HighLow21

Almost upvoted this. Hahahaha :-) Happy holidays Justin.

I guess maybe my reason for posting this is that until now, I hadn't played with Robots. I was under the assumption that they were reasonably good with some quirks. I'm finding them extremely erratic and disappointing.

Double and lead a trump. If 6♣ makes and it's worth a bit more than 4♥ doubled, that's too bad, but I would think we can take 4♥ down 4 more often than 6♣ is good. All in all, a good, close problem. Then again maybe I'm showing my biases as an IMPs player. I hate risking going minus when we are guaranteed a good score by not stretching... and I'm fully prepared to abandon 120-odd potential points to gain a sure 800+.

Yes and on the run of the black winners East gets squeezed in the reds. Can West do something with the A♠--namely, can he open with a club and either (1) win the second spade and return a heart (assuming one round hasn't been cashed) or (2) duck the first 2 spades to avoid rectifying the count?

Right... what exactly is wrong with Gerber?

Isn't that the exact basis upon which to decide to make a slam try? You're not afraid of any opening lead, so the 5 level must be safe. West should have realized this. Another slam try is in order. 5♣ by West. You have great spades, a fit, all prime cards, 14 HCP, and 5 losers. C'mon. East, meanwhile, has to be afraid of clubs. He can't push forward because he is subminimum--even though the diamond situation is ideal.

I find the number of hands on auctions like this where bidding 4♠ over 4♥ to be a bad move are very few. And by "bad move" I mean more than a small swing. Very few. Like maybe 10% of the time. About 50% of the time it's a great move, 20% of the time it's indifferent or a slight positive, and 20% it's an IMP swing of around -3 to -5. I can live with those odds, because there will be plenty of times when passing 4♥ is a disaster.

Why is game unlikely when you have an independent spade suit and 4 losers? All partner needs is one of the following for you to have a good shot at game: A. Kx(x)♠ B. K ♥ or QJ ♥ to any length C. Q ♦ to any length D. A ♣ or KQ ♣, especially on a club lead. Literally any of these makes game likely. In terms of the rebid -- do you really want to languish in a diamond part score or play in 5♦? My rebid is double if it suits your partnership, otherwise 3♠.

Pass and double. Two good advertisements in my view for the idea that the 5-level belongs to them not us. Hand I: It's unlikely that 5♥ is making while 5♠ is also. True, 5♠ down 1 is a good sacrifice when 5♥ is making. But it's not a good one if there's a chance to set 5 hearts. It's hard to tell who makes what, but I'm going to guess that the club suit will be an important factor. It's very possible that 5♥ is down and 5♠ is also. No need to gild the lilly by doubling because if 5♥ makes you have a guaranteed bottom. Hand II: I do not see how anything but double is correct here. 4♠ will make some of the time, but it will be from freak distribution and it will be rare. Much more rare, in my guesstimation, than any bad result for your side if you pass or bid anything else.

Agreed, that's a fair point.

So BBO has now started automatically replacing players with Robots when they leave the table. I find that they are awful. Anyone care to opine? - They bid when they shouldn't. Especially in competitive auctions, they show no judgment. - On defense I find their signalling and random play of cards to, again, show no judgment. - It seems there are very few situations in which you can make a penalty double opposite a robot. They stick the double rarely, and when they do, they often defend in ways that allow the contract to make. - Their declarer play is worse than average on BBO, and to me that is saying a lot. Has anyone else had this experience? And is there a way to turn them off entirely, say when hosting a table?

Here's one: Ax♠ KJ9x♥ AKxxxx♦ x♣

Given that this is "Interesting Bridge Hands" and not "Advanced" or "Expert" I would play this as penalty and wouldn't want a partner doubling here for takeout unless we had specific agreements.

It's a binomial problem. 16 cards contain points. What is the odds that your side will get each card? You have 26 cards out of 52. After the first point-containing card, place the second one. Assuming you get the first card, you have 25 slots left for 51 cards remaining, so 25/51. Let's start with Aces. The probability that your side has all 4 Aces, then, is 26/52 * 25/51 * 24/50 * 23/49. 3 Aces would be as follows: there are 4 different ways to get 3 of the 4 aces, so 4 * 26/52 * 25/51 * 24/50 * 26/59. (Why is the last numerator 26 rather than 23? It is because we are assuming the opponents get the 4th ace, not your side). Similarly, there are 6 different ways you can get 2 of the 4 Aces, and it's calculated in the same way. It gets more complicated when you start adding Kings, Queens and Jacks. But overall, there are a certain number of ways the cards can be distributed among the 4 hands, and a certain probability for each distribution of cards. You'd want to calculate the probability of each and then add up all the different scenarios that result in your partnership having each point count total: 0 points: only one way to do this. The opponents get all 16 point-containing cards. 1 points: only 4 ways to do this. You or your partner have 1 jack and that's it. 2 points: already there are 10 different ways to do this. 4 ways for your partnership to have exactly one Queen, and there are 6 different sets of 2 jacks your partnership could have. 3 points: 4 ways to have 3 jacks; 16 ways to have one jack and one queen; and 4 ways to have one king = 24. As you can see, this math gets extremely complicated very quickly. I'm sure it has been done, and I just spent a minute browsing Richard Pavlicek's site to find out if he's published this (I'd be surprised if he hasn't), but there's the math behind it. This is how you would calculate the probability of any number of HCP, 0-40, for the partnership on a given deal. (Note: here is the probability chart for a given PLAYER on a single deal, but the probabilities are much different from those of a partnership: http://www.bridgehands.com/P/Probability_HCP.htm) Let's just say, hypothetically, that this table tells you you have a 30% chance of getting 23 HCP or more (this shouldn't be far off). You can calculate the probability of getting fewer than a certain number of 23-HCP hands as follows: Take the probability of getting the good (23-HCP) hands upfront (30% apiece), and then the "bad" (22 or lower) hands after that (70% apiece). Multiply the probabilities together. ==> For example, getting 3 good hands out of 26, we would calculate 30% * 30% * 30% * 70% * 70% * 70% * 70% * .... = 0.000738%. Call this statistic "Initial Probability." This is the odds of getting exactly the following sequence: GOOD then GOOD then GOOD then 23 BAD hands. But we don't care what ORDER we get the good vs. bad hands in; we just want to compute how likely it is to get exactly 3 GOOD hands, and 23 BAD hands, in any possible sequence. Since each sequence is equally likely, all we have to do is take this Initial Probability, and multiply it by the NUMBER OF DIFFERENT ORDERS of possible combinations of 3 Good and 23 Bad hands. I'll spare you the details on this, but there is a factorial formula called "combinatorics" that will tell you the answer: the number of possible ways to order x objects, where x can be of the value "a" or "b" only, is: (x!) / (a!*(x-a)!) where: a = the number of "good" hands b = the number of total hands x! = the product of all integers from x to 1. Thus the right combinatoric coefficient for 3 good hands and 23 bad hands is: (26!)/(3! * (26-3)!) = 26!/(3! * 23!) = [(26 * 25 * 24)/(3 * 2 * 1)] * (23!)/(23!) = 26 * 25 * 4 = 2,600. And the probability of getting exactly 3 good hands and 23 bad hands out of 26, assuming good hands occur 30% of the time, is 2,600 * 0.000738% = 1.92%. Now, we need to compute this for other outcomes BELOW 3 good hands. In other words, we want to know, "How frequently will we get 3 OR FEWER good hands?" Thus we need to add the probability of getting 0 good hands, 1 good hand, and 2 good hands out of 26. This will tell you how rare your outcome was. In 26 boards, you would expect to get, on average, 30% * 26 = 8.4 of these deals (i.e., 23+ HCP hands). The odds of getting 0 hands that are 23 HCP or more is 1*(1 - 30%)^26 = 0.00094% The odds of getting 1 hand that is 23 HCP or more is 26*(1 - 30%)^25*(30%)^1 = 0.10% (26 is the number of ways you can be dealt 1 good hand and 25 bad ones) The odds of getting 2 hands that are 23 HCP or more is 325*(1 - 30%)^24*(30%)^2 = 0.56% (325 is the number of ways you can be dealt 2 good hands and 24 bad ones) The odds of getting 3 hands that are 23 HCP or more is 2600*(1 - 30%)^23*(30%)^3 = 1.92% (2600 is the number of ways you can be dealt 3 good hands and 23 bad ones) The sum of all these is 2.60%. In other words, IF we assume that 23-HCP hands occur 30% of the time, then on a board of 26 hands, we will get 3 or fewer of these good hands 2.6% of the time (or about 1 time in 40). That's fairly rare, but nothing egregious. However, that 30% was an assumption on my part. The correct answer for how frequently you will get dealt 23+ HCP hands is likely to be different, and the right answer will depend on the EXACT probability of getting dealt 23 HCP or more. I am virtually certain that the correct answer will be between 30% and 40% of the time, so I have provided a table for you, below, to tell you how rare your event was: 23-HCP Prob Your Event 1 chance in 30% 2.60% 39 31% 2.03% 49 32% 1.57% 64 33% 1.21% 82 34% 0.93% 108 35% 0.70% 142 36% 0.53% 188 37% 0.40% 252 38% 0.30% 339 39% 0.22% 460 40% 0.16% 628 As you can see, as the probability of getting dealt a 23-HCP hand varies from 30% up to 40%, the odds that you will get only 3 23+ HCP hands in your partnership (or fewer) starts to plummet. Assuming the probability is 40%, then you will be dealt 3 or fewer 23+ HCP hands (out of 26) only once in 628 boards -- i.e., once in a blue moon. Any feedback from anyone out there would be most appreciated.

Pass. No other choice makes any sense to me.

Exactly -- the key is to not release the space ace until you see what happens on the first round of spades.

(1) Housing insurance is different -- your decision to opt out of housing insurance doesn't affect the rates for MY insurance. It's different for car insurance or health insurance. (2) That's exactly what Obamacare is designed to do -- all citizens must have insurance that meets certain minimum requirements, and Obamacare creates a market for doing so. What's the problem?

First hand: partner should bid game 100%. Second hand: what an awful overcall.

3♥ and 3nt

That's the key point and it's covered well in the book "What's the Matter with Kansas?" Year in, year out, lower and middle class whites in many states in the Great Plains, Rockies and South have been voting against their own economic interests. They have been placing ideology ahead of what is best for them. It's not inexplicable--but the explanations are cringeworthy. I'm not going to get into them here. Suffice it to say, the Republican governors who rejected medicaid expansion do not need to answer to their constituents for their decision because the voter base agrees with them ideologically anyway. To them, government is bad, plain and simple. And trust me -- I don't care what explanation they give at the podium, or on the campaign trail, or how they spin it on Drudge or Fox News or Rush -- they rejected it to create an unseemly gap in Obamacare. The gap known as "too rich for Medicaid and too poor for Obamacare." This is where Obama miscalculated. He never anticipated that so many states would reject what is essentially free money to help the working poor within that state. But they did, "because Obama." Issue after issue has demonstrated that it doesn't matter what Obama does--whatever he does, these people will oppose, obstruct, and sabotage him. All in the name of trying to make him fail and thereby help the Red team regain political power. They may even succeed. This is why I will never vote Republican again. The political leaders in that party are willing to harm American citizens deeply for the SOLE purpose of regaining power.

Definitely not rich, at least by the standards of where a higher income tax rate would start applying if I were drawing the tax map (or if Obama were). But yes I own a cell phone and an adorable little 2013 MINI Convertible and I recognize how wealthy that makes me. I've always said that if you have cable you're rich. You can afford to junk $100+ per month on home entertainment.

Then she has Medicaid and she's good to go. If she's in one of the states that rejected the expansion, she's still uninsured and ineligible for subsidies. The real rationale for rejecting this expansion was to make Obamacare fail, truth be told. This is a prime example of Republicans sabotaging government so they can continue to run on the platform of "government is bad" and be re-elected. I've never known anywhere else that someone can interview for a job with the attitude of "this job is lousy, so hire me!" and get hired with such alacrity.

Ya gotta bid those clubs. 25% is plain unlucky but you're never going to overcome that bad luck without North bidding.

Plaw low in dummy and take Q♦. Play the K♥ -- it would be weak defense to take the A♥ first round -- then play low to dummy. You can force an entry to dummy. I believe the goal should be 4 clubs, 4 red suit tricks and 1 spade. If N has singleton A♥ you are probably home free. If he returns a diamond finesse and take the A♦ if the finesse loses, then force a club entry. If he returns a spade you should probably throw him back in in spades and rely on a minor suit finesse for a dummy entry.

As long as entries are OK the correct play is to play the Ace first. Then later lead low to the QT9. This helps you avoid a guess whenever the finesse loses the first time and it never loses relative to the double finesse since you have 9 cards to the AQT9.