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                        Four Losers Equals Ten Tricks!

How can that be? Simple mathematics suggests that if you have four losers in your hand that you only have nine winners. Sometimes, it pays not to be a mathematician! Remember, though, the number “nine” as we will return to it at the end of this article.

For now, our focus is on the number 10! We need 10 tricks to make 4. Here goes:

 

2 showed 5 hearts and 4+ minor, less than opening strength. West leads 8. (Trumps break 2-1).

In case you are still counting, you have a losing heart (the 5- card suit is with East), two losing clubs and the A. There’s not much you can do about the last of these, though I do remember once making a 3NT contract for 9 tricks when the A was missing for all 13 tricks (it was a defender’s card)….later found under the previous table, presumably a useless discard!

So, one of the other three losers has to vanish, though probably not in the way of that A. These opponents are too clever to fall for that!

You have to create some form of end-play. A different throw- in was executed at the two tables, one on each opponent and both equally successful.

So, you win the opening lead in hand and produce the half standard deceptive play of J, knowing that West cannot afford to play their ace for fear of crushing their partner’s singleton king. One day they will. Actually, it matters not whether they win the ace or not.

So, the J wins … and you need a piece of good fortune but you have nothing else to play for. You are going to cash your two diamond tricks and then exit…but before you do that, just play an innocent looking club to the ace. Both opponents follow low. Then, A and K. The order does not matter as you are going to continue with two more rounds of hearts, exiting to East and await developments. If East started with 5 hearts, 1 spade, 4 diamonds and Q xx, then this is not your lucky day. However, that was not the case:

 

East cashed the K with resignation and played a diamond. Ruff and discard and contract made. South could anticipate that East did indeed have at least one of the missing club honours as with all three, West would surely have led K. Indeed, South would be successful any time East has 5 diamonds and any honour doubleton in clubs.

If East does unblock the K, or indeed any club honour they happen to hold, South can still survive. They will cash the second round of hearts, play off their two diamond winners and exit with either a club or a trump. Either way, West will win, cash three black suit winners…but now it is the heart loser which disappears as West must continue with a minor suit.

Indeed, that is how the same contract was made at the other table where the initial lead was Q. East did unblock K as declarer won the first trick with the ace. However, a heart to hand was followed by the 10, ducked. Now the diamonds were cashed, second high heart played and was followed by a club exit. Four losers again became three. It would not have helped West to win A and exited as they could only cash one club winner.

An “impossible” lead

West can defeat 4 but only with a seemingly “impossible” low club initial lead. East unblocks K and West takes the first round of spades, cashes their two club winners and always has a safe trump exit, even if South had played off their heart and diamond winners first.

After any other lead, the contract is there to be made. Did you make it?

At the start of this article, I mentioned the number “nine”. “Nine” is the number of spades you hold in our problem for tomorrow. Quite simply, playing Teams at adverse vulnerability, what is your choice of opening bid with the following:

 An everyday kind of shape! Do not reach for your textbook for the recommended opening bid. It has still to be written! 

Over to you? Wednesday brings the answer...well, what happened!

Richard Solomon

 

 

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