[MickyB]

AQ98

opposite

xxx

for three tricks. Obviously, when you play small to the eight, it loses to a minor honour.

More below in hidden text, includes spoiler, and also includes the "unusual" part of the question:

Spoiler

I played small to the eight then small to the nine, losing to JTx offside. Partner suggested that I should have played the queen on the second round. This line of play rings a bell in my mind, but I don't think it's this combination that it applies to - can anyone recall a similar combination where you should play like this, where the restricted choice argument is negated by other considerations?

This post has been edited by MickyB: 2007-December-10, 08:00

[FrancesHinden]

I think your counting has gone a bit wrong although I agree with the answer.

Playing to the 8 then then 9 wins, when to the 9 then to the Q loses, when LHO started with

K10xx or KJxx (6 cases)
K10xxx or KJxxx (2 cases)

Playing to the 8 and then to the Q gains, when taking two deep finesses loses, when LHO has

Kxxx (1 case)
Kxx (3 cases)

the cases are not all equally likely, but it still seems to be right to finesse twice

[Edmunte1]

Suppose that we lost fist trick to the ten. Now:
a) Winning cases for playing to 9

KJxx -Tx - 3 cases (from 15) =24.25/5 =4.85%
KJxxx -T - 1 case (from 6)=7.25/6 =1.21%

b ) Winning case for playing to Q
Kxx vs J10x - 3cases (from 20)= (35.5*3)/20=5.32%
Kxxx vs. J10 -1case (from 15)=24.25/15 =1.62%

But we ignored our general strategy, when an small honour appears to finesse for the other (in other words we ignored the restricted choice principle, in all 4 cases in situation b ) East could have won with the other honour). So in long term we will win twice for the case a) (when East wins the T and J) and only once for the second case, so if we add all other cases when we win in both situations a) and (KT, KJ, KTx, KJx, KJT, KJTxx, KJTxxx) playing for
line a) will win in 50% of cases instead of
line b )will win in 44.82% of cases

[dake50]

Give me something more. LHO an aggressive bidder who would seldom pass KJxxx. Did he have a cheap chance to bid but didn't?
I really want to 3-6% reduce KJxxx, so the choice becomes a pure 'table feel' guess.

[Fluffy]

About the similar combo, I think I read the perfect play with KQ109 opposite xxxx was low to the king, then low to the ten.

[P_Marlowe]

Hi,

to answer your question:

Treat the situation similar as a double finesse
situation.

With kind regards
Marlowe

[Edmunte1]

Fluffy, on Dec 10 2007, 10:24 AM, said:

About the similar combo, I think I read the perfect play with KQ109 opposite xxxx was low to the king, then low to the ten.

Yes, as i was saying, today is raining in Madrid

[MFA]

MickyB, on Dec 10 2007, 08:29 AM, said:

Partner suggested that I should have played the queen on the second round. This line of play rings a bell in my mind, but I don't think it's this combination that it applies to - can anyone recall a similar combination where you should play like this, where the restricted choice argument is negated by other considerations?

Perhaps

AQ98xx
xx

for 5 tricks?

We start with small to the 8, H (= J or T).
What next?

Well, if west plays the other H on the second round, we'll have to finess the Q to cater to his KHx (4 cases).
This means that a good player in west will always play the H from Hxx to do us in. There are only two Hxx cases, so there's nothing we could do about that - we are bound to finess.

Then, what if west plays small on the second round? On the surface, playing for KJ/KT (rise with A) rather than for JT (play Q) looks 2:1 due to restricted choice.
But as we saw, a competent west player always plays the H from Hxx on the second round.
So we'll play east for JT tight and hook the Q, in spite of what restricted choice might tell us.

[Echognome]

What am I missing here? Seems that playing to the 8 and to the 9 plays for West to have either the T or J, whereas playing to the Q only caters to West having the K. Seems a 2 to 1 proposition.

Isn't this similar to AJ9 opposite xxx? We finesse the 9 to cater for lefty to have the KT or QT rather than for lefty to have both the K and Q.

[han]

Jlall, on Dec 10 2007, 09:11 AM, said:

I got 3.5:7 and you got 4:8, it's the same thing I just did it assuming RHO won the ten already and then took half. You can look at it either way.

edit: ok nm I was thinking there were 4 small ones out not 3 I see what you mean now lol. Don't post hungover

Does that mean that from now on you will only post when you are drunk?