Main     Analysis 7A47 by Richard Pavlicek    

Echelon Play at IMPs

The following play problem was recently presented by Leonard Helfgott, New Jersey, though the actual hands occurred about 25 years ago. After a simple auction, South is declarer in 3 NT with eight top tricks. The diamond suit will always provide at least two more but might require giving up the lead twice, which declarer cannot afford after the attacking spade lead. The club suit might also provide a ninth trick, so the goal is to find the best way to combine chances in these two suits, sometimes called an “echelon play,” to achieve the maximal success rate.

3 NT South S A 5
H K 8 4
D 10 9 5 3
C A J 9 4
Both Vul

West

Pass


North

3 NT


East

All Pass


South
1 NT



 
Table


 
Lead: S 3 S K 6
H A Q 7
D A J 8 4 2
C K 6 3

A preliminary analysis reveals two viable plans: (1) Win the D A and lead another diamond, then fall back on the club finesse if diamonds do not behave, or (2) Win the D A, then if no honor drops, win the C A, C K and lead a club toward the J-9. Either plan is strong and will succeed a great majority of the time. But which is better?

Regardless of the plan chosen, declarer should win the S K and cash two top hearts (ace and king) in case there is a revealing show-out, then lead the D 10 from dummy to the ace. Assuming everyone follows suit and no diamond honor drops, you are now at the crossroads: Continue diamonds? Or switch to clubs?

Problems like this are difficult to analyze, because the probabilities of the division of any particular suit are dependent on the distribution of other suits, a virtual merry-go-round. The best way to get an accurate assessment is to consider hand patterns that fit the scenario. First, let’s assume spades are 5-4 (either way), not guaranteed because West could have led from three, but practical for analysis. From winners cashed, hearts must be 4-3 or 5-2 (either way) and diamonds 2-2 or 3-1 (either way). Further, West’s choice of a spade lead should preclude holding a longer suit. Based on these conditions, my Hand Pattern Analyzer lists 18 possible distributions with the relative percent chance of each.

Hand Pattern Analysis for West-East 9=7=4=6
Known: Ws4-5 Wh2-5 Wd1-3 Ws)Wh Ws)Wc
Contracts compared Vulnerable: 3N 3N
18 Distributions Sorted by Frequency

WeightTricks WonWestEastFactorsPercentMake PctTrick AvgIMP Avg
1100B95323 24423 1126 35 6 2013.3310010011.009.002.000
2100CC5 B94 B865422 34324 1126 35 6 1510.0010060.0011.339.605.730
3100CC B924324 15422 3126 35 6 1510.0010010011.3310.001.330
4100BB2 A9 795413 44333 0126 35 4 208.8875.001009.7510.000.253.25
5100BB2 A9 794333 05413 4126 35 4 208.8875.001009.7510.000.253.25
6100CC5 BB10 AA 994 794 786 5332 24414 3126 35 4 156.6666.6780.009.5310.0002.33
7100CC5 BB10 996 A94 7A55314 44432 1126 35 4 156.6683.331009.9710.330.132.17
850B94423 15323 2126 35 6 206.6610010011.009.002.000
9100CC B925224 34522 3126 21 6 156.0010010011.3310.001.330
10100BB2 A9 795233 24513 4126 21 4 205.3375.001009.7510.000.253.25
11100CC5 BB10 996 A94 7A54234 15512 6126 21 4 154.0083.331009.9710.330.132.17
1250CC5 BB10 995 794 7864432 15314 4126 35 4 153.3366.6780.009.509.9702.33
1350CC5 BB10 996 A94 7A54414 35332 2126 35 4 153.3383.331009.9710.330.132.17
14100CC BB5 99 89 8845431 44315 4126 35 4 62.6758.3366.679.679.7501.00
1540CC5 BB10 AA 994 794 7865512 64234 1126 21 4 151.6066.6780.009.5310.0002.33
16100CC BB5 994 AA A95215 64531 4126 21 4 61.6010010010.2510.170.080
1740CC B9 BB45521 64225 3126 21 6 60.9610010011.1710.830.330
1840BB 885530 74216 6126 21 4 10.1150.0050.009.509.5000
Known18 cases         Totals 2.312.343970764100.0086.8692.7410.439.811.271.35

The Factors cell in each row, when multiplied out, shows the number of specific layouts for that case. For example, in Case 1 spades can be distributed 126 ways (9c5), hearts 35 ways (7c3), diamonds 6 ways (4c2) and clubs 20 ways (6c3) to produce 529,200 specific layouts.

An adjustment should be made, analogous to restricted choice, in cases where West has equal length (4-4 or 5-5) in the majors. On average he would probably select a spade lead only half the time, so the weight of these cases is halved (50 instead of 100). Also, when West is 5-5, I reduced the weight even further to 40, because he might have bid on some of those hands — subjective of course, but the amount is surely more accurate than no adjustment at all. Weight adjustments are reflected in the Percent column and all totals.

To compare two plans, it is necessary to determine the tricks won by each plan in each case. Cases 1 and 8 are simple: Plan 1 always wins 11 tricks (diamonds 2-2) and Plan 2 always wins 9 (fourth club sets up). Tricks won by each plan are given respectively in hexadecimal (to ensure a single digit for both) followed by the portion to which it applies. A portion of 1 is presumed if no number is given.

Case 2 is not as easy. When West has C Q-10 or Q-x (5/15 of the club layouts) both plans win 12 tricks, because Plan 2 can safely switch to diamonds when the C Q drops, so the first entry is CC5. When West has C 10-x (4/15 of the club layouts) Plan 1 wins 11 tricks, but Plan 2 wins only 9, so the second entry is B94. Finally, when West has C x-x (6/15 of the club layouts) Plan 1 still wins 11 tricks, but Plan 2 fails (8 tricks) so the last entry is B86. Note that the sum of portions (5+4+6) equals the number of club layouts.

Case 6 is more difficult, because with diamonds 3-1 there are two equally likely occurrences: an honor drops, or it doesn’t; so the 15 club portions must be considered twice, effectively 30 portions in all. For the 15 club layouts with a stiff diamond honor, both plans succeed, winning 12 tricks if the C Q drops (5 layouts) or 11 tricks if it doesn’t (10 layouts), so the first two entries are CC5 BB10. When no diamond honor drops, both plans win 10 tricks when West has C Q-10 (1 layout) but only 9 tricks when West has C Q-x (4 layouts) so enter AA 994. Now for the bad news: When West has C 10-x (4 layouts), Plan 1 not only fails but is down two (C A is lost as E-W scarf up the rest), while Plan 2 survives (794); and when West has C x-x (6 layouts) both plans fail but Plan 2 by only one (786). Check the sum of portions (5+10+1+4+4+6) to verify it equals 30.

Results clearly show Plan 2 to be better. The totals of the Make Pct column show that Plan 1 makes 3 NT 86.86 percent of the time, which is certainly good, but not as good as Plan 2 with 92.74 percent, almost 6 percent higher. The most meaningful column, however, is the IMP Avg which indicates the average number of IMPs won per board if every possible layout were played once. Note that these are IMPs won (not net IMPs) so the crux is the difference in the two averages, which shows that Plan 2 gains 0.08 IMPs per board.

The Trick Avg column tells a completely different story, which is pertinent to matchpoints. As might be expected when overtricks are crucial, Plan 1 is the big winner.

Vulnerability matters

What I found most interesting about this problem is that the proper play depends on the vulnerability. As given with North-South vulnerable, Plan 2 is clearly superior as shown. When nonvulnerable, however, the IMP results change considerably, because game swings are lessened, and overtricks remain the same. By my account, with North-South nonvulnerable Plan 1 averages 1.19 IMPs per board, and Plan 2 only 1.03, which means Plan 1 now gains 0.16 IMPs per board, or twice the amount lost when vulnerable. Quite a turnaround.

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© 2016 Richard Pavlicek