The renowned American bridge player, analyst and teacher Richard Pavlicek has a very interesting challenge running on his website - www.rpbridge.net, 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: http://rpbridge.net/8f17.htm
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:
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:
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
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: http://rpbridge.net/8f17.htm
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:
♠ -
♥ -
♦ 42
♣ 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
♥ 6543
♦ T642
♣ T83
| ♠ AJ872 ♥ QT ♦ K93 ♣ AJ4 | | ♠ Q53 ♥ AKJ2 ♦ A85 ♣ K65 |
♠ KT9
♥ 987
♥ 987
♦ QJ7
♣ 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*
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*
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.
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