The Improbable Heart Game…And How It Becomes Possible! ♦ ♣
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♠ ♥ The Improbable Heart Game…and How It Becomes Possible! ♦ ♣ On Thursday, 19th October 2017, Board #12 proved challenging for defenders and declarers. The printed hand records suggest that ten tricks can be made yet only five of fourteen declarers managed that feat! And, in truth I am not surprised. Why? Because upon first look it is not clear how to make ten tricks. Declarer can count five trump tricks, one top spade and two top diamonds. But, that brings the total to only eight tricks. Yes, an extra trick can be created in spades BUT that still leaves declarer with two losing clubs and two losing spades. So, how can ten tricks LEGITIMATELY be manufactured by declarer? First of all I will venture some possible explanations as to how ten tricks were made at some tables. 1) In all cases North was declarer and the obvious lead is the ♣K. How the defense proceeds after that hinges on what West does at trick one and / or trick two and then what East or West does at trick three. Best is for West to overtake at trick one with ♣A and return a club to East’s ♣Q after North follows suit. It’s at this point that I would venture a bet that some East’s fell from grace by leading ♣10 at trick three. Declarer will ruff, draw trumps and then subsequently cash the ♣9, which by now, has been established as a winner! This can be avoided if East leads a small club, which West can ruff and North will over-ruff …BUT East will still hold the master club (♣10) and dummy will have the singleton ♣9 which is NOT a winner! 2) An alternative explanation as to how the contract might have succeeded could be that East switched to a diamond after winning the second trick with ♣Q. Declarer could win with ♦J, draw three rounds of rumps, cash ♦K and then exit with a spade to dummy’s ♠10 or ♠Q. Yes, West will win with ♠J (or ♠K) but he/she is then end-played and must do either of the following: a. Play a diamond to dummy’s ♦A. This will give declarer a ninth trick: 5 hearts, 3 diamonds and 1 spade. The tenth trick can then be created in the spade suit if declarer finds the right play to the second round of that suit! b. Return a spade which declarer can let run to dummy’s ♠10 or ♠Q. This automatically gives declarer the two extra tricks for the contract, i.e. the ♠10 (or ♠Q) and ♦A 3) Another possibility is that West switched (fatally!) to a spade after winning the second trick with ♣A. Declarer will win this trick with ♠10 and can subsequently promote the ♠8 in his hand thereby garnering an otherwise impossible three tricks in the suit! 4) And yet another plausible explanation is that West discarded a spade thereby establishing declarer’s fourth spade as a winner. But, declarer will still need to find the winning play that will give him / her three tricks in the suit! And yes, that is the same play that is referred to in 2) a. So, how can declarer make the contract irrespective of what the defenders do? Well, this is really a double-dummy problem, i.e. have a look at all four hands and see can you spot the winning plays! Given that the diamond suit should only produce two tricks it is obvious that the spade suit must provide the three tricks necessary to make the contract. So, let us examine the spade layout and see can we identify how to make the requisite three tricks. Open Pairs, October 19th 2017 - Board #12 A 8 4 2 N K J 7 5 W E 9 6 S Q 10 3 First of all, it should be noted that with ♠KJ75 sitting over dummy’s ♠Q103, that standard plays will be unsuccessful. For example, cashing the ♠A and then leading towards dummy’s ♠Q10 will result in East/West making three tricks. Similarly, leading low from the North hand twice towards South’s ♠Q103 will only produce two tricks because of the fact that the ♠9 is doubleton with East and hence although West will win two tricks with ♠K and ♠J North’s ♠8 will win the fourth round of the suit. But, in this instance, declarer will lose four tricks, 2 clubs and 2 spades! So, how can we avoid losing two tricks in the suit given that East/West will have already scored two club tricks? The solution involves declarer making two spade tricks whilst subsequently ruffing the fourth round in dummy. So, win a trick at first opportunity and immediately (without drawing trumps!) play a low spade towards dummy’s ♠Q103. When East plays low declarer can try the ♠10 in the hope that East holds ♠Jx or♠Jxx. When the West wins with the ♠J declarer knows that East does not hold ♠Jx or J♠xx BUT there is still a layout that will enable declarer to win the second and third round of spades AND then to ruff the fourth round in dummy! However, in this instance declarer does not get good news irrespective of whether he plays the ♠10 or ♠Q from dummy. Declarer should now play for the only reasonable layout that will enable him/her to make two spade tricks and that involves a smother play! At this point declarer holds ♠A84 in hand opposite ♠Q3 in dummy and is missing ♠K975. How can declarer manufacture two tricks without first losing one trick? Clearly one option, albeit not a great one, is to cash the ♠A and hope the ♠K drops. But, another and more likely option, particularly if West is suspected to hold length in spades, now presents itself. We simply place East with a now singleton ♠9 which will put West under pressure from a ‘smother play’. Declarer needs to lead the ♠Q from dummy, which will pin East’s now solitary ♠9. Obviously declarer lets the ♠Q run if West does not cover and of course wins with the ♠A if West does cover with ♠K thus promoting declarer’s ♠8 as a winner. The same applies if declarer played the ♠Q from dummy on the first round and it was beaten by West’s ♠K. Now declarer leads the ♠10 on the second round, again pinning East’s ♠9 and thereby promoting declarer’s ♠8 if West covers with the ♠J. Playing West for the length in spades is more easily done if East has already been shown to have started with five clubs and therefor West is known to hold only two clubs. BUT, declarer is not out of the woods yet! After playing the second round of the spade suit, irrespective of whether West covers or not, declarer must now play precisely TWO rounds of trumps thereby denuding East of all trumps whilst leaving one trump in the South hand. Now, and only now, declarer should play the third round of spades (the winning ♠A or ♠8) and then play the fourth round, which will be ruffed in dummy. Yes, I found the smother play in the spade suit but unfortunately I had already played three round of trumps and so I failed by one trick. And, what are the chances of making three tricks in the spade suit without ruffing the fourth round in dummy? Not good. i. A singleton ♠K in the West hand. Cashing the ♠A and then finessing against East ♠J will produces three tricks. ii. Defender’s spade holdings breaking 3-3 and where the honours (K and J) are favourably placed, i.e. East holds ♠Kxx or ♠Jxx AND where declarer guesses correctly or where a guess is not needed (East holds ♠KJ, ♠KJx or ♠KJxx(xx). And a correct guess might be necessary where East holds ♠K9 and has risen with the ♠K when declarer leads towards dummy on the first round of the suit! When the ♠9 falls on the second round declarer will have to guess whether East started with ♠KJ9 or simply ♠K9… So, if you found that winning play and made ten tricks without assistance from the defenders then I salute you. Paul J Scannell, October 25th, 2017 .