5D contract

[Trumpace]

[hv=d=n&v=n&n=skqxxxhaxdkqcqjxx&s=sahxxdatxxxxxcxxx]133|200|Scoring: Rubber

Contract 5D. Lead HK.[/hv]

 

You somehow reach a 5 diamond contract and LHO leads the heart king.

 

Plan the play.

 

(Adv/+ please refrain from spoiling it too early)

[MattieShoe]

First instinct was:

take with A♥

unblock ♠

low trump to the board

lead low ♠ and trump

low trump to board

run KQx♠ sluffing three losers

 

But that has problems with a 3-1 or 4-0 split in diamonds or a horrific spade split... There must be something better...

[gwnn]

I give up. can you give us a hint?

[vuroth]

The hand is easy if ♦ are 2-2, which is < 50%.

 

If I unblock ♠s, cash the KQ♦ then discard a ♥ and a ♣ on the KQ♠ (then ruff a ♥ back to hand), I can pick up ♦s 3-1 if ♠s are 4-3. ♠s are 4-3 over 50% of the time, so that's good.

 

If RHO has short ♠s and 3♦s, I'll be able to overruff the ♠. I can then try a ♣ towards the Q. I think I can promote the queen unless east has the AK♣, which gives me a 76% chance of success.

 

If ♦s are 4-0 with east, then I'll follow something more like MattieShoe's line - after one round of trumps, I'll ruff a spade back, return to dummy, and play on in spades. If east trumps in, my trump loser goes away. If he doesn't, I get some round suit discards. I just have to hope, again, that spades are 4-3 so that they're running.

[vuroth]

Math question in hidden. Not really relevant to the problem

 

 

Assuming my post above is factually correct, and regardless of whether it's the best line, I think I make:

 

1. on trump 2-2

2. on spades 4-3, trumps, not 2-2

3. east has 2 spades, 3 or 4 diamonds.

 

Probabilities:

1. 41%

2. spades is 63%. 63% of non-2-2s is 63x59 = 37%

3. half of all spade 5-2s is 16.5%. Half of all diamonds not 2-2 is 30% 76% of this (club honours) ====>~3.8%

 

So my chances of making are something like 41 + 37 + 4 = 82%

 

Regardless of whether or not my line is right, is that how the math works?

 

 

Edited October 5, 2009 by vuroth

[vuroth]

I can pick up ♦s 3-1 if ♠s are 4-3.

Actually, I can handle any trump split if ♠s are 4-3.

[Simplicity]

Math question in hidden.  Not really relevant to the problem

 

 

Assuming my post above is factually correct, and regardless of whether it's the best line, I think I make:

 

1.  on trump 2-2

2.  on spades 4-3, trumps, not 2-2

3.  east has 2 spades, 3 or 4 diamonds.

 

Probabilities:

1. 41%

2. spades is 63%.  63% of non-2-2s is 63x59 = 37%

3.  half of all spade 5-2s is 16.5%.  Half of all diamonds not 2-2 is 30%  76% of this (club honours)  ====>~3.8%

 

So my chances of making are something like 41 + 37 + 4 = 82%

 

Regardless of whether or not my line is right, is that how the math works?

 

Regarding the maths of your line:

 

You will make:

 

1. When diamonds are 2-2

 

2. When diamonds are 3-1, and the long trump hand has at least 3 spades (he wont be able to ruff as you discard your losers)

 

3. When East has 3 diamonds, 2 spades and there is a club trick

 

4. When West has 4 diamonds and 4 spades (setting up a long spade to discard 1♥, 2♣

 

5. When East has 4 diamonds and spades are 4-3

 

So our probabilities are:

 

1. P(♦ 2-2) = 40.69%

 

2. P( ♦3-1 & long hand has 3♠) = 49.74 x 73.22 = 36.41%

 

3. P(East 3♦, 2♠ and not ♣AK) = 24.87 x 20.89 x 73.33 = 3.81%

 

4. P( West 4 ♦ and 4♠) = 4.78 x 21.13 = 1.01%

 

5. P( East 4 ♦ and ♠4-3) = 4.78 x 56.35 = 2.69%

 

Total = 1+2+3+4+5 = 84.61%

[bill1157]

since all you need is 2 pitches on the spades: HA, SA,DK,DQ,SK,SA pitching heart and a club.

 

Bill

[jdonn]

I am almost certain this exact play problem has been posted on the forums before. Not that I have the slightest chance of being able to find it. Where did you get this hand?

[Trumpace]

This was dealt by BBO's dealer recently.

[woefuwabit]

♥A, ♠A, ♦K

 

If east shows out, you need west to have 4 ♠s to make this contract by discarding a heart and 2 clubs. Cash ♠KQ discarding a heart and club, ruff the 4th spade to hand, reenter with ♦Q and discard another club on the last spade.

 

If west shows out, you need spades 4-3 and ♣A or K onside. Cash the ♠K discarding a heart, then cash the ♠Q. If east discards you are down. Discard a club. Ruff the 4th spade, reenter dummy with ♦Q and play the 5th spade to discard another club. If at any time east ruffs your spade, you overruff, draw trumps and play clubs twice towards the QJ.

 

Otherwise continue by cashing ♦Q. If diamond breaks 2-2, you are home with a heart and a club discard.

 

If diamonds break 3-1 either way. Cash ♠KQ discarding a heart and a club. If ♠Q gets ruffed by east, overruff, draw trumps and play clubs twice to the QJ. If ♠Q gets ruffed by west you are down. Otherwise you make the contract by ruffing a heart in to draw trumps.

 

 

Some may consider abandoning spades 4-3 if west has 3 card diamonds, instead discarding the hearts, ruffing a heart to hand, drawing trumps and play clubs twice towards the QJ. This is the inferior play. The probability of east holding AK of clubs is ~24%, while the probability of east holding 5 card spades à posteriori is 10/18 * 9/17 * 8/16 = 14.7%

[kenberg]

First three trick as the Wabbit says. If E has the four diamonds, cash the KQ of spades discarding a heart and a club. If everyone follows you are home regardless of how the clubs lie. Play another spade, ruffed. Back to the board in diamonds and lead the fifth spade just as wabbit says. But now pitch another club regardless of whether E ruffs. You will lose one club and one trump.

 

My first though was that only a miracle would save you if trumps were 4-0 but that is not really the case. I haven't calculated the odds of success but they seem to be high.

[woefuwabit]

Doesn't matter if you choose to discard or overruff if east ruffs, either case you lose 2 tricks. (2 clubs if you overruff)

[kenberg]

Doesn't matter if you choose to discard or overruff if east ruffs, either case you lose 2 tricks. (2 clubs if you overruff)

Indeed you are right! My error. I was right that when E holds four diamonds then a 4-3 spade split suffices, we don't care about the club AK. But overruffing or pitching is irrelevant, as you say.

 

So let's see. First three tricks; ♥A, ♠A,♦K.

If both follow to trick three, then second diamond is cashed.

 

If diamonds are 2-2. Success.

 

If diamonds are 3-1 and the person with thee diamonds holds three or more spades, success.

 

If E holds three diamonds and ruffs the third spade, overruff and hope W holds one or both of the high clubs. With both he might have played one at trick 1, so this is not so likely. Even if we rule out AK being with W, he should hold the A or K two thirds of the time..

 

If W holds four diamonds and four spades, success

 

If E holds four diamonds and three or four spades success.

 

Does that cover it? There are some cases where we need to ruff a heart back to hand after taking the pitches. Presumably if hearts split 8-1 we would have heard about it.

[woefuwabit]

What I meant was if E has 4 diamonds and 2 spades, you can still make the contract with A or K of ♣ onside. He will ruff the 3rd spade, so you overruff, draw trumps, and play club twice towards the QJ.

[kenberg]

IF he ruffs, yes. I think if the defender is up to pitching a card, declarer will score three spades, one heart, six diamonds. Down 1. Put another way, after taking the ace of spades and pitching a heart on the king of spades, pitching one club on the Q of spades doesn't help him. If spades are 4-2 then he can set up another pitch bu ruffing the fourth spade and getting back to the board.

 

Probably most players, quite likely including myslef, would ruff that third spade if they could, But they shouldn't.