Board 9, 18th Feb. 2020: "The importance of play at trick one" |
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Mostly this deal was played in 3NT or 4♥ . At one table North made 3NT + 3 for a complete top. How was that possible?
East led off the ♦ 4, being his 4th highest of his "strongest" suit. Declarer played the ♦ 3 from dummy and West played the ♦ 10. This is a blunder by West. Now declarer cannot be prevented from making 12 tricks with good play as follows: Naturally declarer wins the first trick with the ♦ Jack, plays 3 rounds of hearts ending in dummy. On the fourth round of hearts from dummy, West will probably discard a club, declarer will discard a diamond and East will probably discard a spade. As declarer has the spade suit under control, he decides to get at least one trick from clubs, by leading one of the club honours from the table. It does not matter whether or not West ducks, so assume West will take the trick with the ♣ Ace and that West will lead back a diamond. Declarer will take the Ace of ♦ , play a spade to dummy's Ace, and lead back a spade, trapping West's Queen. Two more spade leads, and now left with ♦ 5 and ♣ 5, declarer can play the 5 of clubs through East's ♣ Jack, ♣ 9 into dummy's ♣ Queen, ♣ 10 for a finesse (as the contract is secured, declarer has nothing to lose by taking the club finesse), which in this case wins.
Had West played ♦ Queen on trick one, then if declarer takes the trick with the ♦ Ace, he will only make 9 tricks, because after 4 Hearts and 4 Spade tricks, West will win the 10th trick with the ♣ Ace, lead the ♦ 10 through declarer, and EW will win 3 diamond tricks: If declarer covers West's ♦ 10 with his ♦ Jack, East will win the last 3 tricks with diamonds. If declarer does not cover, West continues with their ♦ 6, and East will win the last 2 tricks, thus holding declarer to 9 tricks.
Had West played the ♦ Queen, then only if declarer ducks this, will he make 10 tricks. Thus the play at trick one can determine whether or not declarer makes 9, 10 or even 12 tricks.
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Board 26, 14th January, 2020 "A difficult endplay" |
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At Badmaster's table, he was South. Despite holding 5 ragged spades to the Queen, Badmaster decided to be cautious and open 1♦ and await developments. When West overcalled 1♠ , Badmaster felt his caution was amply justified and was very glad he had not opened 1♠ !! North overcalled 2♥, East passed, South raised to 3♥ and North bid game in Hearts. East led off the Ace of Spades, declarer played low from dummy. East continued with the ♠ 10, declarer played low from dummy, West followed low and declarer discarded a ♣ from hand. East switched to a low ♦ , won by the Ace of ♦ in dummy.
Prospects for 10 tricks are not very rosy. It is reasonable to assume that West has 10+ high card points for the vulnerable 1♠ overcall, this would leave East with 6- high card points, 4 of which have already been accounted for with the opening Ace of ♠ lead. Thus it is likely that East has at most one Queen or one Jack, and West has the rest. So declarer should play West for the Queen of ♥ .
Declarer plays the Ace of ♥ from dummy, notes East drops the 10 of ♥ . Alas this betokens that West is likely to have 4 hearts including the ♥ Queen. Declarer leads dummy's 9 of ♥ , to finesse against West. This wins and East shows out in Hearts. Now declarer knows that West has 5 spades, 4 hearts and 1 diamond and 3 other minor cards. East had 2 spades 1 heart and 10 minor cards. Declarer plays dummy's last heart and finesses the ♥ 8, and cashes the ♥ King felling West's ♥ Queen. What a bother to get that dame!
After the 7 rounds of 2 spades, 1 diamond and 4 hearts, declarer plays the ♦ Queen from hand, noting that West follows suit. Declarer now plays the Ace of ♣ , again noting that West follows suit and indeed sees West play the 10 of clubs.
With 4 rounds to play, declarer has a diamond, ♣ Jack and 7, and a trump, East has probably 1 diamond and 3 clubs, dummy has ♠ Queen, ♠ 8, ♦ King and ♣ 9, while West has ♠ King and ♠ Jack and ♠ 9, and either a club or a diamond. Suppose now declarer exits with a club, which must be won by either East or West. Now East cannot have both ♣ King and ♣ Queen, as then West would have overcalled 1♠ having 7 high card points, a GROTESQUE possibility. It follows that West has one of the honours which is now a singleton, and East has the other honour.
Should East win declarer's club exit (perforce with the ♣ King to West's ♣ Queen....we know this is NOT the lie of the cards seeing all 4 hands, but declarer sees only 2 hands), then Declarer's ♣ Jack will be high, and he will win the last 3 tricks, with the ♣ Jack (if East returns a ♣ ), ♦ King in dummy and the 5th trump. Or if East should return a ♦ , declarer will win with the King in dummy, get back to hand from dummy by leading a ♠ from dummy and trumping it in hand, and win the last trick with the now promoted ♣ Jack!
In the ACTUAL lie of the cards, West is forced to win the 9th trick with the ♣ King, and now has only 3 spades from which to lead. If West leads ♠ King, declarer will ruff with the last trump, enter dummy with the diamond to score ♦ King and the now promoted ♠ Queen. If West decides to lead a lower spade, declarer will let it around to dummy's ♠ Queen, cash the ♦ King and win the last trick with their remaining trump!
Badmaster thinks this was a most difficult game to make and is in awe of Brigid McKenna, who was the only delcarer to bid and make game in hearts. Congratulations Brigid on masterly declarer play!
ADDENDUM: On trick 2, West may overtake Eas'ts 10♠ with West's Jack ♠ . In this case declarer has no option but to ruff, immediately play the four rounds of trumps as described above. At this juncture the situation is as below with North to play:
North: 3 diamonds and 4 Clubs; East: 3 Clubs and 4 diamonds ; South 2 Spades, 3 Diamonds and 2 Clubs; West: 3 Spades, 2 Diamonds and 2 Clubs.
Now North leads the Ace ♣. If wicked West dumps his ♣ King (in order to avoid being put on lead), then declarer must continue with 3 rounds of diamonds, ending in dummy. Now dummy's ♣ 9 is led, West plays the ♣ 10, North plays low. If East wins with the ♣ Queen, they (East) now lead from their ♣ 8,6 into North's ♣ Jack ,♣ 7 giving North the last 2 tricks. So East allows West to win with the ♣ 10. Now West has to lead from their ♠ King, ♠ 9, into Dummy's ♠ Queen and ♠ 8, so the best West can do is take the ♠ King and concede the 10th trick to dummy's ♠ Queen.
In this addendum one or other of East West is endplayed!
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Board 8, 26th Nov. 2019: "Create a doubt for declarer" |
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At many tables this hand was played in 6 Spades by North. Once trumps are drawn, the only loser is the Ace of Diamonds. North may not wish to think of a 4-1 or even worse, a 5-0 Spade division, so with eternal optimism and calm serenity North will proceed to draw trumps. No problem.
If however, on the first round of Spades, West were to play the ♠ 10, what would declarer think? Badmaster would not like to be in declarer's seat if on the first round of Spades, one of the 10 or Jack of Spades appeared from West, as declarer would have to consider the possibility that this is a singleton and then try countermeasures against East's presumed spade holding of the other ♠ Honour, 7, 5, 2. Moreover North may realise that while not very likely, at 28.26% a 4-1 Spade division could not be dismissed. It would be a terrible dilemma for North.
If West plays the 10 on the first round of Spades and North thought this was a singleton (and thus East has 4 Spades to the Jack), one countermeasure would be for North to finesse Dummy's 9 of Spades on the 2nd round, play the ♠ Queen, return to hand and catch East's ♠ Jack with the top remaining Spade. But now suppose West had the ♠ Jack, 10 doubleton and thus on the first round HAD to play one of the honour cards, then this line of play would cost the contract! These and other scenarios may be playing out declarer's mind. It would be horrible!
Thus West has nothing to lose by playing the ♠ 10 on the first round, instead of the ♠ 7, and creating a lot of doubt in declarer's mind, possibly sending declarer on the wrong path altogether.
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For this reason, 6NT is an impregnable contract. Indeed at the table where that was bid, East led off a low ♦ . Little (!) did East realise that this should have cost them their Ace of ♦ as follows: Declarer (North) wins the ♦ lead, tests two rounds of Spades noting that everyone follows suit. Then, if North has thespian tendencies, she can table her hand with a theatrical flourish and claim the remaining 10 tricks, consisting of 3 further ♠ tricks, 4 top ♥ tricks and 3 top ♣ tricks, for a clean sweep of 13 tricks in jig time, say 15 seconds perhaps!
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Board 9, 26th Nov. 2019: "4 Spades is a very difficult contract" |
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The bidding is suggested by Badmaster. It has the advantage that after North's 2♣ bid, South has a better idea of the shape of North's hand. At most tables the contract was indeed 3NT. A few tables wound up in 4 Spades. At one such table North opened 1♣ , South (after announcing that the 1♣ could be short) responded 1♠ , North bid 2♣ , South 2♠ , North then bid 3♦ and South bid 3♠ (on the basis that three Spades would likely make, and be a better score than 3 Diamonds), North closed the auction with 4 ♠ (????). With a void in Spades, North should only do this holding Ace and King in 2 suits, i.e. 4 top tricks in outside suits.
In a 3NT contract, the only major(!) worry would be that both top Hearts would be in the East hand (a 24%) possibility a priori; in which case declarer would lose 3 (and possibily 4) Heart tricks, a Diamond trick and a Club trick. However declarer has a good (76%) chance that the top hearts will be split or that West has both top Hearts. After playing 2 rounds of Spades, declarer will rejoice at the appearance of the ♠ Jack, and run off 4 more spade tricks, and then play on Diamonds. All except one of the 3NT contracts were made.
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Now to analyse the 4 Spades contract: This is very tricky to play, and Badmaster has established (see the Addendum) that the a priori probability of making the 10 tricks with Spades as trumps is just below 30%. Considering that there is a 76% chance of getting 9 tricks in 3NT, 4 Spades is a very inferior contract.
After South recovers from the shock of Dummy's void in Spades, it is evident that there are 2 losers in Hearts (the Ace and the King) and most likely a loser in Diamonds (the King). Therefore South CANNOT afford to lose a trump trick. This requires the 7 outstanding spades to be distributed no worse than 4-3 or 5-2 AND that the ♠ Jack be in the shorter holding. MOREOVER, as in the 3NT contract, South will also have to hope that the top Hearts are split (which as we recall from above has a 76 % chance).
Thus on a club lead from West (very likely a singleton), South must win with Dummy's Ace of clubs. Now South MUST return to hand via the Ace of ♦ , start leading trumps and lie back and think of Ireland! Lo and behold, on the 2nd round East's ♠ Jack appears. Now South can draw West's remaining 3 Spades, lead a Heart towards dummy's ♥ QJx, East will win. If East returns a club, South will have to ruff with their last ♠ , and hope that East hasn't the King of Diamonds. South will get 6 Spade, 1 Heart, 2 Diamond and 1 Club trick. This is by no means an easy or obvious plan to play the hand, and really N-S would be far better off in 3NT.
Addendum (for those of some intellectual endowment only!): From page 20 of the book "Bridge Odds for Practical Players) by Hugh Kelsey & Michael Glauert (or Wikipedia if you prefer), there is a 62.18% chance of a 4-3 split of seven outstanding Spades. However in only 15 of the 35 possible cases will the ♠ Jack be in the shorter holding, yielding 3/7 x 62.18% = 26.65%.
There is a 30.52% chance of a 5-2 split of the seven Spades. However in only 6 of the 21 cases will the ♠ Jack be a doubleton, thus yielding 2/7 x 30.52% = 8.72%. Adding 26.65% and 8.72% gives 35.37%. Now we must eliminate the deals which have 6-1 and 7-0 splits, (because declarer is bound to lose a trump trick as well in these cases) which is 7.3%. So the actual probaility is 35.37 divided by 92.3 which is 38.15%.
In addition the top 2 Hearts must not both be in East's hand, a 76% chance. Finally we must take the product of these two probabilities (because declarer needs both events to occur), viz. (.76 x .3815) which gives 28.994%, or just under 30%. This 30% chance of success, only with perfect play, is not an enticing prospect. If you got this far, consider yourself to be of some intellectual endowment!!!
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"The Squeeze is on" Bd 13, Nov 12th 2019 |
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The "analysis" of the Claddagh Club's resident "Badmaster" runs as follows:
As there are 11 winners in the combined hands, perhaps a squeeze is on the cards (outrageous pun!). There is a possibility of getting a 4th trick in Hearts or a third in Spades if either suit breaks 3-3. What to do if both suits break 4-2, or worse even?
Anyway East led off a spade. North should duck, and East will probably switch to a diamond. Say North wins in hand, They should play a top Spade and test two rounds Hearts to learn that everyone follows suit.
Assuming 4-2 breaks in both majors (otherwise a 12th trick is a cinch), then if one hand has the remaining 4 major cards, that hand will only have 4 minor cards, (because a diamond has already been played at trick 2), so North can play off 4 clubs and the remaining top diamond and watch (with exquisite but concealed pleasure) East discard a heart or a spade and, hey presto, the 12th trick will materialise!
Should both outstanding Hearts and both outstanding Spades be in separate hands (and then East and West have 6 major cards each, and thus 7 minor cards each), then the only hope is that Clubs are not 3-3, as then Diamonds would be 4-4 and "Badmaster" thinks the slam can't be made. So North must cash the 4 club tricks, and, mirabile dictu, East turns out to have only 5 minor cards, and thus after the 4th top club and the second top diamond, East, with weeping and wailing and gnashing of teeth, has to part with one of their beloved major cards. Then North can pounce for the 12th trick, making the small slam. Congratulations to those who got 12 tricks!
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