Two Chances Are Better than One

One of the conceptual difficulties one encounters in the bridge literature is that one suit is often treated as being independent of the other suits. So we encounter lists of percentage plays in a suit taken in isolation. This can be helpful, but it gives the wrong impression from a theoretical perspective. Suits are intertwined. Here is an example in which an expert asked, ’is there a clear-cut solution?’

 

♠ K97	♠A86532		1NT (15-17)	2♣ (transfer)
♥ KJ64	♥AQ9		2♠	4NT (RKCB)
♦ A9	♦72		5♦	5♥
♣ AJ53	♣K9		6♠	Pass
7 losers	6 losers

 

The bidding is unknown. We assume responder evaluated his hand according to the losing trick count as 6 losers opposite a strong 1NT opening bid will often produce a slam. The 5 controls are worth an equivalent of 17 HCP, provided opener can come up with 3 spades to the king. As things turned out the 1NT bidder was happy to bid the slam after an encouraging 4NT RKCB. The lead was a not unexpected ♦K. Declarer won and played 2 rounds of spades hoping for a 2-2 split. The opening leader showed out on the second round, so the RHO had a sure trump trick. Unlucky? The problem was how to arrange a diamond discard on a club or heart winner before the RHO won his trump trick.

There are 3 possible plays that achieve the happy result: 1) play the ♣K and finesse the ♣J, discarding a diamond on the ♣A; 2) play 3 rounds of hearts and discard a diamond on a winning heart; 3) play 3 rounds of clubs hoping the queen falls and the RHO can’t ruff. Viewed in this light one might be inclined to hope the hearts are favorably split, but that is the wrong approach. Well, perhaps the club finesse is a reasonable shot, as the LHO has already shown up with the ♦KQ. Wrong again. Let’s see why #3 is the correct approach, as noted by Tim Bourke.

 

The less often you lose, the more often you win.

Rather than search for the winning percentage, it is often easier to look at the losing percentage. The location of the ♣Q depends on the number of clubs held in the defenders’ hands, and there is no indication that the RHO has many more clubs that the LHO. We conclude the finesse has roughly a 50% chance of losing.

Playing off the hearts will work as long as the RHO has 3 or more hearts. Let’s look at the combinations available in hearts taken in isolation:

 

Split		0 – 6	1 – 5	2 – 4	3 – 3	4 – 2	5 – 1	6 – 0
Combinations		1	6	15	20	15	6	1	Total = 64

 

The 3-3 split stands right in the middle and is included in the winning category, so the LHO will hold 3 or more hearts more than 50% of the time. That observation alone makes the heart play superior to the club finesse. The LHO will follow to 3 rounds of hearts for 37 out of 64 combinations, which translates to roughly a 40% chance of failure. The playdown of the hearts clearly has a lesser chance of failure than the club finesse.

The chance of failure by playing 3 rounds of clubs hoping to drop the ♣Q can be calculated in the same way. A full calculation is burdensome at the table, so we consider only a few of the more even splits to get a rough estimate. Whereas the even-numbered heart distribution was shaped like Mt Fuji with one peak in the middle, the odd-numbered club distribution is shaped like Table Mountain with an extensive central plateau.

 

Split		2 – 5	3 – 4	4 – 3	5 – 2
Relative Weights		60	100	100	60		Total = 320

 

The third round of clubs will be ruffed in the ratio of 60 to 320, roughly 19% of the time. This represents clearly a lesser risk than playing on hearts. Success may depend on what transpires during the next phase of play.

The Chance of Success

With regard to the first 2 strategies, either one wins or one loses on the completion of the play in the suit. However, with regard to playing 3 rounds of clubs, there are card combinations for which declarer neither wins nor loses: the queen doesn’t drop, but the RHO doesn’t ruff. Failure has been avoided, so there is now an additional chance to win, namely the hearts may be played down with the hope that the RHO holds at least 3 hearts. It has become a case of 2 chances are better than 1.

A represents the number of combinations where the initial club play works.

B represents the number of combinations where the initial club play fails.

C represents the number of combinations without a resolution.

A + B + C represents the total of all combinations allowed.

We start with a division of sides for the defenders of 4=6=9=7. The lead is the ♦K. Early in the play West shows out of spades so we assume he began with ♦KQ and a singleton spade, whereas East has 3 spades and at least 1 diamond. This leaves us with 10 vacant places in the West hand and 9 in the East hand. However, West had to discard something on the second round of spades. What information has that sent us, and by how much does it affect the odds?

If West were to discard a heart, declarer would be happy enough to play on clubs as planned. If West discards a low club, that might raise concerns that West is long in clubs (5) and, therefore, that East has only 2 clubs. If West discards a diamond, declarer proceeds happily in his plan in a neutral state of mind. In theory one assumes the discard was a random selection from the unseen cards in West’s hand. The appearance of a diamond, heart, or club doesn’t materially affect the odds, so we stick with the vacant places 9 and 10 for the sake of convenience. On that basis we do our probability calculations.

 

Probability of initial failure	16.8%		(15631 combinations)
Probability of success	36.5%		(33682 combinations)
Probability of neither	46.6%		(43065 combinations)

 

The club plays have left us with combinations where the club splits of 3-4 and 4-3 represent over 80% of the remnant combinations. We are now working on the high plateau of card distributions in accordance with the general tendency of proportionally more even splits as the cards are played out without incident. It is matter of calculating with high expectations for how many of these combinations are the hearts evenly split as well. This is tedious to do by hand, but I did it. The probability that the heart play will succeed is 67.4%. So the probability of success overall is given by the following:

Probability of success = 0.365 + .466 x .674 = .679

Probability of failure = .321

A rough approximation is suggested:

Probability of success = 1/3 + (1/2) x (2/3) = 2/3

Probability of failure = 1/6 + 1/6 = 1/3

If one employs the a priori odds for the club splits, the probability of failure in the first phase is less, close to 1 in 7.

The full deal was as follows:

 

		♠	A86532
		♥	AQ9
		♦	72
		♣	K9
♠	4			♠	QJ10
♥	8532			♥	107
♦	KQ1093			♦	J854
♣	Q94			♣	10872
		♠	K97
		♥	KJ64
		♦	A9
		♣	AJ53

 

Again thanks go to Tim Bourke for bringing an interesting problem to my attention along with the correct solution. He is developing a computer program that should prove very useful for calculating compound probabilities for complex situations. In the end one hopes to develop from such results insights which will serve as good guidance at the table.