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Ilan Wolff

Ilan Wolff

Ilan Wolff

Ilan Wolff

Ilan Wolff

Ilan Wolff

Ilan Wolff

Ilan Wolff

Ilan Wolff

How many hands are there with exactly AKQJ of clubs? Well the first four cards are given to us, so we need to choose nine more. However there is a constraint. None of them can be clubs. Therefore we're choosing 9 cards from 39 options. 39C9=211915132

How many hands are there with AKQJ of clubs and more than four clubs? The first four cards are still given to us, but this time we can choose any of the 48 remaining cards, so the number we want is 48C9=1677106640. However this includes all of the hands with exactly four clubs, so we must subtract those out. 48C9-39C9=1465191508, the total number of hands with AKQJ of clubs and greater than four clubs. Dividing the total number of hands with greater than four clus and AKQJ of clubs by the total number of hands with AKQJ of clubs (48C9-39C9)/48C9=87.4%. Therefore he has greater than four clubs 87.4% of the time and exactly four clubs 12.6% of the time.

Ilan Wolff

Ilan Wolff

Ilan Wolff

2S relays 2NT, then suit bids by opener are non-forcing.

2NT by opener is non-forcing.

Suits at the three level are forcing.

Actually having agreements for this situation makes the 2H second negative much more effective

Ilan Wolff

Ilan Wolff

Ilan Wolff