i have some incomplete thoughts on C, strictly fwiw.

I often express skepticism of LOTT but let's try it out here. We are looking at 7 clubs, the most likely holding for partner is 2 clubs but 3 is certainly possible. I have not calculated it all out but think of it this way Perhaps everyone else has two clubs. Or maybe they are distributed 1 to one player, 2 to another, and 3 to the third. In the first case pard has2. In the second case he will have 3 a third of the time, 2 a third of the time, and 1 a third of the time. Of course there are other possibilities but this suggests he will have 2 or 3 a decent portion of the time.

My initial thoughts were, and still are, that making 3C is a decent bet. The problem is that we might not be playing 3C passed out. If the unseen 6 clubs are split2-2-2 then maybe everyone passes and I make 3C, but that's asking for a bit.

When pard has three clubs then the opponents, together, have 23 non-clubs and so there is a good chance that they have a 9 card fit somewhere. Even when pard has only 2 clubs there is a fair chance they have a 9 card fit somewhere. If so, LOTT says that there are 18 total tricks.

I do think that LOTT is oversold ( last heard, Mike Lawrence agrees, that smart fellow). It's more like ona average the total;tricks and total trump are the same, sometimes one is ;arger, sometimes the other, and it balances out to being about equal.

Still, I have never claimed LOTT is irrelevant, It is just not as easy to apply as its supporters often claim.

I'm still thinking about this.