Small odds problem

[1eyedjack]

[hv=pc=n&s=saj73hkq8dakjt4cq&n=skt62ht95d5cakt98&d=e&v=n&b=2&a=2hdp3sp4np5dp5sppp]266|200[/hv]

 

2H is standard weak 2.

4N is standard Blackwood.

 

Criticism of the bidding, while probably fully justified, is not solicited. This is a card play problem, and the bidding is only published in that it might influence the defensive play and odds of remaining cards.

 

I wish I could work out how to insert the play of the cards into the hand diagram, but up to the decision point the play goes:

 

♥A, ♥8, ♥4, ♥5

♥3, ♥Q, ♠4, ♥9

♣2, ♣8, ♣5, ♣Q

♠A, ♠5, ♠2, ♠8

♠3, ♠9, ♠?

 

 

Not an exact science, I know, but how do you rate the relative odds of (a) ♠K or (b) ♠T, for picking up the suit without loss?

[helene_t]

It is 8 sec vs Q8 by East, he has fewer empty spaces for the queen so I go for the finesse.

[WellSpyder]

It is 8 sec vs Q8 by East, he has fewer empty spaces for the queen so I go for the finesse.

Sounds right to me. To be more precise, East has 7 (hearts+low spades) and West has 4 (hearts+low spades), leaving 6 and 9 other cards respectively. So I think the odds are more or less 9/15 on the finesse.

[FM75]

The relative odds are not really material, so much as which is more likely here, IMO. For the odds to matter you would need to include the forms of scoring, etc.

 

For the purpose of picking up the suit, at trick 5, you need only know which hand is more likely to hold the missing queen, not by how much.

 

Information available:

East preempted with H-AJ7632, 5 hcp and by trick 5 a small club and a small spade and 5 other cards.

The locations of the 2 pointed Queens and the club Jack are not known.

The ruff at trick 2 eliminated 16 of 32 possible spade distributions by defenders. It is also in the suit we would like to have used vacant spaces to figure the current probability of the location of the honor. The spade 5 is also missing and is equivalent to the 4 in this problem if West held it.

 

 

The most precise vacant spaces information known was that at one point W had 12 and E had 7. Ignoring the ruff, but including the club trick, that dropped to 11-6. Including the known play of the 4 spades, changes it to 8-5, but that excludes the restricted choice impact. So on a pure distribution probability you would play the spade ten, because the odds are better than 1:1 that west holds the Q.

 

What we do not know, from the original information provided is what East's minimum limits are for a weak 2 hearts bid - though they are clearly lower than Frank Stewart's :).

 

If E would not preempt with less than 10 hcp on that heart holding, it is 100% that E holds the Queen. If he needs a minimum of 7 it is a worse than a coin toss because he must have at least one of them. If E needs 7-10 for his bid, the odds favor playing the King. It might even be true if he needs 6, because that can only be satisfied by holding at least one of the other missing honors.

 

So to quote Dirty Harry..."Do you feel lucky?"

 

Now to apply selection bias. At the table, declarer needed to play the King. If he didn't, then this topic probably would not have been posted. :)

[bigbenvic]

Would there be any benefit in not playing trumps at trick 4, rather A and ruff a ♦ then AK of clubs? If East ruffs, you over ruff (need to save the Ace!) if not then you get a more accurate picture of the hands and shapes to make a decision?

 

You only lose when East has a stiff Diamond, finding a side suit 6-1 (or 7-0) would count as bad luck in my book!

 

On the hand, I'm expecting east to ruff a club (the ♥3 seemed to request a ♣ shift) and now the hand makes, if he pitches instead, I'll cash a third one, dumping the ♦'s and I'll be back to guessing his shape.