Main   Analysis 8A07 by Richard Pavlicek

# Draw Trumps or Crossruff

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 VulWestPassPassPassAll Pass North1 3 5 EastPassPassPass South1 4 NT6 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

CaseWestEastFactorsPercentPlan A SuccessPercentPlan B SuccessPercent
14342224515 10 70 213.9062(4·21+6·6)(15·70)2.232115 10 70 61.1161
24333225415 10 56 355.2083(4·35+6·15)(15·56)3.4226(7·35+3·15)(15·56)4.3155
34324226315 10 28 352.6042(4·35+6·20)(15·28)1.9345(4·35+6·15)(15·28)1.7113
43352323520 10 56 214.1667(4·21+6·6)(20·56)2.381020 10 56 61.1905
53343324420 10 70 358.6806(4·35+6·15)(20·70)5.7044(4·35+6·15)(20·70)5.7044
63334325320 10 56 356.9444(4·35+6·20)(20·56)5.158720 10 56 356.9444
73325326220 10 28 212.0833(4·21+6·15)(20·28)1.726220 10 28 60.5952
82362422515 10 28 211.5625(4·21+6·6)(15·28)0.892915 10 28 60.4464
92353423415 10 56 355.2083(4·35+6·15)(15·28)1.711315 10 56 152.2321
102344424315 10 70 356.5104(4·35+6·20)(15·70)4.836315 10 70 356.5104
112335425215 10 56 213.1250(4·21+6·15)(15·56)2.589315 10 56 213.1250
124252233515 10 56 213.1250(4·21+6·6)(15·56)1.785715 10 56 60.8929
134243234415 10 70 356.5104(4·35+6·15)(15·70)4.2783(9·35+1·15)(15·70)6.1384
144234235315 10 56 355.2083(4·35+6·20)(15·56)3.8690(3·35+7·15)(15·56)3.1250
154225236215 10 28 211.5625(4·21+6·15)(15·28)1.2946(3·21+7·6)(15·28)0.7812
163262332520 10 28 212.083320 10 28 60.595220 10 28 60.5952
173253333420 10 56 356.9444(4·35+6·15)(20·56)4.563520 10 56 152.9762
183244334320 10 70 358.6806(4·35+6·20)(20·70)6.448420 10 70 358.6806
193235335220 10 56 214.1667(4·21+6·15)(20·56)3.4524(3·21+7·15)(20·56)3.3333
202263432415 10 28 352.604215 10 28 151.116115 10 28 151.1161
212254433315 10 56 355.2083(4·35+6·20)(15·56)3.869015 10 56 152.2321
222245434215 10 70 213.9062(4·21+6·15)(15·70)3.2366(6·21+4·6)(15·70)2.7902
Totals5644800100.0000378756067.0982375676066.5526

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. TopMain