Crocodile or not

[jdonn]

As was pointed out to me by a prominent poster on this thread in private e-mail correspondence, I missed an important point in the analysis of this problem. And MFA, while noting that this point exists, seems to base at least part of his analysis without taking the point into consideration.

 

The distribution of all four hands is known to the declarer and to the defenders.

 

Since RHO is known to hold a doubleton, declarer should not consider finessing the Q with an original holding of AQx. By playing A and then low, if the K is onside it will fall on air.

 

Playing A and then low also has the advantage of creating a nearly impossible problem for the defense to solve regarding the blocking of the suit.

 

As the problem states that declarer knows that RHO has a doubleton, he should also be able to make the play of A and another. So the clue that declarer might have finessed the Q if he held AQx is not a clue at all.

 

Hopefully, this will conclude this discussion. I have contributed to the confusion on this issue as much as anyone.

You know that was mentioned by David C in an earlier post in the thread. In fact that point was his entire post.

[dburn]

As the prominent poster mentioned by ArtK78, I should like to apologise for my discourteous assertion that he was writing nonsense.

 

On the question of agreements, there is this to be said:

 

East must play an honour from Qx, Jx and QJ. If a partnership has no agreement, then West should never play the king on the second round of the suit. The defenders will get the position right every time East holds Q2 or J2, and get it wrong every time East holds QJ - that is, they will succeed in 66.67% of cases and cannot do better with any agreement in place.

 

Of course, they might not do worse. Suppose that you have an agreement in place that East will always play the queen from QJ. And suppose that we replay the scenario of the original problem (with no inferences available to the defenders) 300 times. Then:

 

On 100 occasions East will play the jack from J2, and West will win the next two tricks when declarer leads low from his hand.

 

On 100 occasions East will play the queen from Q2, and on 100 occasions he will play it from QJ. Since West must guess which holding East has, and since he will guess correctly half the time, the defenders will prevail in 100 of those 200 cases. This means that defenders who have an agreement will succeed exactly as often as defenders who have none: two times out of three.

 

If one is to rely (as of course one always does at the table) on inferences from the bidding, it is likely to be superior always to play the jack from QJ. It is more likely that West will be able to infer from the bidding which of East and South has a missing queen than a missing jack.

 

However, players who have recently learned that they should play randomly from QJ when declarer leads the suit are not likely to welcome the news that after all, they shouldn't.

[whereagles]

isn't this a bit of a guess?

[ArtK78]

isn't this a bit of a guess?

More than a bit. But you try to do the best that you can based on the information available to you at the time.

[MFA]

@dburn

 

How about the restricted choice that applies when declarer chooses his x from Axx? :blink:

East could have QJ, Q2, Q3, J2, J3, so the odds of QJ is only 20%.

 

But surely, if west decides to pay off to QJ tight, and to QJ tight only, he can and will do so whatever agreements and/or logic he think applies in this situation. He can take his 80% and run, so to speak.

 

I find it more interesting to try to enable E-W to improve on the 80%. To do so takes some clues about declarer's hand, obviously, which we don't have in this exercise. But in real life, with bidding and 8 preceding tricks and all, we are bound to have some clues.

 

If we then, at the critical point, don't know for sure whether p will play Q or J from QJ, we are strugling with an a priori 80-20 that clues have to overcome, regardsless of p played Q or J. So getting it right, when we need to make a Croc, will be hard.

 

But if we realize that bridge logic dictates (which is my claim) that p should play J from QJ, we are better off. When the Q shows, he is 100% to have Qx. When the J shows, it is now only a 67-33 proposition that clues have to make up for. We will be able to make an educated Croc much more often.

 

Well, nuff said from me. This discussion has suffered badly from people not discussing the same thing at the same time. (At least, this is what I hope has gone wrong).

[dburn]

@dburn

 

How about the restricted choice that applies when declarer chooses his x from Axx?

East could have QJ, Q2, Q3, J2, J3, so the odds of QJ is only 20%.

 

But surely, if west decides to pay off to QJ tight, and to QJ tight only, he can and will do so whatever agreements and/or logic he think applies in this situation. He can take his 80% and run, so to speak.

 

I find it more interesting to try to enable E-W to improve on the 80%. To do so takes some clues about declarer's hand, obviously, which we don't have in this exercise. But in real life, with bidding and 8 preceding tricks and all, we are bound to have some clues.

 

If we then, at the critical point, don't know for sure whether p will play Q or J from QJ, we are strugling with an a priori 80-20 that clues have to overcome, regardsless of p played Q or J. So getting it right, when we need to make a Croc, will be hard.

 

But if we realize that bridge logic dictates (which is my claim) that p should play J from QJ, we are better off. When the Q shows, he is 100% to have Qx. When the J shows, it is now only a 67-33 proposition that clues have to make up for. We will be able to make an educated Croc much more often.

 

Well, nuff said from me. This discussion has suffered badly from people not discussing the same thing at the same time. (At least, this is what I hope has gone wrong).

Indeed - I seem to have been guilty of keeping up with the d'Alemberts. Sorry about that.

 

We fix the North cards as 8765 and the West cards as K1094. Then, from West's point of view there are indeed five possible equally likely layouts (since everyone knows the distribution and everyone knows South has the ace). East's holding is QJ, Q3, Q2, J3 or J2. South is about to play ace and a low card, whatever his holding and whatever the defenders do.

 

Assume that the position recurs 60 times, and assume that East will choose randomly from QJ and South will choose randomly from 32. Then:

 

West will see Q-3 (East plays the queen on the first round, South wins the ace and plays the three on the second round) three times when East has QJ and twelve times when East has Q2. Similarly, he will see Q-2 three times when East has QJ and twelve times when East has Q3, and so on for J-3 and J-2. With everyone playing at random, then, the defence will beat the contract four times out of five (48 winning cases to 12 losing ones) if West plays the nine on the second round, and not two times out of three as I foolishly stated earlier.

 

Now suppose that East always plays the jack from QJ. Then:

 

West will see Q-3 only when East began with Q2, and Q-2 only when East began with Q3. The defence will prevail in all 24 of those cases.

 

West will see J-3 six times when East has QJ and twelve times when East has J2. Similarly, he will see J-2 six times when East has QJ and twelve times when East has J3. Unless the clues from the bidding and play make it more than twice as likely that the jack really is from QJ than from Jx, West is still better off playing the nine on the second round. The same applies, of course, if East always plays the queen from QJ.

 

I am not sure whether this justifies the assertion that West will be able to make an "educated" play of the king "much more often" than he could otherwise have done. Still, it only has to work once to show a profit (provided that it doesn't then fail).

[nige1]

Dealer: North Vul: None Scoring: IMP

♠ 8764 ♥ J ♦ [space] ♣ [space] ♠ KT9x ♥ [space] ♦ [space] ♣ x ♠ ?? ♥ [space] ♦ xxx ♣ [space] ♠ A?? ♥ T6 ♦ [space] ♣ [space]

South, declarer in a heart contract, needs four of the five tricks. The distribution is known to all players, so East knows West has four spades but not what they are. South leads a spade from the North hand. East plays a quack and South wins the trick with the ♠A. A low spade is led, what card do you play as West if (i) East had played the ♠J (ii) East had played the ♠Q

Suppose partner played ...

• ♠Q. Then IMO ♠9=10, ♠K=2.
  Here Partner can work out that false-carding is likely to give problems to me rather than to declarer. Even without an agreement, I feel that partner should play the J from doubleton QJ. Then ♠9 is obvious. And If partner would play the Q (or play randomly) from QJ doubleton, MFA seems to be correct that Qx is twice (or four times) as likely as QJ.
  
• ♠J. Then IMO ♠9=10, ♠K=5.
  If ♠J is what he would play from QJ or Jx doubleton, then MFA's argument still applies.