Saturday, October 30, 2010

Little Deuce Coupe - a challenging card problem

The renowned American bridge player, analyst and teacher Richard Pavlicek has a very interesting challenge running on his website -, which I dearly recommend as a knowledge source for any bridge player (it has helped me develop aspects of my game countless times).

In this post I will present his challenge "Little Deuce Coupe" and my solution. I thank Mr. Pavlicek for his contest and look forward to seeing the best answer to it - be it mine (hopefully :) ) or some other contestant's.

The problem is: construct a complete bridge deal (all 4 hands) in which, ON A GIVEN LEGAL DEFENCE (not necessarily what we call "good" or "practical" defence), SOUTH CAN MAKE 3NT and ALL FOUR DEUCES WIN A TRICK, when the following conditions are also met:

- No revokes, leads out of hand etc. Other than that, you can force all the plays of all four hands to be exactly what you want them to be.
- West has EXACTLY five spades
- No other hand has a suit over four cards

There are probably multiple solutions, therefore Mr. Pavlicek adds that "the fewer high card points (HCP) for North-South, the better. Correct solvers will be ranked by the total N-S HCP, fewest being best."

Until Monday 01.11.2010 you can enter YOUR solution here:

My solution and reasoning:

- ALL FOUR DEUCES must win a trick. But a deuce can only win a trick when three hands do not possess that suit anymore. So the first deuce that wins a trick is the one that started out in a "lengthy" suit - a four or five card one given the problem's description.
- I chose to attribute the 2 of spades to West, with 4 other spades. I also assume it's the first 2 cashed.
- I then constructed a 4 card ending in which two deuces win a trick for NS. The lead is in North:

              ♠ -
              ♣ T3
       ♠  A
    ♦ -    
    ♣ AJ4
  ♠ -
  ♥ K
  ♦ -
  ♣ K65
              ♠ -
              ♣ Q972

When North plays the 2, both E and W throw high clubs, while S discards the 7. Then the T is overtaken with the Q while EW play the 6 and the J, the 9 wins trick 12 and the 2 is won at trick 13.

- Notice that both E and W have the VERY important "spare cards" - the A and the K. Both cards can be kept in when discarding on the diamond, and allow the 2 to win the last trick.
- These "spare cards" were protecting the corresponding deuces (otherwise these deuces could not have won a trick). I already assumed the 2 to have been with W, and I could now place the 2 with E.
- Therefore E had the length - strictly 4 cards. To promote and cash a deuce when holding 4 cards, one must cash 4 tricks, which would mean at least 5 tricks for EW (the 2 plus 4 hearts), incorrect for "3NT". But if everyone else has time to discard hearts, three rounds of hearts can suffice.
- Hearts must then be discarded on spades (the lengthy suit). When cashed, the 2 must "squeeze" the same suit of NS, the one protected by E and his deuce, namely hearts. This would be enough so that only 3 rounds of hearts are required to promote the 2.
- So 3 tricks in hearts and the dreaded  2 = 4 tricks. EW cannot have any other winners. So spades must bring 3 winners to NS.
- The idea came to me that it would be nice for NS to play hearts voluntarily in order to promote the 2. This would make communication simpler (I reasoned this by trial and error mostly).
- For the rest, I kept subtracting honor cards from NS.  

Here is the integrated deal and line of play that came out of it. Please excuse EW for their poor, but legal, defence :).

The full deal:

               ♠ 64
               ♣ T83
       ♠  AJ872
    ♥ QT     
    ♦ K93    
    ♣ AJ4
  ♠ Q53
  ♥ AKJ2
  ♦ A85
  ♣ K65
               ♠ KT9
               ♣ Q972

The card play (* = the trick was won by this card; the next trick will be initiated by the same hand)

Trick       W      N      E       S 
  1         J     6    Q    K*         
  2         8    4    5     T*      
  3         7    6    3     9*      
  4         Q*  5    J     9            
  5         2*   4    A     8      
  6         T     3   A*    7     
  7         K     8   2*    Q   
  8         9     6    8      J*  
  9         3     T*  5      7  
  10       A    2*  K      7  
  11       J     T   6      Q* 
  12       4     3   5     9* 
  13       A     4    K     2* 

All four deuces win a trick - at trick 5,7,10,13. EW only gets tricks 4,5,6,7.

Notice that NS needs only 8 points for this to happen. How's that for an inspired card play? :)

I hope my solution proves to be among the best. I spent almost 3 hours dealing with various other paths, but none came out as good as this one.

I dare you to go explore this problem yourself, and to leave your comments on my analysis.

I also recommend you these books to develop your card reading and positioning sense - whether you use them for humor or actual technique development.

Use the links below or explore Hef's Bridge Attic, an powered store on which I bring you the most important and valuable bridge products out there - bridge books, playing cards, bridge software, bridge apparel.

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