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#22
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Posted 2007-August-14, 14:50
Ralph, these odds alone don't matter at all, as I tried to explain above. Btw, if partner was always going to lead a club, then the odds are much more in favor of the doubleton than what your computation is saying, as the lead of the J rules out 3/4 of the singletons but only 3/6=1/2 of the doubletons.
The easiest way to count losers is to line up the people who talk about loser count, and count them. -Kieran Dyke
#23
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Posted 2007-August-14, 15:17
cherdano, on Aug 14 2007, 03:50 PM, said:
Ralph, these odds alone don't matter at all, as I tried to explain above.
What you were trying to explain above was, that Line A wins over Line B when two conditions co-exist: that (1) clubs are split 2-2 and (2) declarer has a stiff spade.
And that B wins over A when clubs are split 3-1 (with partner having the one).
And that B is more likely that A.
But that is not pertinent to the issue of, how likely it is that the clubs are in fact divided 2-2, compared to how likely it is that they are divided 3-1, with partner holding the 1. To figure out the "how" you need math, n'est ce pas?
♣♦♥♠ Philosophy consists very largely of one philosopher arguing that other philosophers are all jackasses. He usually proves it, and I should add that he also usually proves that he is one himself. H.L. Mencken. ♣♦♥♠