[joshs]

You are palying 2/1, your opps are playing precision:

P-P-4H-?

Scoring: IMP

---

a. what do you bid?

Lets say like me, you made an agressive x. and the auction continued:
P-P-4H-x
P-4S-P-P
5H-P-P-?

b. What now?

[kfgauss]

This depends a lot on your style of doubling/removing doubles of 4H. If partner likes to take out your takeout doubles, I'll pass now, but if partner thinks he needs reasonable shape to take out your doubles of 4H, then 5S is very tempting, but I'm still not sure I could bring myself to bid it.

[Part a also depends on this consideration: if partner likes to take them out, sure I'll double, but if not, I would have passed. I suppose this means that in no case would I now be bidding 5S, as I either would've passed initially or I'd pass now.]

Andy

[pclayton]

Pard heard my double of 4♥ and passed the shoe to me. I don't think I have anything really special so I'll pass.

It sounds like RHO is really heavy for 4♥.

I'm guessing both 4♠ and 5♥ are -1.

[inquiry]

Is this really a problem? Pass, pass, pass. Sure they may not make it, sure you might make something, but you have done enough.

[Al_U_Card]

Sometimes a pass is just that. You have 6 losers for a spade contract. To make 5S pard needs a 7 loser hand........and if he has it, why didn't he open? They may make 5 or 6 or 7 on the lead, they might go down too, once you have pushed them up a level, sit back and enjoy the fruits of your labor.

[Free]

I would also start with a double, but now I'm passing.

[mcphee]

Isn't partners pass forcing?

[cade909]

Pass.

[Free]

mcphee, on Jun 10 2006, 03:40 PM, said:

Isn't partners pass forcing?

Why should it be forcing? He never showed any values, and neither did you (extra's that is)...

[Winstonm]

If I were to bid, and it is borderline, I'd bid 5C. Double has a huge risk of catching partner on no good suit and leaving in the double.

[Cascade]

mcphee, on Jun 11 2006, 02:40 AM, said:

Isn't partners pass forcing?

Yes.

It forces me to pass.

I am minimum for my double and partner did not want to take any action - double or 5S - so pass seems automatic (forced) for me.

[han]

Agree that pass now is automatic, I think that the previous call (pass, double, 5C?) was much more interesting.

[joshs]

I guess I was the only person who thought this was an interesting problem.

lets have a vote:
a. How many tricks do we expect them to take? 9 tricks or less? 10 tricks? 11 tricks? 12 tricks? (Give probabilities)
b. What distribution would you give on how many tricks we can take?

[mike777]

joshs, on Jun 12 2006, 02:59 PM, said:

I guess I was the only person who thought this was an interesting problem.

lets have a vote:
a. How many tricks do we expect them to take?  9 tricks or less? 10 tricks? 11 tricks? 12 tricks? (Give probabilities)
b. What distribution would you give on how many tricks we can take?

Assuming pard is 4=2=4=3 shape with 7-8 working hcp (whcp) then I expect to take 9 tricks in spades. Assuming pard has 9-10 whcp I expect to take ten tricks.

Assuming Pard has only 2 clubs then add one more trick.

I expect the opp can make ten (65%) or eleven (35%) tricks in Hearts

[hatchett]

I would pass. I have done a good thing by doubling and pushing to the 5 level, why should I try and win the board twice in the same hand?

[Blofeld]

joshs, on Jun 12 2006, 02:59 PM, said:

I guess I was the only person who thought this was an interesting problem.

lets have a vote:
a. How many tricks do we expect them to take? 9 tricks or less? 10 tricks? 11 tricks? 12 tricks? (Give probabilities)
b. What distribution would you give on how many tricks we can take?

My gut feeling on this runs something like:

a.
9 or fewer 10%
10 40%
11 40%
12 10%

b.
9 15%
10 60%
11 20%
12 5%

But then my gut feeling was also to bid 5♠ now. Very notably partner didn't pass 4♥x, and our hand has a higher than typical ODR for the auction.

[inquiry]

joshs, on Jun 12 2006, 02:59 PM, said:

I guess I was the only person who thought this was an interesting problem.

lets have a vote:
a. How many tricks do we expect them to take? 9 tricks or less? 10 tricks? 11 tricks? 12 tricks? (Give probabilities)
b. What distribution would you give on how many tricks we can take?

A sucker bet... on any given hand, anything could make from 5♥ for them to 5♠ to you, or nothing can make. Just depends upon where the cards are. From your Response, I would guess that one side makes 11 tricks and the other 10 so the "correct" bid is 5♠ no matter which side wins 11. So what? Should that be the case, it makes 5♠ "the winning bid" but that outcome doesn't make double or 5♠ the "correct bid". And even in that case, if you bid 5♠ you might get it wrong, maybe it was far better to double, (neither side can amke 11 tricks), but if you doulbe, you miss your 5♠ when you make or you play 5♥X when they make.

Without your follow up question, if I had to bet, I would think this is a 19 trick hand. or maybe a 20 trick hand. One thing seems clear, you will not run into a horrible spade split or they would have let you play 4♠. So partner likely has five, or if only four, they are spliting 3-2, and with five, no worse than 3-1.

Without your follow up, my expectation would be for them to win 10 tricks. I expect we can win 10 tricks. I could be off by one trick for either side (we win 11, they win 10, or vise versa, or we win 9 they win 10 or vice versa).

Bidding 5♠ wins if either side can make 11 tricks, loses if neither side can make 11 tricks, loses if it lets them bid 6♥ that makes, loses big if we can make only 8 tricks. Remember, you forced your partner to bid spades, how would you feel if he tabled four small spades to the Jack? (well you tabled your hand in spades and saw that).

[joshs]

mike777, on Jun 12 2006, 03:15 PM, said:

joshs, on Jun 12 2006, 02:59 PM, said:

I guess I was the only person who thought this was an interesting problem.

lets have a vote:
a. How many tricks do we expect them to take?  9 tricks or less? 10 tricks? 11 tricks? 12 tricks? (Give probabilities)
b. What distribution would you give on how many tricks we can take?

Assuming pard is 4=2=4=3 shape with 7-8 working hcp (whcp) then I expect to take 9 tricks in spades. Assuming pard has 9-10 whcp I expect to take ten tricks.

Assuming Pard has only 2 clubs then add one more trick.

I expect the opp can make ten (65%) or eleven (35%) tricks in Hearts

I don't believe that is a possible hand. Lets say you held that hand, wouldn't you have x'ed 5H?

[mike777]

AQxx..xx...Kxxx..xxx
or
AQxx..xx....Qxxx...xxx

AQxx..xx.....Qxxx....Jxx

AQxx...xx.....Kxxx....Jxx Ok partner cannot have this hand, NV, We opened it in first and second seat.

Maybe I should x but no I would not.

[joshs]

inquiry, on Jun 12 2006, 04:02 PM, said:

joshs, on Jun 12 2006, 02:59 PM, said:

I guess I was the only person who thought this was an interesting problem.

lets have a vote:
a. How many tricks do we expect them to take?  9 tricks or less? 10 tricks? 11 tricks? 12 tricks? (Give probabilities)
b. What distribution would you give on how many tricks we can take?

A sucker bet... on any given hand, anything could make from 5♥ for them to 5♠ to you, or nothing can make. Just depends upon where the cards are. From your Response, I would guess that one side makes 11 tricks and the other 10 so the "correct" bid is 5♠ no matter which side wins 11. So what? Should that be the case, it makes 5♠ "the winning bid" but that outcome doesn't make double or 5♠ the "correct bid". And even in that case, if you bid 5♠ you might get it wrong, maybe it was far better to double, (neither side can amke 11 tricks), but if you doulbe, you miss your 5♠ when you make or you play 5♥X when they make.

Without your follow up question, if I had to bet, I would think this is a 19 trick hand. or maybe a 20 trick hand. One thing seems clear, you will not run into a horrible spade split or they would have let you play 4♠. So partner likely has five, or if only four, they are spliting 3-2, and with five, no worse than 3-1.

Without your follow up, my expectation would be for them to win 10 tricks. I expect we can win 10 tricks. I could be off by one trick for either side (we win 11, they win 10, or vise versa, or we win 9 they win 10 or vice versa).

Bidding 5♠ wins if either side can make 11 tricks, loses if neither side can make 11 tricks, loses if it lets them bid 6♥ that makes, loses big if we can make only 8 tricks. Remember, you forced your partner to bid spades, how would you feel if he tabled four small spades to the Jack? (well you tabled your hand in spades and saw that).

I don't think the cards are randomly distributed here. I think you have a lot of information about all 3 players hands...

There has already been one VERY suprising turn of events here. You xed 4S, preying to god that if partner sits, that they are actually going down. Partner, thankfully bid 4S. And then passed 5H. I would have bet good money that partner was going to x 5H. This tells you a lot of information.

I believe on this auction they are making 5H over 80% of the time and probably over 90% of the time. The main risk of bidding again, is not a phanton sac going for a number. It is pushing them to a making 6H.

My matrix would have been:
Making 6: 15%
Making 5: 75%
Making 4: 10%
Making 3 or less, close to 0

This is based on partner not xing 5H when I have shown a LOT more defense than I actually have.

As to what we are making. I find it hard to beleive that partner would not have xed even with a 42(43) 7 count. I expect most of the time partner has a 5'th spade, and usually some shape. So the question is: are we close to making? If I think they are [probably] making and we are close to making its clearing a good save at imps, unless it pushed them to 6H which is also making.

Anyway, those were my thoughts when I gambled a 5S "save". Which part of the argument do you guys disagree with?

As to the events matrix:
5S wins BIG if either side (or both) can make 5.
It loses BIG if it pushes them to a making 6
It loses big if it goes down 2 xed when they weren't making
It loses just a little if both sides were down 1.
Note that the consequences assocaited with each of these combinations are different and I don't think the relative probability of each of these are anywhere close to equal....