A puzzle

[lamford]

Declarer bids and makes game. In the play, all four 2s win tricks. How many overtricks did declarer make? (With acknowledgement to Martin Taylor).

[Lovera]

But all of these 2 are in the declarer line ?

[broze]

I can think of one where ...

 

 

declarer makes 3N exactly so I guess that must be it! Zero overtricks.

 

[nullve]

But all of these 2 are in the declarer line ?

If not, then here's a possible hand:

 

 

 

[hv=pc=n&s=skhqdjcat98765432&w=sqhjdat98765432ck&n=sjhat98765432dkcq&e=sat98765432hkdqcj&d=n&v=0&b=1&a=p3nppp&p=cackcqcjc2d3h3s3djdadkdqd2h4s4c3hjhahkhqh2s5c4d4sjsasksqs2c5d5]399|300[/hv]

I don't know yet if the puzzle really has a solution, but if it does, then it has to be 0 overtricks (in either 3N or 4M). (If either 3N or 4M makes with overtricks, then either 4N or 5M also makes (with one overtrick less), in which case there's no solution. And if 5m makes, possibly with overtricks, then the hand can be transformed (e.g. by swapping suit symbols) into one where 5M and therefore also 4M makes (with an extra overtrick), so again there's no solution.)

 

 

[stupbr]

Contract 3NT. Sorry it does not seem that for a 3NT contract, logistics can work out. Removed my original post of one overtrick comment.

Question is - if 3nt is the contract, making, and all 2’s are winners, then there will be 16 tricks minimum before considering discards and not able to come up with any layout where it can happen. Still thinking. Thanks.

[nullve]

 

 

When the last 2 is played, the remaining players have to discard from suits that

 

* only they have left (or else the 2 in that suit could not have been a winner)

* they've already cashed the 2 in (which is only possible after the other players have shown out).

 

So each opponent had to get in on a non-2 at some point to be able to cash a 2. That's at least 4 tricks for the defenders, so the only possible game contract is 3N=.

 

My previous post shows that such a 3N= hand can be constructed, so the solution is 0 overtricks (in 3N).

 

 

[Cyberyeti]

I presume you are intending this to be without infractions as there are plenty of solutions involving revokes and transferred tricks I suspect

[Lovera]

[hv=pc=n&s=s654hak32d654ca63&w=sqjth987dqjt92c87&n=sak32h654dak3ck54&e=s987hqjtd87cqjt92&d=n&v=0&b=1&a=1np3nppp]399|300|I had thought this one for +1 trick more.[/hv]

[stupbr]

Please keep this puzzle open until a complete solution is posted- with play. Thanks.

[annne]

None

[annne]

No analysis given for obvious reasons

[nige1]

 

[hv=pc=n&s=SAJ8762H765D76C76&w=SKTHAK2DKQJT9CAKQ&n=S543H43D543C85432&d=s&a=3NPPP&p=SKS3SQSASJSTS4S9S8CQS5C9S7CKH3CTS2CAD3CJC7D9C3H8C6DTC8H9C5HTH5DJC2HJH6DQH4HQH7HAH2D4D8D6DKD5DAD7D2S6HKC4]399|300|

Hit Next

for a variation on nullve's solution.[/hv]

 

[barmar]

My puzzle is why this can't happen online?

[GrahamJson]

I reckon none.

[GrahamJson]

Obviously declarer can’t cash all four Twos, so one has to be for the defence. For that to happen the defence has to lead a suit, get in twice to continue it then once more to cash the two, making four tricks in all. Meanwhile declarer must have something like 4432 opposite 3343.

[stupbr]

Solution does not exist. Will post the proof later today.

 

I am wrong - solution exists and posted on this thread.

[Vampyr]

 

[hv=pc=n&s=SAJ8762H765D76C76&w=SKTHAK2DKQJT9CAKQ&n=S543H43D543C85432&d=s&a=3NPPP&p=SKS3SQSASJSTS4S9S8CQS5C9S7CKH3CTS2CAD3CJC7D9C3H8C6DTC8H9C5HTH5DJC2HJH6DQH4HQH7HAH2D4D8D6DKD5DAD7D2S6HKC4]399|300|

Hit Next

for a variation on nullve's solution.[/hv]

 

 

This requires the opponents to cooperate by discarding winners.

[nige1]

This requires the opponents to cooperate by discarding winners.

I can't construct a hand where 4 deuces can win tricks but declarer makes exactly 9 tricks. For example, in Nullve's construction, defenders can take 13 tricks.

 

Given the constraint that 4 deuces must win tricks, however, the jettison plays are probably necessary in my construction. And they're fun.

Edited December 26, 2017 by nige1

[m1cha]

[deleted]

[Vampyr]

Simple solution to a great puzzle.

 

 

[/size]

None. (Any denomination.)

 

:

 

If the puzzle had asked for tricks, not overtricks, an answer wouldn't have been possible without specifying the denomination.

 

No need to proove that a solution exists since proof wasn't asked for. But congratulations to those who did find a proof. :)

 

If a solution exists, the answer obviously cannot be anything but 'none', and it is not necessary to worry about those distracting powerful 2s.

 

 

 

"Any denomination" does not work, since major and minor games require different numbers of tricks. But anyway this was not one of the constraints of the puzzle.

[m1cha]

"Any denomination" does not work, since major and minor games require different numbers of tricks.

True but I'm still thinking if it matters. Anyway, I have deleted my former post because I came to the conclusion that the OP probably doesn't imply what I thought it would imply.

[Vampyr]

True but I'm still thinking if it matters. Anyway, I have deleted my former post because I came to the conclusion that the OP probably doesn't imply what I thought it would imply.

 

I think it kinda does, because if you can make 11 tricks in a trump contract, there obviously isn't a unique answer.

[lamford]

When the last 2 is played, the remaining players have to discard from suits that

 

* only they have left (or else the 2 in that suit could not have been a winner)

* they've already cashed the 2 in (which is only possible after the other players have shown out).

 

So each opponent had to get in on a non-2 at some point to be able to cash a 2. That's at least 4 tricks for the defenders, so the only possible game contract is 3N=.

 

My previous post shows that such a 3N= hand can be constructed, so the solution is 0 overtricks (in 3N).

This is a good answer. 4M= and 3NT= are the only possible contracts. The problem with 4 major seems to be that the defence have to make one deuce by ruffing and one length trick, but declarer must have won a length trick with a deuce in two other suits so the partner of the defender winning the length trick with the deuce cannot have any suit to discard! So I think 3NT= is the only solution.

[Lovera]

This is a good answer. 4M= and 3NT= are the only possible contracts. The problem with 4 major seems to be that the defence have to make one deuce by ruffing and one length trick, but declarer must have won a length trick with a deuce in two other suits so the partner of the defender winning the length trick with the deuce cannot have any suit to discard! So I think 3NT= is the only solution.

My solution allows an overtick cutting comunication in club and unblocking the major suits (3+3+2+2=10). Is it valid ?

[lamford]

My solution allows an overtick cutting comunication in club and unblocking the major suits (3+3+2+2=10). Is it valid ?

I don't think it does unless you can give an order of play.