JLOL Posted January 28, 2009 Report Share Posted January 28, 2009 A lot of intermediates have this problem, and it might not be so easy to spot where their logic is wrong. Say you have AJ82 opposite K543 as declarer. You lead the king and it goes 6 2 9. You now lead up to dummy and LHO plays the 7. This means that the suit could be divided: Q76 T9QT76 9T76 Q9 T9 is less likely because of restricted choice (they could play the T or the 9 the first round). Q9 is more likely than stiff 9, so I should go up with the ace. Why is this wrong? Adv/exp hide your answers. Quote Link to comment Share on other sites More sharing options...
awm Posted January 28, 2009 Report Share Posted January 28, 2009 I think the first trick should be 6 2 9 and not 6 3 9, since otherwise you played two cards out of the same hand. Quote Link to comment Share on other sites More sharing options...
JLOL Posted January 28, 2009 Author Report Share Posted January 28, 2009 I think the first trick should be 6 2 9 and not 6 3 9, since otherwise you played two cards out of the same hand. edited thanks Quote Link to comment Share on other sites More sharing options...
Echognome Posted January 28, 2009 Report Share Posted January 28, 2009 Without trying to give anything away and at the risk of asking a question that sidetracks the issue, how many tricks do we need? Or are we playing for max expected tricks? Quote Link to comment Share on other sites More sharing options...
JLOL Posted January 28, 2009 Author Report Share Posted January 28, 2009 Without trying to give anything away and at the risk of asking a question that sidetracks the issue, how many tricks do we need? Or are we playing for max expected tricks? You can always make 3 tricks from this position no matter what you play, so I'm asking about the fourth. Quote Link to comment Share on other sites More sharing options...
IdiotVig Posted January 28, 2009 Report Share Posted January 28, 2009 I'll give it a whirl: Let's the say the player with the 9 has Q9. He knows that if he drops the 9 on the first trick, declarer will work out to fly ace next round. So, when the 6 appears, dropping the Q is a likely possibility to steal a trick when declarer hooks the 8 on the next round. If his partner doesn't have the 7 and the 6, this play doesn't really cost, as declarer was likely to play up anyway. So, when we see the 9 appear on the right, it's less likely than expected it's from Q9 - it's probably a stiff, or occasionally played from T9 - and declarer should hook the jack when the 7 appears. Quote Link to comment Share on other sites More sharing options...
JLOL Posted January 28, 2009 Author Report Share Posted January 28, 2009 So, when we see the 9 appear on the right, it's less likely than expected it's from Q9 - it's probably a stiff, or occasionally played from T9 - and declarer should hook the jack when the 7 appears. If this is true then shouldn't declarer play the 8 on the second round? As you said "it's probably a stiff." (Not saying that this is the correct or incorrect answer btw). Quote Link to comment Share on other sites More sharing options...
TimG Posted January 28, 2009 Report Share Posted January 28, 2009 How does the defender know that declarer doesn't have the Ten? Quote Link to comment Share on other sites More sharing options...
IdiotVig Posted January 28, 2009 Report Share Posted January 28, 2009 Er, sorry, yeah. Meant to write "hook the eight" and not "hook the jack" at the end. Quote Link to comment Share on other sites More sharing options...
Winstonm Posted January 28, 2009 Report Share Posted January 28, 2009 If I am understanding this correctly it is because with 96, 97, or 976 the choice is not really restricted - the defender could have played the 9 first time with any of these holdings. Restricted choice means no options available - the choice of what to play is truly restircted Quote Link to comment Share on other sites More sharing options...
TimG Posted January 28, 2009 Report Share Posted January 28, 2009 This means that the suit could be divided: Q76 T9QT76 9T76 Q9 Isn't 76 QT9 also a possibility? Quote Link to comment Share on other sites More sharing options...
JLOL Posted January 28, 2009 Author Report Share Posted January 28, 2009 Er, sorry, yeah. Meant to write "hook the eight" and not "hook the jack" at the end. ok, If the defender is playing the Q from Q9 then why should the declarer hook on the 8 on the second round. Also, if declarer is hooking the 8 on the 2nd round when the 9 is played then why should the defender play the Q from Q9? Also, as TimG said how does the defender know his partner has the T? Also if declarer can have 5 of the suit playing the queen from Q9 is a pretty big disaster as well. Quote Link to comment Share on other sites More sharing options...
JLOL Posted January 28, 2009 Author Report Share Posted January 28, 2009 If I am understanding this correctly it is because with 96, 97, or 976 the choice is not really restricted - the defender could have played the 9 first time with any of these holdings. Restricted choice means no options available - the choice of what to play is truly restircted The partner of the person who played the 9 has played the 7 and the 6. Quote Link to comment Share on other sites More sharing options...
JLOL Posted January 28, 2009 Author Report Share Posted January 28, 2009 This means that the suit could be divided: Q76 T9QT76 9T76 Q9 Isn't 76 QT9 also a possibility? Then your play does not matter Quote Link to comment Share on other sites More sharing options...
mikeh Posted January 28, 2009 Report Share Posted January 28, 2009 There is this thought: as lho, if you hold 1076, what card do you think you should, at least quite often, play at trick 2 of the suit? Consider how you would have played with Q106.. you would perforce have played the 10.. so with 1076, you would be well advised to play the 6 then the 10 on the second round. When you don't, then declarer has the inference that you are more likely to hold Q76 or Q1076.. in either of which cases, your play is forced. Thus, this is really a neat double restricted choice analysis, and Justin has done an excellent job of making it seem to be all about rho's restricted choice possibilities. If you can be counted on to play the 10 at least a good part of the time (from 1076), then the fact that you didn't tilts the percentages against the corresponding Q9 holding in rho.. it is no longer more probable (all other distributional clues aside) than the stiff 9. Edit, out of a desire for completeness: we also have to figure in the odds of rho playing the 9 from 97.. but a good player will often do this, especially if he thinks that you are intermediate... you might play the A... and a poor player may do it in order to give count.. I mean, who lies? :) At least, that's my take :) Quote Link to comment Share on other sites More sharing options...
gwnn Posted January 28, 2009 Report Share Posted January 28, 2009 (edited out) Quote Link to comment Share on other sites More sharing options...
helene_t Posted January 28, 2009 Report Share Posted January 28, 2009 I don't see the problem. There are two possibilities for the finesse and one for the drop, wtp? Even considering restricted choice it is still 60/40. Quote Link to comment Share on other sites More sharing options...
JLOL Posted January 28, 2009 Author Report Share Posted January 28, 2009 I don't see the problem. There are two possibilities for the finesse and one for the drop, wtp? Even considering restricted choice it is still 60/40. Which two possibilities? Quote Link to comment Share on other sites More sharing options...
helene_t Posted January 28, 2009 Report Share Posted January 28, 2009 oh sorry I am smoking weed, just realize that there is one case for the deep finesse and one for the jack and one for the drop. Quote Link to comment Share on other sites More sharing options...
Lobowolf Posted January 28, 2009 Report Share Posted January 28, 2009 oh sorry I am smoking weed, just realize that there is one case for the deep finesse and one for the jack and one for the drop. If you're smoking weed, you should play the 8. It's deep. Quote Link to comment Share on other sites More sharing options...
JLOL Posted January 28, 2009 Author Report Share Posted January 28, 2009 Mikeh has given the correct answer btw. Quote Link to comment Share on other sites More sharing options...
vuroth Posted January 29, 2009 Report Share Posted January 29, 2009 Brilliant. I love that, as I get deeper in to this game, there are still horizons a long way off to be conquered. Thanks, J. Great post. V Quote Link to comment Share on other sites More sharing options...
Simplicity Posted January 29, 2009 Report Share Posted January 29, 2009 I don't think Mikeh's post is the entire picture. It is absolutely correct that if we never play the ten from QT76 and sometimes play from T76, then a finesse of the 8 is correct as there is the inference that the ten might have been played from T76. However suppose that you actually hold QT76, then you know looking at dummy that if you follow the never play the ten strategy declarer will deduce that the percentage play is to finesse the 8 and therefore you are always going to lose. Look at the implications if we suppose that we play the T a fraction y from QTxx and a fraction z from Txx. Then ignoring shape considerations if the play proceeds K629, 37.. The 8 is correct if LHO QT76 = (1-y) combinationsThe J is correct if LHO Q76 = 1/2 combination (restricted choice on T9)The A is correct if LHO T76 = (1-z) combinations However if the play were to proceed K629, 3T... Then: The J is correct if LHO is QT6 or QT76 = 1/2 + y combinations (assuming RC on 97)The A is correct if LHO is T76 = z combinations The natural strategy suggested in Mikeh's post (y=0, z=1/2) presents declarer with a guess when the ten is played and makes the 8 the correct play if the 7 is played. However we could adopt an alternate strategy of: y=1/2 and z=1 ( play T half the time from QTxx and always from Txx) Here we again have a guess when the T is played, but we also have equal odds of the Jack or 8 being correct when the 7 is played. So whats the best bet for declarer when the 7 is played? - Proabably the 8 when LHO has little clue z rates to be greater than equal to y when LHO is fairly decent he might tray making z=1/2+y, in which case probability the 8 is right = (1-y)/((1-y)+1/2+(1-z) = (1-y)/(2-2y) =1/2 Whats the best strategy in defence? Well if you're an evil genius and you decide that everyone else is playing some kind of z= y+1/2 strategy, you might try inserting the T always from QT76 safe in the knowledge that declarer will view this as a guess. Meanwhile you can now just play low from T76 and watch declarer get it wrong every time. Sometimes bridge is just amazing Quote Link to comment Share on other sites More sharing options...
Finch Posted January 29, 2009 Report Share Posted January 29, 2009 If you want to avoid your brain emitting steam, curling up and crawling down the back of your neck while you are at the table, it's a good idea to i) Think about these things offline (e.g. here), andii) Learn the most common suit combinations Quote Link to comment Share on other sites More sharing options...
vuroth Posted January 29, 2009 Report Share Posted January 29, 2009 iii) never play against Frances ;) Nice, Frances. Now, I'm afraid to play rubber bridge even! :P Quote Link to comment Share on other sites More sharing options...
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