Realization of these 2 ideas:
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Abbr.: Enq.= Enquirer ¦¦ Res. = Responder ¦¦ AsKs = sum of Aces and Kings
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1. ..... (look at the acticle)
2. If the Enq. will know the AsKs of the P., (s)he has 2 possibilities to do that:
4 ♣ or 4 ♦. This depends on his own AsKs:
When Enq. has 3 or 4 AsKs, the question is 4 ♣,
when (s)he has 5 (or even more), the question is 4 ♦.
The advantage of this question choice is:
The Res. knows rather exactly too, how many controls the Enq. has (the minimum exactly).
The Res.´s answers after 4 ♣:
4 ♦: 0 - 2 AsKs (no slam possible) minimum together 3
4 ♥: 3 AsKs and neighter Single nor a void min. together 6
4 ♠: 3 od. 4 AsKs and either a Single or a void min. together 6
4 NT: 4 AsKs and neither a Single nor a void min. together 7
5 ♣: 5 AsKs 8
The Res.´s answers after 4 ♦:
4 ♥: 0 or 1 AsKs and neither a Single nor a void min. together 5
4 ♠: 1 or 2 AsKs and either a Single or a void min. together 6
4 NT: 2 AsKs and neither a Single nor a void min. together 7
5 ♣: 3 AsKs 8
The rebids of the Enq.: .......
(Look at the article)

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