Necessary Shape

by Matthew Kidd


This is board 20 from the July 19, 2013 Friday evening STAC game at Adventures in Bridge with the hands rotated to make declarer South in 4♥. But I’ve swapped the ♠A and ♠K to make a spade lead a real possibility without otherwise changing the analysis.

On a non-spade lead, declarer wins, tests trump and quickly pitches two spades on dummy’s diamonds to make the contract easily. This line occurred five times. After a low spade lead, the defense must immediately cash their other spade winner or suffer the same fate. So let’s say the defense cashes a second high spade. Now what?

Declarer will win the next trick in hand and try a high trump. Despite the 4=0 split, declarer may make the contract. The only hope is to endplay West in trump. This only works if declarer can shorten his trump down to West’s length and cash all his side suit winners first without one getting trumped. West must have the necessary shape, exactly 3=4=3=3 or 3=4=4=2 and declarer must guess which one to try for. Being the only hope, one might as well play for it.

Say West returns a club at trick three. Declarer cashes the top trump, unblocks diamonds, ruffs a diamond winner in hand to begin shortening his trump holding, and returns to dummy with the ♣A. The moment of decision has arrived. To finish shortening his trump holding, declarer must trump a club if he thinks West is 3=4=3=3, or trump a diamond if he thinks West is 3=4=4=2. If he gets it right, declarer cashes the ♠Q and leads low from his remaining ♥KT9 into West’s ♥QJ7. Whether West ducks or wins, he gets only one trump trick.

Which West distribution should declarer try for and what are his chances of either occurring? When hearts split 4=0, the probability that West will be 3=3=3 in the other suits is (7,3) × (7,3) × (8,3) ⁄ (22,9) = 13.8% because there are 22 non-hearts between the defenders and collectively the defenders hold seven spades, seven diamonds, and eight clubs. Here (N,K) means the number of ways to choose K objects from N objects and is computed as N! ⁄ (K! × (N-K)!) where N! = N × (N-1) … × 3 × 2 × 1. Similarly, the probability will that West will be 3=4=2 in the other suits is (7,3) × (7,4) × (8,2) ⁄ (22,9) = 6.9%. Together this give declarer about a 20% chance when trump are 4=0. It’s definitely better than nothing.

Although the math suggests playing West for the flatter 3=4=3=3 shape, you should almost certainly base your decision on how the opponents discard on clubs and diamonds. If they give count, you know what to do. If the defender’s count is contradictory, trust East because West may well see what is coming and realize it is time to deceive. But if East is a really poor player, you’ll have to decide whether East isn’t signalling properly (or at all) or West is deceiving. Conversely, if East is a very strong player, he may guess his partner’s situation and give false count consistent with his partner, the best defense. No one ever said it was an easy game.

Declarer might also be tipped off by the fall of the ♣Q. This may seem like a restricted choice situation where West either holds ♣Q6 or ♣QJ6. However it is different from the classic restricted choice example where a defender drops the queen or jack offside when four cards are out and declarer must guess if it is singleton or from QJ tight, finessing or playing for the drop accordingly. In the classic situation, the odd are nearly 2:1 in favor of the singleton if the defender correctly randomizes his play from QJ, though the actual odds are 11:6 because any given two card holding is a slightly more likely than any given singleton. But Q6 vs. QJ6 with the defenders holding eight cards is quite different. The odds are 16:11 in favor of the Q6 assuming West randomizes his play of the queen or jack on the second round.

Ah but wait—if you see the queen, does West really have to have QJ6 if he has three? West has a good count on the hand by time he discards the second club. In fact if the defenders exited with a spade at trick three, West knows declarer is exactly 3=6=2=2. That means West can safely throw the queen on the second round from Qxx. And he should do so while signalling four diamonds to paint the picture that he is 3=4=4=2 whenever he is 3=4=3=3. So the apparent fall of the ♣Q does not help declarer decide.

By the way, on the actual hand where the ♠A and ♠K are swapped, one defender did plunk down the ♠A at trick one. Declarer wasn’t up to the endplay and finished down one.

Finally, although declarer has 20% a priori chance when hearts are 4=0, his odds have already gone up by time he discovers the bad break. That is because each defender has shown up with at least two spades. For simplicity, suppose the defense exits with a spade at trick three. West is then known to have three or four spades with the three spade case favored 5:3 (62.5%). When West has three spades, he will have 3=4=3=3 or 3=4=4=2 shape 58.74% of the time. That puts declarer up to 63.5% × 58.74% = 37.34% of pulling off the endplay by time he discovers the bad break. One could bid a vulnerable game at teams with those odds!