Analysis 7A56   Main

The Truth Will Out

by Richard Pavlicek

The following declarer-play problem appeared recently in a bridge magazine. After the auction shown, South is declarer in 6  with the K lead (East follows) won by the ace. The contract is laydown with an even heart division, so declarer starts with the A (both follow) then a heart to the king.

IMPs Both Vul	A K 7 5 4 K 7 3 2 J 5 K 6		West Pass Pass Pass	NORTH 1 4 5 Pass	East Pass Pass Pass Pass	South 2 4 NT 6
Lead: K
6 South	Q 10 A 8 6 5 4 A 10 8 4 A 10

No luck, as an opponent shows out on the second heart. The magazine analysis goes on to state:

“If East is the one with the remaining heart winner, you have to hope that East has J-x-x-x as well. If he does, after finessing the 10, you can cash the Q, cross to the K and play the A-K and the last spade to discard three diamonds. East can ruff the fifth spade, but it’s too late to cash a diamond.”

Surely the spade finesse is the best chance to make 6 , and the correct play at total points, but the form of scoring was IMPs. Therefore I wondered if the recommended play was truly correct. The obvious caveat is that your chances of success are poor, and most of the time you will finish down two instead of one. Further, if East has a singleton diamond, you can make 6  against any 4-2 or 3-3 spade division.

When East turns up with three trumps, there are two reasonable lines of play:

A. Finesse the 10. If it wins cash the Q and A* then cross to the K and lead good spades.
 *overlooked in magazine but clearly correct in case East has no more diamonds

B. Win the Q, K (unless J fell) and A, then cross to the K and lead spades from the top.

Note that in Line B, if East has a singleton diamond you will always succeed (barring 5-1 spades) because he must yield a ruff-sluff if he ruffs any spade, or if not then, when you exit with a trump.

Problems like this bug me, as I can’t have peace of mind until I find an answer. Alas, this one seemed out of reach to solve intuitively, and certainly a three-piper for Sherlock Holmes. Fortunately, many years ago I created a Hand Pattern Analyzer, so I fed it the data to obtain a listing of all possible distributions (excluding a spade void, 8+ clubs or a singleton K lead) and the percent chance of each. Results are tabulated below:

#	Wt	West	East	Tricks	Factors	Percent	A Pct	B Pct	A Avg	B Avg	A IMP	B IMP	A MP	B MP	A TP	B TP
1	100	3154	3325	ab3 bb3	20 4 10 126	15.36	0	0	10.50	11.00	0	1.50	0.50	1.50	-150	-100
2	100	3145	3334	ab3 bb3	20 4 10 126	15.36	0	0	10.50	11.00	0	1.50	0.50	1.50	-150	-100
3	100	4144	2335	ab4 bb2	15 4 10 126	11.52	0	0	10.33	11.00	0	2.00	0.33	1.67	-167	-100
4	80	2155	4324	ab2 cb4	15 4 10 126	9.22	66.67	0	11.33	11.00	11.33	1.00	1.33	0.67	887	-100
5	100	4153	2326	ab4 bb2	15 4 10 84	7.68	0	0	10.33	11.00	0	2.00	0.33	1.67	-167	-100
6	100	2146	4333	ab2 cb4	15 4 10 84	7.68	66.67	0	11.33	11.00	11.33	1.00	1.33	0.67	887	-100
7	100	4135	2344	ab4 bb2	15 4 5 126	5.76	0	0	10.33	11.00	0	2.00	0.33	1.67	-167	-100
8	100	3136	3343	ab3 bb3	20 4 5 84	5.12	0	0	10.50	11.00	0	1.50	0.50	1.50	-150	-100
9	90	3163	3316	ac3 cc3	20 4 5 84	4.61	50.00	100	11.00	12.00	0	8.50	0.50	1.50	615	1430
10	80	2164	4315	ac2 cc4	15 4 5 126	4.61	66.67	100	11.33	12.00	0	5.67	0.67	1.33	887	1430
11	100	5143	1336	ab5 bb1	6 4 10 84	3.07	0	0	10.17	11.00	0	2.50	0.17	1.83	-183	-100
12	100	5134	1345	ab5 bb1	6 4 5 126	2.30	0	0	10.17	11.00	0	2.50	0.17	1.83	-183	-100
13	100	5152	1327	ab5 bb1	6 4 10 36	1.32	0	0	10.17	11.00	0	2.50	0.17	1.83	-183	-100
14	80	4162	2317	ac4 cc2	15 4 5 36	1.32	33.33	100	10.67	12.00	0	11.33	0.33	1.67	343	1430
15	40	1156	5323	ab1 bb5	6 4 10 84	1.23	0	0	10.83	11.00	0	0.50	0.83	1.17	-117	-100
16	40	1165	5314	ac1 cb5	6 4 5 126	0.92	83.33	16.67	11.67	11.17	14.17	2.83	1.67	0.33	1158	155
17	50	2137	4342	ab2 cb4	15 4 5 36	0.82	66.67	0	11.33	11.00	11.33	1.00	1.33	0.67	887	-100
18	100	4126	2353	ab4 bb2	15 4 1 84	0.77	0	0	10.33	11.00	0	2.00	0.33	1.67	-167	-100
19	50	1147	5332	ab1 bb5	6 4 10 36	0.66	0	0	10.83	11.00	0	0.50	0.83	1.17	-117	-100
20	100	5125	1354	ab5 bb1	6 4 1 126	0.46	0	0	10.17	11.00	0	2.50	0.17	1.83	-183	-100
21	50	3127	3352	ab3 bb3	20 4 1 36	0.22	0	0	10.50	11.00	0	1.50	0.50	1.50	-150	-100
Totals and Averages					656,256	100.00	18.40	10.69	10.66	11.11	2.14	2.25	0.60	1.40	129	63

Factors: Number of combinations for each suit distribution between West and East. For example, Case 1 allows 20 ways (6c3) for a 3-3 spade split, 4 ways (4c1) for a 1-3 heart split, and 126 ways (9c4) for a 4-5 club split. Diamonds require special treatment, since West is presumed to hold K-Q for his lead, so there are only 10 ways (5c3) to compose K-Q-x-x-x opposite x-x. Multiplying the numbers shows how many layouts are represented. Case 1 comprises 20×4×10×126 = 100,800 layouts, and the column total shows that all 21 cases comprise 656,256 layouts.

Tricks: Number of tricks won by each line according to the J location. For each group of three characters, the first shows the tricks won by Line A, and the second by Line B, both in hexadecimal. The third number shows the applicable portion (number of sixths). For example, cb4 means 12 tricks for Line A, 11 tricks for Line B, 4/6 of the time. Note that the portions are identical to the spade divisions.

Weight: An attempt to obtain realistic results by reducing the influence of patterns with which an opponent may have bid. For example, with 6-5 in the minors (Case 15,16) I estimated West would bid 60 percent of the time (probably unusual 2 NT) so the fact that he passed reduces the weight of these patterns to 40 percent. Similarly, with seven clubs (Case 17,19,21) I judged West would bid half the time; and with six diamonds (Case 9,10,14) or 5-5 minors (Case 4) 20 percent of the time (but only 10 percent if 3=1=6=3). While my numbers are arbitrary, failure to apply weights would surely be off track.

Percent: The frequency of each West-East pattern after weight adjustment relative to all, which dictates a column total of 100 percent. Note that patterns are listed in order of frequency.

The last 10 columns compare Line A and Line B in five categories: Make percent (how often 6 makes), average number of tricks won, average IMPs won, average matchpoints won (counting 2 for a win, 1 for a tie) and average total points won.

The crux of the problem is the IMP comparison. Each row shows the average number of IMPs won per layout. Case 1 (100,800 layouts) is obvious, because Line A never wins, and Line B wins 3 IMPs on half the layouts, or 1.50 IMPs per layout. Case 4 favors Line A, winning 17 IMPs 2/3 of the time (11.33 avg) while Line B wins 3 IMPs 1/3 of the time (1.00 avg). Note that averages are for IMPs won, so the net gain or loss would be found by the difference.

My suspicion proved to be right. At IMP scoring, Line B is the better play, winning an average of 0.11 IMPs per deal, albeit not a windfall. At matchpoints, Line B is a huge winner. Only at total points is Line A the superior play, and greatly so at that.

Vulnerability matters

Curiously, if North-South were nonvulnerable, the bottom line becomes:

Factors	Percent	A Pct	B Pct	A Avg	B Avg	A IMP	B IMP	A MP	B MP	A TP	B TP
656,256	100.00	18.40	10.69	10.66	11.11	1.76	1.63	0.60	1.40	113	60

Only the IMP and TP columns are affected by vulnerability. Notice the turnaround at IMPs: Line A is now superior, winning an average of 0.14 IMPs per deal. Evidently, the sacrifice of only 2 IMPs when the spade finesse loses (versus 3 IMPs vulnerable) justifies playing for the best chance to make.

Another consideration

The preceding analysis assumes 6 is reached at the other table, which is an accurate assumption in expert competition. After the 4  raise, South’s cards are golden, so it’s hard to imagine a reputable player not driving to slam. Nonetheless, if you thought the slam would not (or might not) be bid, you should take the best chance to make (Line A) because down one or two will not matter. If only 11 tricks can be made, you will lose 13 IMPs regardless.

Analysis 7A56   Main

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© 2024 Richard Pavlicek