Hand #21 was the kind that appears in books—a biddable grand slam that required only twenty-six high-card points.
Board #21
North dealer
North-South vulnerable
| | North
♠ A
♥ K Q 7 6 4 3
♦ 10 3
♣ A K 9 5
| | West
♠ 8 7 6 3 2
♥ 10
♦ A K J 8 7
♣ J 4
| | East
♠ K 5 4
♥ 5
♦ Q 9 6 5 4 2
♣ 10 8 2
| | | South
♠ Q J 10 9
♥ A J 9 8 2
♦ ——
♣ Q 7 6 3
| |
|
| | | | |
| South | West | North | East |
| | 1♥ | 2♦ |
| 4♦* | P | 4NT | P |
| 6♦** | P | 7♥ | P |
| P | P | P | |
* Splinter bid showing 10-12 points, at least four hearts, and shortness (singleton or void) in diamonds.
** Odd number of key-cards and a void in diamonds.
I was sitting South. North's hand only had sixteen points, bur it had tremendous offensive power. In fact, because it has four quick tricks (the KQ counts as one) and four losers (one in hearts and clubs, two in diamonds), it meets Marty Bergen's four-by-four criterion for a 2♣ bid. Ken chose to open 1♥. My recollection is that East overcalled 2♦, but I might be wrong. If she did, she must have been inspired by the favorable vulnerability.
My hand seemed perfect for a splinter bid of 4♦. It is worth noting that if East had decided to overcall 3♦, I probably would have made the same bid, but it would have been more difficult for Ken to interpret. I might have just been showing game-going values with support in hearts. Without the jump it would not necessarily show shortness in diamonds.
West passed. If I held her hand, and East did overcall in diamonds, I would have confidently and without hesitation used the LAW to bid 5♦. I wonder if anyone did this. In this case, it would have kept East-West from using Blackwood.
Ken bid 4NT. We play 1430, but in this case it did not matter. I bid 6♦, which Ken correctly interpreted as an odd number of key cards and a void in diamonds. If I had had zero or two key cards, I would have bid 5NT.
Ken was missing only two key cards, and one was in diamonds. So, he knew that I had to have the ♥A, and he could trump the diamonds in my hand. The only question marks were his two club losers. He knew that I had at least ten points, and no wasted values in diamonds. So, it looked likely to him that we could take all thirteen tricks.
I had what I said that I had. However, when the dummy is exposed it is not easy to count thirteen tricks from North's point of view. Can you do it? One declarer only took twelve tricks.
The key is to do a "dummy reversal", which means to plan the strategy as if the dummy were the master hand. Success might still depend on a 3-2 split in clubs (a 68 percent chance), but the path is easy to see—just ruff three spades in North's hand. As it turns out, the ♠Q sets up after the third round of spades, and so it is not necessary to rely on the club split. So, this was a really good bid. The chance of scoring thirteen tricks must be above 80 percent. It only fails when both spades and clubs misbehave.
My recollection is that East led a low spade, which violated a few principles of leading, but it did no harm because North had to play his ace. Ordinarily, it is best when defending against any notrump slam or a grand slam in a suit contract to make as safe a lead as possible. In this case that would mean a heart or a diamond. Underleading an unprotected honor might give away the setting trick.
Is there anything that East-West can do to avoid writing that 2210 on the scorecard? Amazingly, the answer, because of East-West's favorable vulnerability, is a resounding "Yes!" Take a look at the "Par Score" for the hand. The best option for East-West is to bid 7♠ over 7♥! It goes down six, but even if it is doubled, it only is -1400. In fact, this course is almost a no-brainer if the other team has a relatively certain grand slam. Even if it went down EIGHT TRICKS, it would be better than giving up a vulnerable grand slam!
That doesn't seem right. Perhaps at some point doubled tricks should be valued at more than 300 points.