VM1973

As we have only one heart, our partner having 4+ does not automatically preclude the opponents from having an 8 card fit. Hypothesis: 4♠ is a bad bid under the following circumstances: A) The hand belongs to our side AND 4♠ does not make. OR B) The hand belongs to our side AND 3NT takes the same number of tricks as 4♠ OR C) The hand does not belong to our side, but 4♥ does not make. OR D) Our partner has 3+ spades. Agree or disagree?

Trying again, I have these numbers for the chance of partner having the following numbers of hearts: 0) 0.25% 1) 2.57% 2) 10.59% 3) 22.85% 4) 28.56% 5) 21.65% 6) 10.10% 7) 2.89% 8) 0.49% 9) 0.05% 10) 0.002% 11) 0.00005% 12) 0.0000003% For the total chance of partner having 4+ hearts as: 63.74% Agree or disagree?

Apparently East will duck the first diamond. That might be the opportunity to play on spades. My line of play would be more like: Win the first one in hand, cheaply. Small to the ♦J winning. Small spade towards the ♠Q in hand. If West wins and plays another club when East wins in he won't have a club to return (unless they're 4-3).

Yes, you're right. That's clearly impossible. I would like to say, though, that I ran gwnn's numbers through the anticipated calculations of how many fast defensive tricks AKQxxxx is likely to take at any given suit contract and still came up with 1.23 fast defensive tricks, on average.

I wouldn't expect partner to bid game holding this hand after 1♦-1♥ 3♥ But then again, I was taught that one should strive to invite, not to accept (at IMPs).

Well it's not impossible that I'm doing something wrong. I'm trying to put English formulae into a Spanish-speaking computer and half the time it gives me the ¿#Nombre? error which means I'm misspelling the name of the formula. If you're doing this on an Excel spreadsheet, I'd be much obliged if you'd email it to me so I can see the Spanish-language translation of the formulae in question.

1. Pass and wait for your partner to reopen. 2. You have 23 zar points (the equivalent of 11.5 HCPs) or maybe 11 if you discount the ♣Jack. Assuming partner will have 12-14 I'd say it's worth an invite. 3. You're probably going to end up in 3NT or 5♣ but it can't hurt to check on the heart situation. Double (negative). 4. I don't know how to advise you on this hand. Constructive bidding in response to overcalls is the weakest part of my game.

A week ago I would have said that you should pass in tempo. However, after everything I've learned about the hands people will bid 4M on, I have changed my mind and I would say that it completely depends on your partner.

It's simple math. There are 39 unknown cards, 13 of which are in partner's hand. The number of hands possible with the following number of spades in partner's hand are: 0) 8122425444 1) 48734552664 2) 1.21836E+11 3) 1.62449E+11 4) 1.21836E+11 5) 48734552664 6) 8122425444 As you can see, the probability of partner having 0 is the same as the probability of partner having all 6 remaining cards. Same with 1 or 5 as well as 2 or 4. Three is the most likely holding. There are 5.19835E+11 possible hands your partner might have. Accordingly the chance of your partner having each holding is: 0 1.56% 1 9.38% 2 23.44% 3 31.25% 4 23.44% 5 9.38% 6 1.56% Now anyone with a calculator can add up 31.25+23.44+9.38+1.56 and reach 65.63% Would it also interest you to know that given this holding the chances of your partner having 5+ hearts is 80.62 percent? In fact your partner's most likely holding given this hand is 6 hearts some 22.56% of the time. So for those of you who looked at that hand and thought "the opponents almost certainly have a heart fit" ... think again.

The hand has maybe 5½ losers. As mentioned it's a bit light, but really what bothers me is the stiff ♠A. I'd much rather have the ♦A. The hand is probably worth a 3½♥ bid but they don't permit ½ bids. At MPs I would only bid 3♥

You did not mention the scoring. You can make game on as little as: ♠xxx ♥Axxx ♦Qx ♣xxxx Both hands have only 5 losers. I would bid 4♥ with both hands at IMPs (or use whatever other tools are available splinters, etc.). Even if you trade the ♦Q for a ♦J you have a reasonable shot at game.

To answer your question, 3NT. Additionally, I should like to point out that your partner will have 3+ card support for you 65.63 percent of the time. Accordingly opening 2♠ in no way rules out reaching 4♠ but also preserves the possibility of reaching 3NT.

According to my calculations the suit combination in question will produce, on average, 1.223 defensive tricks. That doesn't count trump promotions, forcing defenses or other possibilities.

Well I've gone through the forum and this is what I've found: Cyberyeti: 4S mck4711: 4S or Pass jmcw: 3S or 4S daveharty: 4S Nigel: 4S / 1S Phil: 4S gwnn: 4S / 3S wyman: 4S VM1973: Pass / 1S mikeh: 4S wank: 7NT the hog: 4S Bill's partner: 3S Bill: 1S ahydra: 4S Mr.Ace: 4S Jlogic: 4S farrnbach: 1S So the score is: 4S = 13 1S = 4 3S = 3 Pass = 2 7NT = 1 Have I missed or misunderstood anyone's opinion?

Driving out the Ace of diamonds has to be the first line of play. You get 3 tricks at the cost of 1. As has already been pointed out, attacking spades will surely result in East winning and leading a club. I recommend leading low to the ♦J at trick 2.

Well your hand is 28 zar points or the equivalent of 14 HCPs. I think you should have opened it 1♥ but now that you've opened it 1NT I think you should bid 2♥

What is "B"?

Well, I think there are some problems with the idea, but I respect him for throwing it out there. It's worth discussing. The first problem is it's the wrong scoring. This idea would be much more fruitful at MPs. The second problem is your partner is quite likely to look at ♠xx and think, "I don't have one of the suits stopped" and bid 4♣ but this could be solved by prior discussion. The third problem is your partner is a passed hand so really I think you should have something outside to make it more palatable.

Sure I could do that, but really that wouldn't tell us much. Looking at the hand in question, for example, let's assume that LHO bids some number of hearts (based on his hand containing 5♥ and 1♠ and outside cards). You could say, "Well, the ♠AK aren't two defensive tricks because he has a singleton..." but don't jump to conclusions. Assuming your partner leads a spade and it goes ♠K, ♠A (ruffed) then the declarer has already been tapped and on discovering the 4-1 heart break he's in a spot of trouble. So you see, even if I did calculate the chance of both honors cashing, it wouldn't mean as much as you think it would.

Well this portion of the thread is really veering quite far from the point at hand and into the realm of epistemology, which is the philosophy concerned with the nature and scope of knowledge. I'm afraid I cannot agree with your assessment. Firstly, I would refer you to http://en.wikipedia.org/wiki/The_Wisdom_of_Crowds and I submit that we would get a better response by just polling a large number of people (let's say the subscriber base of Bridge World) and getting their opinion and averaging it as compared to consulting an expert. Secondly, I disagree that your experience with 100,000 or so hands (or having seen this hand or hands like it before) counts for much for reasons adequately documented on my blog for example http://scienceisbs.blogspot.com/2011/07/big-bang-theory.html and independently at http://faculty.unlv.edu/beisecker/Courses/Phi-101/Induction.htm Personally I would enjoy arguing the finer points of the Hempel's paradox (see http://en.wikipedia.org/wiki/Raven_paradox ) but I rather suspect you wouldn't so I will simply close my epistle by relating this story. I was at the Bridge Academy in Tarzana one day and there was some sort of hubbub about a director's call and the right decision and an appeal the details of which I never found out. But suffice it to say the auction had gone: 1 minor-Pass-Pass-1NT and then something had gone awry, but the the guy in charge there (named Jay Brown) asked a visitor to the club what he thought that bid should mean. He replied that it should show 15-18 with a stopper in the suit. Now I will tell you that I think that's a pretty lousy opinion and I expressed that at the moment he said it to which Jay said, "You can't criticize his opinion. That's (insert some name of some guy who was apparently an expert on something but I don't remember his name and it didn't mean anything to me anyway)!" Just because someone is an expert doesn't make them right.

Well, having never played ACOL I can't give much opinion. Still, it seems strange to use 2♣ checkback on hands that contain 5 spades as well as hands that do not.

I've never played with robots, so I'm going to be speculating aloud. We note that the opener has led the 9 of clubs. Now I don't know what that means because I don't know what robots lead, but I'm going to guess that it might be: A) Stiff B) High from doubleton C) Top of nothing (3 cards) For those that know that robots play MUD or something, they will have different calculations. If we assume that LHO has on average 2 clubs that means RHO has on average 5. The chances of the queen being onside under those circumstances are only 42.1 percent and assuming that you do win in dummy and lead a spade and LHO produces a low one then your odds of the queen being onside have dropped to 38.9 percent (as we have eliminated the possibility of stiff ♠Q with RHO). Now according to my calculations some 9.44% of the time RHO will have Qxxx and another 1.14% will have Qxxxx (but I will warn you that my calculations have been off before). The upshot is that I think you should rethink the assumption that finesses are automagically 50 percent.

That's the kind of outside the box thinking that I can respect. Good for you.

I am on another forum in which a person (named Foadi) is unsatisfied with all choices in all polls - even those that include none of the above, other, something else, etc. and for that reason it has become customary to include the 'foadi' option in all polls.