A Little Night Arithmetic

by Jeff Rubens

I have been asked to comment on six similar declarer-play endgame problems (the nonbidding and early play must be accepted), which I will treat in what seems to be increasing order of complexity (and suggest further twists).

Deal 1:

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

South plays in seven diamonds. West leads the ten of diamonds to dummy's ace. On a diamond to South's king, East throws a club. Declarer plays spade ace-king-low; West ruffs; dummy overruffs. Declarer draws the last trump (East throws a club), unblocks the king of clubs, crosses to the ace of hearts, discards a spade on the ace of clubs, returns to the king of hearts, and leads the last trump, on which all three others discard clubs. When South leads a heart, West follows with the last missing low heart. Assuming that the (unspecified) bidding, lead and defense would have been the same in all relevant cases, what are the chances of success of declarer's two plays?

Among spades (2=4), diamonds (4=1), and the subsuit of low hearts (3=2), the West=East known cards are 9=7. Even with unspecified bidding, the defenders can see at an early stage that the exact clubs they hold are irrelevant, so their plays in that suit may be taken as random. West has four places open for the jack of hearts, East six; the drop is a 3-to-2 favorite. [One can get the same answer by counting specific layouts. If West has the jack of hearts, he started with three clubs, which can be dealt C(9,3) = 84 ways; if West has no more hearts, he started with four clubs, which can be dealt C(9,4) = 126 ways. The raito of 84 to 126 is the same as the ratio of 4 to 6.]

Deal 2:

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

South plays in seven diamonds. The early play is the same as in Deal 1, except that on the second round of hearts, East plays the jack or ten.

The calculations are the same except that the chance that West has no more hearts must be halved, because East might have played the other known-to-be-equivalent card from jack-ten. So the (reduced) numbers of places open are four and three, making the finesse a 4-to-3 favorite. [To do this by counting cases—original cases is usually easier, and does not require attributing any particular behavior to the opponents, so I recommend that approach in general: If West has the remaining heart honor, there were 2 ways he could have been dealt heart honors and C(9,3) = 84 ways he could have been dealt the three clubs to fill out his hand, thus 2 * 84 = 168 relevant hands. If West has no more hearts, there was only one way he could have been dealt his heart honors (the null set) and C(9,4) = 126 ways he could have been dealt his clubs. the ratio of 168 to 126 is the same as that of 4 to 3.]

Deal 3:

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

South plays in seven diamonds. The early play is the same as in Deals 1 and 2, except that on the first two rounds of hearts, East plays two cards from among the jack, ten and nine.

Here, East had three ways to show collectively two of his hearts if he started with jack-ten-nine-tripleton, so the reduced numbers of places open are four and two; the finesse is a 2-to-1 favorite.

Deal 4:

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

South plays in seven diamonds. The early play is the same as in Deal 1, but the queen of clubs does not appear prior to trick 11.

Everyone can see that the queen of clubs is a distinguished card in its suit. In fact, one trick before the critical decision, when the North-South cards looked like:

 NORTH ♠ — ♥ Q 10 ♦ — ♣ J SOUTH ♠ 7 ♥ 4 ♦ 5 ♣ —

and declarer led the last trump, if West then held jack-low of hearts and queen of clubs, the defense was doomed. It is potentially relevant that East, discarding after dummy, had he held the club queen and a low club in the three-card ending, could, without concluding matters, have discarded either club. Thus he *may* have had a strategic decision to make: whether or not to show South who held the queen of clubs. South, recognizing this, must take into account whether East's club play to trick 11 is the queen or not. East has two possible strategies: PlayQ (play club queen if having the option) and KeepQ (do not play club queen if having the option). South has four possible strategies: Hook (always finesse), HookIfQ (finesse only if East shows club queen), HookIfNoQ (finesse only if East doesn't show club queen), and Drop (never finesse). There are 210 ways that the 10 "unknown" cards (HJ, 9 clubs) can have been distributed originally; 28 of those (where West has HJ CQ) have been eliminated by the failure of the show-up squeeze, leaving 182 cases *prior to East's discard at trick 11, and they are all potentially relevant, because their numbers may affect East's play to that trick*. Repeating earlier information and introducing abbreviations, those 182 cases are:

p: HJ|CQ [56] {East has the option.}

q: CQ|HJ [56] {East can't play the club queen.}

r: —|HJ CQ [70] {East is forced to play the club queen.}

This matrix shows how many cases out of 182 South will succeed in different combinations of applied strategies:

PlayQKeepQ
Hook56 (p)56 (p)
HookIfQ112 (pq)56 (q)
HookIfNoQ70 (r)126 (pr)
Drop126 (qr)126 (qr)

South should pick Drop. East can amuse himself by guessing the relative likelihoods of South's blundering into HookIfQ or HookIfNoQ and act accordingly. Deciding in advance that you will play for the drop will succeed 126 times out of every 182; it is awkward to phrase this in terms of the chance that the drop will succeed at the point of final decision, because that might depend on how East will handle various holdings prior to that point.

Deal 5:

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

South plays in seven diamonds. The play goes as in Deal 2, except that the clubs appear as in Deal 4. The general discussion of Deal 4 applies, but here (analyzing through original cases), the count of layout p is doubled from that in Deal 4, because there are two possibilities for the subsuit of heart honors (whereas in layouts q and r, there is only one such possibility). This makes the game matrix:

PlayQKeepQ
Hook112 (p)112 (p)
HookIfQ168 (pq)56 (q)
HookIfNoQ70 (r)182 (pr)
Drop126 (qr)126 (qr)

Or, removing the dominated Hook, which a capable declarer will presumably not use:

PlayQKeepQ
HookIfQ168 (pq)56 (q)
HookIfNoQ70 (r)182 (pr)
Drop126 (qr)126 (qr)

Now I wave my magic wand (in other words, I conceal the ugly details of how I reached the next port of call) and claim that the solution to this game is that its value (the most that declarer can guarantee and the least that the defense constrain declarer to, on average, with best strategies all around) is 126; South does best to pick D; East does best to choose PlayQ anywhere from 1/2 to 5/8 the time. Let's verify that this is correct. South can guarantee 126 by picking Drop. If East picks PlayQ with probability one-half, South' HookIfQ wins (1/2)*168 + (1/2)*56 = 112, while South's HookIfNoQ wins (1/2)*70 + (1/2)*182 = 126. If East picks PlayQ with probability five-eighths, South's HookIfQ yields (5/8)*168 + (3/8)*56 = 126, while South's HookIfNoQ wins (5/8)*70 + (3/8)*182 = 112. Clearly, East would permit more than 126 if he moves lower than one-half or higher than five-eighths (for South, albeit with risk, might pick his more-successful sometimes-hook choice at one-half or five-eighths respectively). If East chooses PlayQ with a probability between one-half and five-eighths, the successes of the sometimes-hook approaches will vary within their limits at the endpoints, never above (or even as high as) 126.

If you decide in advance to decided always to use one particular tactic (sometimes called a "pure" strategy), you could use this to calculate the overall chance of success. As in Deal 4, stating how likely it is that the choice at the point of decision will work is different in nature. Among situations where declarer decided in advance to use a mixed strategy (that is, to assign a positive probability to at least two available tactics), there are some where the probability of success will depend on how the opponents act and sme where it will not.

Deal 6:

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

South plays in seven diamonds. The play goes as in Deal 3, except that the clubs appear as in Deals 4 and 5. The general discussion of Deals 4 and 5 applies, but here (analyzing through original cases), the count of layout p is tripled from that in Deal 4, because there are three possibilities for the subsuit of heart honors (whereas in layouts q and r, there is only one such possibility). This makes the game matrix:

PlayQKeepQ
Hook168 (p)168 (p)
HookIfQ224 (pq)56 (q)
HookIfNoQ70 (r)238 (pr)
Drop126 (qr)126 (qr)

Here, the restricted-choice aspect is so strong (relatively speaking) that it is row Drop that is dominated, leaving the reduced matrix:

PlayQKeepQ
Hook168 (p)168 (p)
HookIfQ224 (pq)56 (q)
HookIfNoQ70 (r)238 (pr)

Continuing as in Deal 5, the value of this game is 168; South can secure this with Hook or gamble in guessing what East would do. (Notice that as the influence of situation p increases through Deals 4, 5 and 6, the effect of such a gamble increases also. In particular, where playing for the drop was a big-enough favorite, South's gambling would be, in the ordinary sense, pointless.) East should choose PlayQ with a probability from the interval [5/12,2/3]. To verify: At 5/12, HookIfQ produces (5/12)*224 + (7/12)*56 = 126, while HookIfNoQ produces (5/12)*70 + (7/12)*238 = 168. At 2/3, HookIfQ produces (2/3)*224 + (1/3)*56 = 168, while HookIfNoQ produces (2/3)*70 + (1/3)*238 = 126.

Those who, like Einstein, would seek generalizing principles must observe with sadness that Deal 6 provides a counterexample to the hypothesis (a generalization of playing the card that one is known to hold) that when an opponent knows that one of your possible holdings is more likely than another, you will gain by playing in a way that reveals information supporting that holding over one that is, from the opponent's point of view, less likely. (In Deals 4,5 and 6, because of the show-up-squeeze possibility, declarer has reason *beyond the places-open evidence* to place East with the club queen. Deal 6 shows that East need not randomize his strategic choices in such a way that favors the cases in which he holds that card.)

More questions could have been asked about this general situation. For example, suppose that North's second-highest club were the ten. Then, declarer would need to consider the possibilities inherent in additional subsituations: West has two club quacks [and not the last missing heart], West has one club quack, West has no club quacks. There are several possibilities for what might have appeared in clubs, quackwise. And similar questions could be asked when dummy has ace-nine-low, ace-eight-low, etc. Hoewever, my computer is tired and wants to go to bed.

ESOTERICA

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