Re: What percentage 12 tricks

On May 8, 11:29 am, "raija d" <musti...@[EMAIL PROTECTED]
> wrote:
> Would you percentage gurus here please tell me what the odds are for 12
> tricks in the following hand and enlighten me as to the formula etc.
used to
> come to the conlusion .  I would very much appreciate it.
>
> Contract is 6NT. Opponents were silent and a small spade is led.
> Unfortunately I don't know the exact spots but apparently no tens other
than
> the ones shown.
>
> Axxx
> A10x
> KQxx
> xx
>
> Kxx
> KQxx
> AJ
> AQ10x
>
> TIA Raija

50-60%.

I'm assuming we begin with a club finesse against the J, and later a
club finesse against the K. If either of these work out we'll be up to
11 tricks and the chance for a 12th trick is, to a first
approximation, the roughly 50% chance of four heart tricks.

Hence we need a majority of the three 50% shots to work out, which is
50% by symmetry (2 or 3 working are equally likely, respectively, as 2
or 3 not working).

To refine this answer a bit, consider the non-club chances more
closely. If the club 10 forces the king then there are significant
squeeze chances in addition to the 50% from the main chance of a heart
break. An exact calculation is tedious at best (many cases to add up,
including consideration of the best technical order of card play), and
impossible without making assumptions about how the opponents will
defend (duck the club king? Win and play spade? club? to wreck the
most squeezes?). Therefore I suggest following Eddie Grove's frequent
example and settling for an approximation of some kind. Suppose that
the hand is 70% rather than 50% if the club J is onside and K offside.

Then a simple calculation -- perhaps simpler than you wanted -- would
be to add up these winning cases:

both clubs onside: 25%
club jack onside, king offside, + heart or squeeze luck: 50% x 50% x
70% = 17%
club jack offside, king onside, + heart or squeeze luck: same again,
17%.

That's 60%.

Charles