should you stop to try and make mahjong and when should you continue? Mrs. Hudson now knows how to… Did you find the solution to the problem?
Mrs. Hudson being East from the very start, had three times chance to win the deal, but instead she continued to win in the 4th run (no flowers in her winning hand).
Question 1: Give an example of Mrs. Hudson’s starting hand and show how she decided to continue game three times.
Question 2: How many declarations (calls) did she make, provided there are no flowers in her winning hand?
Question 3 (for experts): Is it necessary for Mrs. Hudson’s hand to be one-suiter?
Question 4 (for experts): Is it possible to find in four mahjong hands of Mrs. Hudson (three missed and one declared) 88-point fan?
The solution
Answer 2. Answering Question 2 first, one can notice that Mrs. Hudson for the first three times could not declare anything else but ‘Kong!’ without giving out a move to the opponents. So, she made exactly four calls: three times ‘Kong!’ and one ‘Hu’.
Is it possible to declare four kongs? No, otherwise after the 3rd kong and just before declaring the 4th, one hand structure would be 4-1 which is not legal for winning (3+2) if she decided to stop.
Answer 1. Here is one of possible Mrs. Hudson’s hand.
1A. Start:
Concealed -- 
+ + 
+ + 
+ + 
-- mahjong – 7 pairs, declaration – ‘Kong!’ on ( is replacement tile).
1B. After 1st kong:
Table -- , concealed -- + + + , declaration – ‘Kong!’ on (
is replacement tile).
1C. After 2nd kong:
Table -- , , concealed -- + + , declaration – ‘Kong!’ on ( is replacement tile).
1D. After the 3rd kong:
Table -- , , , concealed -- + .
The solution above shows only one of possible transformation chains in which Mrs. Hudson has 3 sets of identical tiles (, and ) in her initial hand.We can show it schematically as:
4442 --> (4)+443 --> (4)(4)+431 --> (4)(4)(4)+32.
Nevertheless, there exist other transformation chains having two or even only one set of identical tiles in initial hand, new kongs in hand appear after replacement tiles, for instance:
4433 --> (4)+4331 --> (4)(4)+431 --> (4)(4)(4)+32,
41333 --> (4)+1334 --> (4)(4)+134 --> (4)(4)(4)+23.
Answer 3. In all above mentioned transformation chains we assumed for tiles in Mrs. Hudson’s hand to belong to one suit, since it is much easy to use specific tile in Chow or Pung.
Can we abandon one-suit restraint?
Yes! The 1st kong might consist of other suit or even of Honors! Reader may replace for any other tile in the above solution to check it.
Answer 4. Based on answer 3 here is possible hand with 88-point fan ‘All Green’:
.
Although, it is highly possible for Mrs. Hudson having such nice hand from start to finish deal immediately assuming that maximum for current deal is achieved.
Comments (1)
1Friday, 11 May 2012 12:31
Sylvain MALBEC
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".