tkass Posted May 5, 2009 Report Share Posted May 5, 2009 qx in front of ak98xx on the queen play the jack appear behind the ak98xx is it a clear restricted choice? some people on web say most of the club player will falsecard the jack with dbl j10 so when the jack appear play the drop, when the 10 appear finesse. i wanna know what is the correct play? thx tkass Quote Link to comment Share on other sites More sharing options...
manudude03 Posted May 5, 2009 Report Share Posted May 5, 2009 J could have easily been from JTx as well (mandatory falsecard), so play for the drop. If I think my opps aren't capable of this, then I finesse. Quote Link to comment Share on other sites More sharing options...
rbforster Posted May 5, 2009 Report Share Posted May 5, 2009 The only winning holding for the hook is J / Txxx when the J falls. Holdings where falsecards don't cost are Jx, Jxx, JTx, JTxx behind (although some of these assume partner would have covered the Q, hence declarer must have the AK). You'll also lose to half the JT tight cases behind when you play for the hook. I think it's clearly restricted choice, but in this case there are lots of potential falsecards from 3-2 splits that make it unattractive hook if you think the falsecards are sufficiently likely. Quote Link to comment Share on other sites More sharing options...
1eyedjack Posted May 5, 2009 Report Share Posted May 5, 2009 manudude is right. The only combinations you need to consider is H singleton, HH doubleton or HHx tripleton offside. There are other possibilities as RobF posts, but in any other case you are either destined to fail or destined to succeed whichever play you make, and most of the time it will be apparent which from the card played to the second trick in front of the Ace. The next question is: what is the frequency with which he should play H from HHx to make the finesse and drop equal odds? (in a vacuum, of course - in reality there will probably be additional information) Quote Link to comment Share on other sites More sharing options...
manudude03 Posted May 5, 2009 Report Share Posted May 5, 2009 A. Txxx/J = 2.823%B. xxx/JT= 1/2 x 3.39 = 1.695%C. xx/JTx= 1/2 x 10.174= 5.087% P(falsecardC)xC=A-B 5.087P=1.128P= 0.2217 or 22.2% This is assuming they will play the J and T at equal probabilities. Quote Link to comment Share on other sites More sharing options...
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