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Hi, When looking for info on the Kokish Rebids after 1♦-2♣, I always find a scheme like this: 1♦ - 2♣ 2♦ = 11+ 5+crd ♦ 2♥ = 4441 2♠ = GF 4crd ♣ 2NT = 15-17 (when playing 12-14 NT opener) 3♣ = 4crd ♣ 3♦ = GF 6crd 3♥/♠ = splinter 3NT = 18-19 (The 2♦ and 2NT rebid can still contain a 4crd ♥/♠) It ends with a comment that it can also be played when playing a 15-17 NT opening. There are some things I don't understand about this: 1. What do you do with a 3442 and 18-19hcp? 2. When playing a 15-17NT opening, the 2NT rebid becomes 12-14 and hence is not forcing. This seems very inconvenient. Any thoughts? Steven

Hi, Anyone playing T-Walsh? 1♣ - 1♦ = 4crd ♥ 1♣ - 1♥ = 4crd ♠ 1♣ - 1♠ = 4crd ♦ After 1♣-1♦: 1♥ = 3crd OR 15-17 4crd 2♥ = 12-14 4crd 3♥ = 18+ 4crd Same structure after 1♣-1♥. What do you think of this? What are your experiences? Steven

Hi, Anyone playing this? 1♥ - (p) - 1♠ = 6-11 0-4♠ F1 * if opener rebids 1NT he shows 11-15 with 4♠ (a Flannery hand!) * any other rebid has the same meaning as after a forcing 1NT answer Consequence: 1♥ - (p) - 1NT = 6+ 5+♠ F1 I have seen this on the CC of Bocchi-Duboin. Do you have any experience with this? How do rate this approach? Do you know whether this convention has a name? Regards, Steven