Tuesday, June 8, 2010

Those tough high-level decisions

One of the trickiest aspects of bridge is when the opponents jam the auction.  Here's an example.  Nobody is vulnerable and you are fourth to speak. You need to have your thinking cap on from the moment you pick up and sort your cards: ♠K9754 JT962 K85 ♣ –.  LHO opens 1♣, partner bids 1♠ and RHO bids 5♣.  Quick, you have 10 seconds to think about this before you start imparting unauthorized information to partner.

You can spend some of the time trying to think about what partner has for his overcall.  But I think that's not going to help much.  Overcalls are, by their very nature, wide-ranging, especially a space-gobbling 1♠ over 1♣.  At all-white, partner could have ♠QJT62 A87 742 ♣32 or ♠AQJT2 AQ7 A742 ♣2 or anything in between.  How can you tell whose hand it is and how high to bid?

Fortunately, you really don't have to.  With all its faults, there is no guide to this sort of situation like the law of total tricks.  So, how many total tricks do we think there are?  We apparently have 10 spades and the opponents surely have 10 clubs.  That's 20 tricks, more or less.  But is it more or is it less?  If it's only 19 tricks, suggesting an "impure" layout, we will be right to bid on only if we can make slam but if we can just make game, then we will be better off doubling (500 vs. 450).  This doesn't really look like a slam situation so we might double (takeout-oriented and maybe partner can pass with wastage in clubs).

What if the total tricks are 20.  This is the toughest problem because if each side can make 10 tricks, we should pass (100 vs. -100).  However, the other divisions of 20 tricks all favor bidding: 450 vs 300, -300 vs -400.

What if the total tricks are 21 (or even more).  Then all situations favor bidding on: 980 vs. 300, 450 vs. 100, -300 vs. -920.  With 22 tricks, bidding at least one more is a no-brainer.

So, it comes back to guessing how many total tricks there are.  Voids tend to increase tricks, short-suit honors tend to reduce tricks.  Again, we don't know partner's hand but if he's a disciplined bidder, he probably won't have Qxx, Jxx or xxx in clubs (three losers) unless he has a very good hand otherwise.  He might have Jxxx or Txxx in clubs but that seems unlikely on the bidding.  So, he might have a wasted A or K in clubs but probably nothing else.  I think in this case, I would estimate more rather than fewer total tricks.  Let's say 21.  Therefore, I would probably bid 5♠.  Even if there are only 20 total tricks, bidding on will be right most of the time, as noted above.

On this particular hand (board 17 from Friday's world-wide pairs), there were 22 total tricks because RHO also had a void (in spades), the opponents had 11 clubs between them and there was essentially no wastage.  In practice, we let them play (and make) 5♣ unmolested for a below-average result.  I'm not saying that pass was wrong, just that I probably wouldn't pass.  It could have been exactly right.  Essentially, RHO made a good bid because it made life very difficult for us.  That's why good players make this kind of bid: it causes problems.

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