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This hand came up on Friday 30th August 2019 and was played by each of the 9 E-W pairs ending up in 7 different contracts: 5♦, 6♦, 7♦ , 4♥, 6♥ , 6NT & 7NT.
With an Acol 2♣ opening, how can 7NT be reached?
Here's my attempt (I've been told of a more advanced method, see foot of page):
| West |
2♣ |
Acol 2♣ |
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| East |
2NT |
Non-Minimum |
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| West |
3♦ |
Showing Diamonds |
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| East |
3♥ |
Showing Hearts |
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| West |
4♥ |
Agreeing Hearts |
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| East |
4♠ |
Cue Bid (First Round Control) |
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| West |
4NT |
Blackwood |
Any form of Blackwood will do |
| East |
5♥ |
Two "Aces" |
Luckily, all forms show two "aces" |
| West |
7NT |
With nothing missing, 7NT |
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The more advanced convention is for opener to bid 4NT directly. Responder then bids according to any aces he holds:
5♣ = none
5♦ , 5♥ , or 5♠ shows the ace bid
5NT = 2 aces.
Opener then bids the final contract. For our hand the bidding goes:
4NT-5NT-7NT
Clearly opener must have a hand that can continue regardless of responer's bid.
(I suggest we all forget this convention as none of us will ever get another suitable hand).
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