[han]

Having said that, your hand is so bad that there is almost no chance that partner does not have a penalty double - what else can he have, given that the opponents have advanced only as far as 2♣?

How about an 8-count with 4 clubs headed by a single honor? Isn't that a reasonably likely hand to hold for partner? That would give the opponents what, 21 HCP and no fit?

[gwnn]

Thank you David for correcting my Swiss cheese-like memory.

[dburn]

Hannie, on Nov 14 2007, 10:53 PM, said:

How about an 8-count with 4 clubs headed by a single honor? Isn't that a reasonably likely hand to hold for partner? That would give the opponents what, 21 HCP and no fit?

Not really. You see, with most eight counts with four clubs to a single honour, partner would have doubled or would have raised hearts (especially if he could double with 3=2=4=4 shape, not being benighted).

One of the reasons mathematicians cannot count is that they are unable to conceive of, or ascribe a meaning to, an indefinite article such as "a"; they persist in regarding it as denoting "one". Wherefore they lose all sense of what is "reasonably likely", and...