Analysis 8A07   Main

Draw Trumps or Crossruff

by Richard Pavlicek

The following play problem was submitted by Bob Munson of Danville, California. It piqued my interest because no particular line of play stood out, and a closer look made the choices even fuzzier. Therefore, it seemed like a good project for a detailed analysis.

South is declarer in 6 after a straightforward auction. South’s 4 NT (Blackwood) with three losing spades is hardly the textbook variety, but with three aces in hand, partner was almost certain to have spade control. The 5  response showed two key cards without the Q, so there was clearly no future beyond a small slam.

6 South	A K 9 2 K 7 6 5 8 K J 8 5			E-W Vul WEST Pass Pass Pass All Pass	North 1 3 5	East Pass Pass Pass	South 1 4 NT 6
Lead: J	8 5 4 A J 9 2 A K 5 2 A 3

West’s lead was Rusinow, which here could be Q-J-x-(x) with length or J-(x) if short; so it has little effect on the analysis. Winning the first trick in dummy was of course a no-brainer. East played the 7, which also reveals little, as it could be from 10-7-(x) if West has length, or Q-10-7-x-(x) if West is short.

There are many reasonable lines of play, some of which are mere transpositions of the order of cashing winners and ruffing, with negligible difference. For this analysis I focused on two plans, starting out the same but diverging at Trick 4 into (A) drawing trumps, or (B) crossruffing. More specifically:

Plan A: A, A, ruff, K, A, ruff, A, [ J if Q fell], finesse J

Plan B: A, A, ruff, A, K, ruff, [ K A if Q fell] else K, ruff, K, J

(Either plan, especially B, has potential detours after a ruff or showout.)

My Hand Pattern Analyzer (HPA) is a useful calculation tool, so I fed it the E-W side holding of six spades, five hearts, eight diamonds and seven clubs. HPA then listed the 266 possible distributions with the percent chance of each. Whoa! This could take forever, so to simplify the task I removed all distributions with a singleton or void, which greatly reduced the scope to only 22 distributions. (Past experience has shown this shortcut to be viable, since the excluded distributions are uncommon, and the difference between two lines tends to remain constant over the entire spectrum.)

For each of the 22 distributions, I had to determine whether Plan A and Plan B would succeed. Only in four cases for Plan B (gold tinted cells) was success deemed to be total. In the majority of cases, a plan was only partially successful, depending on high-card locations.

For example, consider Case 1 where West is 4=3=4=2. Plan B doesn’t fare well, because declarer will be overruffed on the third round of clubs; hence it only succeeds when West has Q-x, so declarer will know to switch horses. Calculating this is simply a matter of adjusting the club factor from 21 (total club layouts) to 6 (club layouts with queen-doubleton). Other suit factors remain the same.

Now consider Case 1 Plan A. Analysis shows that declarer will succeed whenever the Q drops, and failing that whenever West has the Q. The  Q will be doubleton in 4 of the 10 layouts, hence declarer will succeed outright in 4·15·70·21 distributions. In the remaining 6·15·70·21 distributions, he will succeed only in 6 of the 21 club layouts, so it becomes 6·15·70·6. Adding the two gives the total successful distributions. (Numbers are shown in factored form to give meaning to their derivation. All arithmetic is done by computer.)

Hand Pattern Analysis for West-East 6=5=8=7
Constraints: Ws2-4 Wh2-3 Wd2-6 Wc2-5

Case	West	East	Factors	Percent	Plan A Success	Percent	Plan B Success	Percent
1	4342	2245	15 10 70 21	3.9062	(4·21+6·6)(15·70)	2.2321	15 10 70 6	1.1161
2	4333	2254	15 10 56 35	5.2083	(4·35+6·15)(15·56)	3.4226	(7·35+3·15)(15·56)	4.3155
3	4324	2263	15 10 28 35	2.6042	(4·35+6·20)(15·28)	1.9345	(4·35+6·15)(15·28)	1.7113
4	3352	3235	20 10 56 21	4.1667	(4·21+6·6)(20·56)	2.3810	20 10 56 6	1.1905
5	3343	3244	20 10 70 35	8.6806	(4·35+6·15)(20·70)	5.7044	(4·35+6·15)(20·70)	5.7044
6	3334	3253	20 10 56 35	6.9444	(4·35+6·20)(20·56)	5.1587	20 10 56 35	6.9444
7	3325	3262	20 10 28 21	2.0833	(4·21+6·15)(20·28)	1.7262	20 10 28 6	0.5952
8	2362	4225	15 10 28 21	1.5625	(4·21+6·6)(15·28)	0.8929	15 10 28 6	0.4464
9	2353	4234	15 10 56 35	5.2083	(4·35+6·15)(15·28)	1.7113	15 10 56 15	2.2321
10	2344	4243	15 10 70 35	6.5104	(4·35+6·20)(15·70)	4.8363	15 10 70 35	6.5104
11	2335	4252	15 10 56 21	3.1250	(4·21+6·15)(15·56)	2.5893	15 10 56 21	3.1250
12	4252	2335	15 10 56 21	3.1250	(4·21+6·6)(15·56)	1.7857	15 10 56 6	0.8929
13	4243	2344	15 10 70 35	6.5104	(4·35+6·15)(15·70)	4.2783	(9·35+1·15)(15·70)	6.1384
14	4234	2353	15 10 56 35	5.2083	(4·35+6·20)(15·56)	3.8690	(3·35+7·15)(15·56)	3.1250
15	4225	2362	15 10 28 21	1.5625	(4·21+6·15)(15·28)	1.2946	(3·21+7·6)(15·28)	0.7812
16	3262	3325	20 10 28 21	2.0833	20 10 28 6	0.5952	20 10 28 6	0.5952
17	3253	3334	20 10 56 35	6.9444	(4·35+6·15)(20·56)	4.5635	20 10 56 15	2.9762
18	3244	3343	20 10 70 35	8.6806	(4·35+6·20)(20·70)	6.4484	20 10 70 35	8.6806
19	3235	3352	20 10 56 21	4.1667	(4·21+6·15)(20·56)	3.4524	(3·21+7·15)(20·56)	3.3333
20	2263	4324	15 10 28 35	2.6042	15 10 28 15	1.1161	15 10 28 15	1.1161
21	2254	4333	15 10 56 35	5.2083	(4·35+6·20)(15·56)	3.8690	15 10 56 15	2.2321
22	2245	4342	15 10 70 21	3.9062	(4·21+6·15)(15·70)	3.2366	(6·21+4·6)(15·70)	2.7902
Totals			5644800	100.0000	3787560	67.0982	3756760	66.5526

There you have it, folks! Drawing trumps (Plan A) wins by a whisker. This could never be calculated at the table, of course, so the crossruff (Plan B) is sound bridge as well. Perhaps the only significant aspect of the analysis is to prove that bidding to 6  was a worthwhile venture, making about two times in three.

Analysis 8A07   Main

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© 2016 Richard Pavlicek