Mrs. Hudson’s Problem
- Details
- Created on Saturday, 09 April 2011 19:12
- Last Updated on Wednesday, 28 November 2012 18:00
- Written by Vitaly Novikov
Mr. Wan Tao, what should I discard?
After Mrs. Hudson was told that "Chi"+"Hua"+"Hua" was not a breed of dog, Wan Tao offered to Mrs. Hudson to play mahjong.
"Me? No way, I don’t know anything!""Thought I might be your personal game consultant. Would you mind, gentlemen?"
"Not at all! It’s very good idea, we like this game and definitely need fourth player due to you cannot always come to play."
So, wall is built, game started. It is difficult for Mrs. Hudson to make decisions: what to discard, how to meld. And the game came to it’s crucial point. Mrs. Hudson has no melds in hand and asks Mr. Tao (that paragraph was edited lately to avoid ambiguity):
“Mr. Wan Tao, what should I discard? I think that there are NO “spare” tiles in hand.”
“I see. There are no spare tiles to throw because you have mahjong in hand!” – answered Wan Tao.
“So, what should I do? I can’t simply win. Let’s see is there any tile to throw not to somebody’s mahjong – so that others would be not against it.”
“Gentlemen”, said wan Tao. “Let me as a very exception look at your hands to choose discard for Mrs. Hudson?”
Wan Tao took a look at all three hands and sentenced:
“Mrs. Hudson, no way, you can choose safe discard. Any tile you might throw would give mahjong to any gentlemen’s hand. So, please, declare mahjong by yourself!”
- Question (for beginners): Please, reconstruct all four hands (there are plenty of solutions, provide at least one). Please, note that Mrs. Hudson's hand is fully concealed one. And she placed a tile in hand so no point for the “wait”.
- Question (for limit-makers): Please, reconstruct all four hands (Mrs. Hudson's hand is fully concealed one -- Author) so that the sum of all four mahjong would score 400+ pts.
- Question (for experts): Please, reconstruct all four hands (Mrs. Hudson's hand is fully concealed one -- Author) so that the sum of all four mahjong would score exactly 32 pts. That is very difficult to do! As bonus feature try to provide solution of Mrs. Hudson’s hand with as few different tiles as possible (there exists solution for 8 different tiles).
Comments (5)
1Sunday, 10 April 2011 16:45
Bert
Which variant they play?
2Sunday, 10 April 2011 21:42
Author
They play Chinese Officials (MCR)
3Sunday, 24 April 2011 12:40
Adrie van Geffen
Holmes: B 1112345678999
Watson: C 1112345678999
Lestrade: D 1112345678999
Hudson: B123 C123 D123 B789 B66
Watson: C 1112345678999
Lestrade: D 1112345678999
Hudson: B123 C123 D123 B789 B66
4Sunday, 24 April 2011 12:42
Adrie van Geffen
Hudson: B999 C999 D999 B111 D11
All terminals: 64
Fully concealed: 2
Triple pung: 16
4 concealed pungs: 64
Total: 148
Holmes: B 23 234 345 456 55
Winning tile: B1
Full flush: 24
4 shifted chows: 32
Concealed: 2
Tile hog: 2
Total: 60
Watson: D1 D2222 D3333 D4444 D5555
Winning tile: D1
4 kongs: 88
All melded: 6
4 shifted pungs: 48
Full flush: 24
Reversable tiles: 8
Total: 174
Lestrade: B19 D19 C19 ESWN grw
Winning tile: any of Hudson's discards
13 Orphans: 88
Total 88
Summed up: 148+60+174+88 = 470
All terminals: 64
Fully concealed: 2
Triple pung: 16
4 concealed pungs: 64
Total: 148
Holmes: B 23 234 345 456 55
Winning tile: B1
Full flush: 24
4 shifted chows: 32
Concealed: 2
Tile hog: 2
Total: 60
Watson: D1 D2222 D3333 D4444 D5555
Winning tile: D1
4 kongs: 88
All melded: 6
4 shifted pungs: 48
Full flush: 24
Reversable tiles: 8
Total: 174
Lestrade: B19 D19 C19 ESWN grw
Winning tile: any of Hudson's discards
13 Orphans: 88
Total 88
Summed up: 148+60+174+88 = 470
5Sunday, 24 April 2011 12:45
Adrie van Geffen
Hudson: B456 B567 C555 D444 B88
Two concealed pungs: 2
All simples: 2
Fully concealed: 4
Total: 8
(note: only 7 different tiles)
Holmes: melded kong red (rrrr) pung green (ggg) D123
In hand B6777
Winning on B5, B6 or B8
Two pungs dragons: 6
Kong dragons: 1
One voided suit: 1
Total: 8
Watson: B1111 (kong) C22234 D56777
Winning on C5 or D4
Concealed hand: 2
Concealed kong: 2
Terminal pung: 1
Two concealed pungs: 2
No honors: 1
Total 8
Lestrade: melded pung white (www)
In hand D678 C345 B56 EE
Winning on B4 or B7
All types: 6
Pung dragons: 2
Total 8
Overall total: 8+8+8+8 = 32
Two concealed pungs: 2
All simples: 2
Fully concealed: 4
Total: 8
(note: only 7 different tiles)
Holmes: melded kong red (rrrr) pung green (ggg) D123
In hand B6777
Winning on B5, B6 or B8
Two pungs dragons: 6
Kong dragons: 1
One voided suit: 1
Total: 8
Watson: B1111 (kong) C22234 D56777
Winning on C5 or D4
Concealed hand: 2
Concealed kong: 2
Terminal pung: 1
Two concealed pungs: 2
No honors: 1
Total 8
Lestrade: melded pung white (www)
In hand D678 C345 B56 EE
Winning on B4 or B7
All types: 6
Pung dragons: 2
Total 8
Overall total: 8+8+8+8 = 32
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".