Double Dummy Corner


Exchange and Win

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Exchange and Win (E&W) problems are a recent invention by Luigi Caroli of Italy.  I find the idea intriguing and beautiful and I am delighted to be able to promote it here.

The basic concept is as follows.  You are shown all four hands and given a target for one side or the other, as in regular DD problems.  Normally, the target is unmakeable at double dummy and your first task is to show why this is the case.  Then you have to find the unique exchange of two cards, x and y, that satisfies all of the following criteria:

  1. x belongs to a player on the losing side ( y can be in any of the other three hands.)

  2. x and y are in the same suit.

  3. After the exchange, the losing side becomes the winning side at double dummy.

  4. The cost of every other exchange that satisfies criteria A-C is more than that of the exchange of x and y.

The cost of exchanging two cards x and y is calculated as follows:

  1. Take the difference in value (ace=14, king=13, queen=12, jack=11, ten=10, etc.)

  2. If the exchange is not between partners, double the cost.

  3. If y (or x, if the exchange is between partners) is a king, queen or jack, double again.

  4. If y (or x, if the exchange is between partners) is an ace, triple the cost.

  5. If x is a king, queen or jack and so is y, add another 2.

  6. If two exchanges are equal in cost under rules 1-5, the one having the lower sum of the ranks of x and y is considered the cheaper.

  7. If they are still equal, then the one involving the lower ranked suit at bridge is considered the cheaper.


  1. If the losing side's ♠2 is exchanged with opponents' ♠A, the cost is (14-2)x2x3=72, the highest possible cost.

  2. If the 9 is exchanged with partner's 10, the cost is 10-9=1, the lowest possible cost, though exchanging the 8 with partner's 9 is cheaper under rule 6 and exchanging the ♣9 with the ♣10 is cheaper under rule 7.

  3. If a jack is exchanged with an opponent's king, the cost is ((13-11)x2x2)+2=10.

  4. If a king is exchange with partner's ace, the cost is (14-13)x3=3.

Luigi Caroli gives the following simple example:

♠ AQ7



♣ none

♠ K84



♣ none

♠ 1065



♣ none

♠ J9



♣ none

South is on lead at no-trumps.  North-South to make 3 tricks.

Three tricks cannot be made as the cards lie.  North's ♠7 could be exchanged with the ♠10 at a cost of (10-7)x2=6, yielding three easy spade tricks.  But that is not the solution because South could exchange the 5 with East's 8, also costing 6 but involving a lower y rank (8 against 10).  But nor is that the solution, because  South could exchange the 3 with West's 6, also costing 6 but involving a y rank of 6.

Can the cost be further reduced?  Try exchanging North's ♠7 with South's ♠9.  Now West covers South's ♠J lead but North wins with the ♠Q and throws East in with a diamond for two more spade tricks.  Now the cost is only 9-7=2.  But even that is not the solution, for North can exchange the ♠7 with West's ♠8, also at a cost of 2 but involving a lower y rank.  West is thrown in with a heart.  North discards the 4 and East is squeezed so that even if West returns the ♠K either North makes three spade tricks or South's 5 is good with a spade entry.

On this miniature, the exchange of the ♠7 with West's ♠8 is the so-called winning exchange.

An E&W problem does not necessarily end with the discovery of the winning exchange.  After the winning exchange is made, the original winning side becomes the losing side.  If some exchange is now available to that losing side that is cheaper than the winning exchange just made, then that is known as a counter exchange.  After that there might even be a new counter exchange for the new losing side, this process continuing until there is no counter exchange available.  (Every counter exchange must be cheaper than the immediately preceding one.)

To an E&W devotee, an E&W problem is presented just as a deal and a target.  If the target can be made without any exchanges, then the task is to find the cheapest exchange, if any, that permits the other side to defeat it.  If there is still no winning exchange, then the problem is solved; otherwise, the solution consists of the winning exchange followed by a finite sequence of zero or more counter exchanges. 

This completes the E&W "story", except that there is one delightful consequence that might not have struck you.  Every target on every deal is an E&W problem with a unique solution!

For a really thorough and beautiful demonstration of this genre, have a look at E&W Problem #17.

DDC Home Problem Archive First E&W problem Most recent E&W problem

And here is a list of links to the entire collection so far, in descending numerical order: 

Hugh Darwen, 2002

Date last modified: 03 June, 2019