Our foursome gathered to play mahjong at Baker Street 221 B. The wall is built and broken, tiles are dealt. And then Dr. Watson says:

"According to newspaper "Mahjong News" rules of mahjong game vary in different countries. For instance, what an amazing rule called the "Charleston" is in American mahjong. The essence of it is that before the game players pass three unwanted tiles to some other player and do that several times."

"Why all this?" asked Mrs. Hudson.

"It's easy!" said Holmes. "You give out unwanted tiles but may received good one exactly to collect mahjong."

"Provided that all tiles will fit to the place," said Inspector Lestrade.

"Why we do not try this rule, so as an experiment?" asked Mrs. Hudson. "For instance, I do not have in my hand any ready three-tile set (Chow or Pung). I would be eager to exchange three unwanted tiles for more suitable ones."

"Strange to say, I also do not have any ready three-tile set in my hand," replied Watson.

"And me! And me!" shouted Holmes and Lestrade.

"Well, well, it is decided!" said Mrs. Hudson sitting at East position. "Gentlemen, please, take three unwanted tiles and pass them to the player to your right. Let's see what happens".

Next moment something unexpected happened. Mrs. Hudson after receiving three different Honor tiles has in hand complete mahjong, while three gentlemen's hands become waiting for mahjong.

"Incredible!"

Question: Please, provide all four hands under conditions:

  • hands before passing of tiles do not have any ready three-tile sets (Chow or Pung),
  • group of passed tiles consisted strictly of three Bamboo, three Characters, three Dots, three different Honor tiles,
  • there are no Flowers in hands,
  • : main (maximum of points) fan for all four hands is strictly the same,
  • : main (maximum of points) fan for all four hands is strictly different, sum of points of main fans of all four hands is maximal.

Note: Waiting hands for determining main fan are considered to be completed by adding appropriate waiting tile.

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  • Guest - Sylvain MALBEC

    Improvement of my previous answer:

    2) Main fan for all four hands is strictly different, sum of points of main fans of all four hands is maximal:
    • Mrs. Hudson (E):
    Starting hand: 13349bam 19dot 19wan ESWN G
    Pass: 334bam
    Main fan: Thirteen Orphans (88)

    • Holmes (S):
    Starting hand: 22446688bam 199wan GG
    Pass: 199wan
    Main fan: All Green (88)

    • Watson (W):
    Starting hand: 1199bam 1144699dot 11wan
    Pass: 446dot
    Main fan: All Terminals (64)

    • Lestrade (N):
    Starting hand: 2233557788dot RGP
    Pass: RGP
    Main fan: Seven Shifted Pairs (88)

  • If Hudson had a normally shaped hand, then she would have needed at least four new tiles for mahjong (one to complete each set). Therefore, her hand is not normally shaped.

    For example, every player may have a seven-pairs hand:

    • East (Hudson) had 44 bamboo, 11 33 77 characters, 167 dots, and singles of north, east, red. She gave away all her circle tiles.
    • South had 167 dots, 44 bamboo, 236 characters, a single of north, and pairs of east, red. He gave away 236 characters. In the end, he is waiting for north.
    • West had 236 characters, 1256 88 bamboo, and pairs of west, white. He gave away 125 bamboo. In the end, he is waiting for 6 bamboo.
    • North had 125 bamboo, 22 55 dots, 88 characters, and singles of north, east, red, green. He gave away north, east and red. In the end, he is waiting for green.


    I will do the other question later.

  • Guest - marco montebelli

    - the same fan is greater honors and knitted tiles or lesser honors and knitted tiles + knitted straight;
    - three honors cause thirteen orphans, fully concealed (92); three bamboos cause all green, seven pairs, half flush, 2xtile hog, fully concealed (126);
    three characters or three dots cause seven shifted pairs, all simples, fully concealed (94)

  • Guest - Anh-Vu

    Hello,

    Since the players exchange 3 tiles, they cannot complete a regular Mahjong hand with 4 chows/pungs/kongs + pair.
    They can only complete special structures such as:
    - Seven Pairs (but not Shifted, otherwise a chow would be present before exchanging tiles)
    - Thirteen Orphans
    - Greater Honors and Knitted Tiles
    - Lesser Honors and Knitted Tiles
    - Kitted Straight : it is possible because it only requires one chows/pung to be completed, therefore this hand can be done after 3 tile exchange.
    The hand of Watson, Mr Lestrade, Mr Hudson and Mrs Hudson can only be one of these hands.

    First case - same main fan:
    Note that same main fan does not mean strictly same tiles in hand. Obviously East has one more tile, and other players can wait on a different tile which complete the same main fan.
    It is possible for the 4 players to have Seven Pairs at same time; or Greater Honors and Knitted Tiles at same time; or Lesser Honors and Knitted Tiles as same time; or even Knitted Straight at same time.
    In the case of Thirteen Orphans, it is possible that East have a complete hand, and other player have a ready hand for Thirteen Orphans right after the tile exchange. However, non-East players would never complete their hand, because East has one pair, and this tile will not be available to other players in order to finish their hand.

    Second case - strictly different main fan:
    There are four players, therefore one player can have Seven Pairs hand, another one can have Thirteen Orphans, another one with Greater Honors and Knitted Tiles. The last player can have either Lesser Honors and Knitted Tiles or Knitted Straight, it does not change the sum of main fans.

    I hope my understanding and my answer is correct.

    Anh-Vu

  • Guest - marco montebelli

    sorry, i considered only maximum of points. so, i confirm the fan for characters, but for dots i propose:
    - full flush, seven pairs, reversible tiles, 2xtile hog, fully concealed (64)

  • Guest - Test

    testing

  • Guest - Anh-Vu Tran

    Hello,

    Since the players exchange 3 tiles, they cannot complete a regular Mahjong hand (at least for East) with 4 chows/pungs/kongs + pair.
    Thanks to the last mystery, we know that non regular hands are:
    - Seven Pairs (but not Shifted, otherwise a chow would be present before exchanging tiles)
    - Thirteen Orphans
    - Greater Honors and Knitted Tiles
    - Lesser Honors and Knitted Tiles
    - Knitted Straight : it is possible because it only requires one chows/pung to be completed, therefore this hand can be done after 3 tile exchange.

    First case - same main fan:
    Note that same main fan does not mean strictly same tiles in hand. Obviously East has one more tile, and other players can wait on a different tile which complete the same main fan.
    It is possible for the 4 players to have Seven Pairs at same time; or Greater Honors and Knitted Tiles at same time; or Lesser Honors and Knitted Tiles as same time; or even Knitted Straight at same time.

    In the case of Thirteen Orphans, we can suppose that East has a complete hand, therefore he has a pair which causes a tile shortage for other players. For example, if east has a pair of east wind, there will remain only 2 east wind for other 3 players. Thirteen Orphans is not possible.

    Second case - strictly different main fan:
    We can notice that east must have a complete hand with 14 tiles, but other player shall only have a ready hand with 13 tiles.
    It means that east must have a non-regular hand as stated in the introduction of this answer. But other players might wait for both non-regular and regular hand, because they can complete 3 chows/pungs.

    Let's try to assign 88 points distinct fans for all players.
    - Nine Gates is not feasible after 3 tiles exchange, because it would require the hand to have at least one chow or pung before exchange.
    - Four Kongs is obviously not feasible, it requires at least one pung before exchange.
    - Seven Shifted Pairs is not feasible, it requires at least three consecutive tiles before exchange.

    It remains only 4 88 fans for 4 players.
    - Thirteen Orphans must be assigned to east because east must have a non-regular hand. East cannot have All Green+Seven Pairs because it would require a chow/pung before exchange.
    - Another player can be waiting for Big Four Wind
    - Another one can waiting for Big Three Dragons.
    - The last player would be waiting for All Green, the only possible green tiles before exchange can be 22446688b + 2 green dragons and the player receives 3 green tiles. This is not possible because there would lack green dragon tile for other fans, 2 green dragons to wait for Big Three Dragons, one green dragon for Thirteen Orphans, we would need 5 green dragons !!

    We cannot assign 4 88 points fans to all players !

    Let's try a slight variation: player is waiting for Little Three Dragon instead of Big Three Dragon. Due to shortage of Green Dragon (All Green has 2 green dragons, Thirteen Orphans has 1 green dragon), this player shall have only one Green Dragon in his/her Hand after exchange and is waiting on the pair of Green Dragon. Therefore the hand has already 4 chows/pungs, this is not possible after 3 tiles exchange !

    So let's try another 64 points fan: All Terminals instead of Big/Little Three Dragons. This player shall have 5 pairs of terminal tile before exchange, and receives tiles which completes 3 pungs. He/she has two pairs and double wait. This is possible and compatible with Thirteen Orphans !

    Note that other 64 points fans are not possible, they would require at least one chow/pung before exchange or there would lack some honor tiles.

    One possibility which maximizes sum of fans:
    - east has a complete Thirteen Orphans hand. The pair is a terminal
    - another player waits for Big Four Wind.
    - another player waits for All Green.
    - another player waits for All Terminals.

    Another possibility which maximized sum of fans:
    - east has a complete Thirteen Orphans hand. The pair is a terminal
    - another player waits for Big Four Wind.
    - another player waits for Big There Dragons.
    - another player waits for All Terminals.

    In both cases the sum of fans is 3*88 + 64

    Lots of explanations for this difficult mystery !

    Anh-Vu

  • In addition to my original answer to the first question, here’s my answer to the second question:

    • The player who received bamboo (e. g. South) clearly has an All Green hand (88 points). He initially had 22446688 bamboo, a pair of the green dragon and three useless tiles in his hand (e. g. 336 characters), and he received, for example, 246 bamboo in exchange for his useless tiles, leaving him waiting for 8 bamboo or a green dragon.
    • The highest-scoring possibilities for Mrs Hudson are Thirteen Orphans (88) and All Honour (64). The highest-scoring possibilities for the players who received characters and dots are Seven Shifted Pairs (88), Thirteen Orphans (88) and All Terminals (64). Since all fan are different and the sum of their values is maximal, one of the three players has Thirteen Orphans, one has Seven Shifted Pairs and the remaining player has either All Honour or All Terminals. Here is an example of how this can be achieved:
    • Mrs Hudson (East) initially had 19 bamboo, 19 characters, 19 dots, a single of each wind, a single white dragon and three single simples (e. g. 246 bamboo). She received another white dragon, a single red dragon and a single green dragon in exchange for her single simples. This completed her Thirteen Orphans hand.
    • One gentleman (e. g. West) initially had 2244557788 in one suit (e. g. characters), as well as three tiles in another suit (e. g. 119 dots). He received 336 in his suit (characters) in exchange for his useless tiles, leaving him waiting for 6 in his suit (characters).
    • Another gentleman (e. g. North) initially had 1199 in two suits (e. g. bamboo and characters), 99 in the remaining suit (e. g. dots) and three useless tiles (e. g. a single of each dragon). He received 119 in the third suit (dots), leaving him waiting for another 9 in the same suit (dots). Of course, Mrs Hudson had the last 9 of that suit in her hand, but this doesn’t matter.

    [*]The total value of all four main fan is 88 + 88 + 88 + 64 = 328 points.
    [/list]

  • Guest - Quentin

    Since no set is present in initial Hands they can only be irregular. Three tiles are not enough to get ready from a Hand without any complete set.

    If all Hands have same major fan, provided the tiles they exchange it can be [13 Orphans, tiles exchanged being 199 or 119 in each suit.
    Total points is 88 * 4 = 352.

    If Hands are ready for different major fan let's give 13 Orphans to one Hand (e.g. the winning one), another Hand ready for Greater Honors and knitted tiles, another one ready for Seven Pairs and last one like Bams 147 Craks 258 Dots 369 SS RR ready for S and R Lesser Honors and knitted tiles. Total points of all waiting Hands 88 + 24 + 24 + 12 = 148.

  • Guest - Cyrille

    Here is a new answer for part 2.

    Hudson : W S N EE(winds) WW RR GG (dragons) + 1 99(C)
    Watson : 1 9(B) 1 9(D) W S N E(winds) W R (dragons) + 33 6(D)
    Holmes : dots 22 44 55 77 88 + 33 6(B)
    Lestrade : 22 44 66 88 (B) GG + WSN

    Hands after Charleston (main fan):
    Hudson : WW SS NN EE(winds) WW RR GG (dragons) (all honors - 64)
    Watson : 1 9(B) 1 9(D) 1 99(C) W S N E(winds) W R (dragons) (13 orphans with G - 88)
    Holmes : dots 22 33 44 55 6 77 88 (pure shifted pairs with 6D - 88)
    Lestrade : 22 33 44 666 88 (B) GG (all green with 6 B 8B or G - 88)

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