Higher Order Throw-Ins

Johan Wastlund

Department of Mathematics
Linkoping University
Linkoping, Sweden
November 2005

An Unusual Endgame

Suppose that a suit is distributed:

	NORTH ♠ x x x
WEST ♠ Q J x		EAST ♠ x x x x
	SOUTH ♠ A K 10

Is it possible for South to get three tricks in the suit by throwing West in two times, forcing him to lead this suit twice? The answer seems to be no: In order to force West to lead the suit, all other suits must be eliminated, and it will not be possible to throw West in a second time. Now consider the following position with West on lead:

	NORTH ♠ x x x ♥ x x x ♦ — ♣ —
WEST ♠ Q J x ♥ Q J x ♦ — ♣ —		EAST ♠ — ♥ — ♦ x x x ♣ x x x
	SOUTH ♠ A K 10 ♥ A K 10 ♦ — ♣ —

If West leads a spade honor, he will be put on lead in hearts and forced to lead spades a second time, and vice versa. Therefore, South can get five of the remaining six tricks.

A throw-in that leads to this type of position can be called a second order throw-in. The gain from the first throw-in is not an extra trick but the possibility of an ordinary "first order" throw-in later.

A Counting Rule

This type of throw-in works because the suits involved combine the following two features:
(i) A threat of scoring an extra trick if the defense leads the suit sufficiently many times.
(ii) A possibility for South of using the suit as an exit by cashing top tricks and playing a low card.
There is a method for evaluating this type of position which is described in the mathematical research paper [8]. This method applies to symmetric positions, where essentially only two players are involved. Each suit is assigned a number that describes how many tricks the suit is worth on average. The connection between numbers and games was discovered by the mathematicians Elwyn Berlekamp and John Conway in the 1970's, while examining go endgames [1]. Conway developed a fascinating theory of the numbers that occur in games [2]. These numbers can be integers, but just like musical notes, they also come in halves, quarters, eighths, and so on. What is new in [8] is the discovery that this theory applies also to some positions in bridge. To find the value of a particular suit, count 1 for each "safe" trick, and 1/2 for each trick that could be produced by an ordinary throw-in, that is, if the opponent must make the first lead in the suit. If the suit has properties (i) and (ii), then add 1/4 if the opponent must lead the suit twice for the extra trick to be produced, and 1/8 if three leads from the opponent are required. In theory, we should count 1/16, 1/32, etc. for even weaker threats, but this will not matter in practice.

To find the total number of tricks that can be obtained in the position, we sum these numbers and round to the nearest integer. If the sum has fractional part exactly 1/2, then it should be rounded in favor of the player who can put his opponent on lead.

Although this sounds like a rule of thumb that sometimes works and sometimes doesn't, it is actually a mathematical theorem. Within its scope of application, it will always give the correct answer.

A Third-Order Throw-In

In the endgames that arise at the bridge table, the quarters and smaller fractions will rarely influence the outcome of the game; but when they do, the result will be quite spectacular. The effect is similar to that of a squeeze in that the extra trick cannot be localized to a particular suit. Instead, it arises through the combination of threats in several suits. The opponent can handle each of the threats separately but cannot cope with all of them at once. Consider the following three-notrump deal with a spade lead from West:

	NORTH ♠ 9 7 5 2 ♥ 6 2 ♦ 8 6 5 3 2 ♣ 6 3
WEST ♠ K J 8 4 3 ♥ 8 7 5 ♦ 9 4 ♣ 7 4 2		EAST ♠ Q 10 ♥ A 10 9 3 ♦ Q J 7 ♣ J 10 9 5
	SOUTH ♠ A 6 ♥ K Q J 4 ♦ A K 10 ♣ A K Q 8

South has eight easy tricks but no entry to dummy. A third diamond trick can be made only if the defense leads the suit twice. Scoring three hearts or four clubs is even more remote, requiring the defenders to lead the same suit three times without declarer's touching it.

Still, as the cards lie, the contract is unbeatable. South wins the lead and plays another spade. If West wins, he cannot cash a second spade without squeezing his partner. Therefore, he can just as well let East win. East is now subject to a quite unusual endplay, a third order throw-in. According to the counting rule, the hearts are worth 2+1/8, the diamonds 2+1/4, and the clubs 3+1/8. This adds up to 7+1/2, and, since East is on lead, South should be able to take eight of the remaining tricks.

Let us see how, in practice, the fractional extra values materialize into a real-world trick. There are two main lines. If East plays a low heart, South wins and plays four rounds of clubs to reach the following position with East on lead:

	NORTH ♠ 9 ♥ 6 ♦ 8 6 5 3 ♣ —
WEST ♠ K J ♥ 8 7 ♦ 9 4 ♣ —		EAST ♠ — ♥ A 10 9 ♦ Q J 7 ♣ —
	SOUTH ♠ — ♥ K Q 4 ♦ A K 10 ♣ —

If East plays another heart, three rounds of diamonds will force him to lead a third one as well, while the diamond-queen shift allows declarer to make the contract by winning and leading the heart king.

If, instead, East tries the club jack at trick three, South wins and plays his hearts from the top. Eventually, this leads to a similar position after East wins the fourth round of hearts:

	NORTH ♠ — ♥ — ♦ 8 6 5 3 2 ♣ 6
WEST ♠ K J ♥ — ♦ 9 4 ♣ 7 4		EAST ♠ — ♥ — ♦ Q J 7 ♣ 10 9 5
	SOUTH ♠ — ♥ — ♦ A K 10 ♣ A K 8

If East leads a club, he is thrown in with a diamond for another club lead, and vice versa.

Further Examples

In theory, the counting rule applies to endgames with only two players involved, but there are several effectively three-or-four-player endgames where it applies in practice. If a suit is distributed:

	NORTH ♠ A x x
WEST ♠ K 10 9		EAST ♠ x x x x
	SOUTH ♠ Q J x

it combines the features (i) and (ii), provided that it can be led conveniently from the South hand. Therefore, in the following three-notrump deal, the spade suit has the characteristics of a combination worth 2+1/4.

Suppose that West (to his regret) leads the club eight to the nine, jack and king.

	NORTH ♠ A 8 3 ♥ 10 8 4 ♦ A K Q 7 3 ♣ 9 3
WEST ♠ K 10 9 5 ♥ Q J 7 2 ♦ — ♣ A Q 10 8 4		EAST ♠ 7 4 2 ♥ 9 6 3 ♦ J 10 8 6 2 ♣ J 6
	SOUTH ♠ Q J 6 ♥ A K 5 ♦ 9 5 4 ♣ K 7 5 2

South continues with a diamond to dummy's ace, and West shows out. There is no way of setting up an ordinary squeeze against West. Moreover, West can lead either of the majors once without giving up a trick. Therefore, South must resort to a strip-squeeze, followed by a second order throw-in. He cashes the remaining diamond honors in dummy. Since West needs to retain three cards in each major, he must throw a club, thereby reducing to the same pattern as South. We assume that West throws his low club, although if he sees the throw-in coming, he should unblock the ten in the hope of finding his partner with the seven. When dummy leads a club, West has nothing better than to cash his three club tricks, reaching this ending:

	NORTH ♠ A 8 3 ♥ 10 8 4 ♦ —- ♣ —
WEST ♠ K 10 9 ♥ Q J 7 ♦ — ♣ —		EAST ♠ 7 4 2 ♥ 9 6 2 ♦ — ♣ —
	SOUTH ♠ Q J 6 ♥ A K 5 ♦ — ♣ —

If West plays a spade, South wins and plays three rounds of hearts to compel another spade lead from West. If, alternatively, the defender tries the heart queen, South wins and leads a spade honor from hand. After cashing two spade tricks by means of a finesse, he puts West in for a second heart lead.

The lack of entries is a problem that can potentially be solved by a throw-in, and there are also situations where a higher order throw-in can arise in this context. Sometimes, fractional values can be assigned to card combinations where one hand lacks entries outside the suit. Consider the following combination:

	NORTH ♠ A x x
WEST ♠ Q 10 x		EAST ♠ x x x
	SOUTH ♠ K J

South, declarer, has no entry to dummy outside this suit. One option is to cash two tricks and then put West on lead. There is also a distant threat, worth a quarter of a trick, of obtaining three tricks in the suit. This happens if West leads the suit twice. The first lead goes to the jack; then, after South unblocks the king, the second lead will go to the ace. To verify that this can actually occur, we combine this suit with another combination, one whose value has fractional part 1/4:

	NORTH ♠ A x x ♥ x x x ♦ —- ♣ —
WEST ♠ Q 10 9 ♥ Q J x ♦ — ♣ —		EAST ♠ x x x ♥ — ♦ A x x ♣ —
	SOUTH ♠ K J ♥ A K 10 ♦ x ♣ —

As the reader can verify, South will get five of the remaining six tricks if West is on lead.

References:
[1] E. Berlekamp, J. H. Conway, and R. Guy, Winning Ways for Your Mathematical Plays I-II, Academic Press, New York 1982.
[2] J. Conway, On Numbers and Games, Academic Press 1975.
[3] J. Kahn, J. C. Lagarias, and H. S. Witsenhausen, Single-suit two-person cardplay, Internat. J. Game Theory 16 (1987),291-320.
[4] J. Kahn, J. C. Lagarias, and H. S. Witsenhausen, Single-suit two-person cardplay, II: Domination, Order 5 (1988),45-60.
[5] J. Kahn, J. C. Lagarias, and H. S. Witsenhausen, Single-suit two-person card play, III: The misere game, SIAM J. Disc. Math. 2 No. 3(1989), 329-343.
[6] E. Lasker, Encyclopedia of Games, Vol. I, Card Strategy, E. P. Dutton and Co., New York 1929.
[7] J. Wastlund, A solution of two-person single-suit whist, The Electronic Journal of Combinatorics, 12 (2005), #R43. See also: Linkoping Studies in Mathematics, No.3 (2005), www.ep.liu.se/ea/lsm/2005/003/.
[8] J. Wastlund, Two-Person Symmetric Whist, The Electronic Journal of Combinatorics, 12 (2005), #R44. See also: Linkoping Studies in Mathematics, No.4(2005), www.ep.liu.se/ea/lsm/2005/004/.

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