BunnyGo

This is a minor complaint, but it has fooled me on many occasions. When presenting diagrams with just dummy and one defensive hand shown, it is common that the defensive hand is presented on the "wrong side" of the dummy. For example this thread and this one both have the South hand "misoriented" with respect to the West hand. Would it be possible to have the bug causing this fixed?

It is always possible to simulate a "bounce." Using Mac's Mail program the "command-shift-B" combination does it (also found under the "message" menu, the option "bounce.") One can google "fake a bounce with gmail" or whatever else you may use for instructions how to do it. However, some people recommend not faking a bounce with spammers, because they generally fake the "from" address and wouldn't ever receive the bounce anyways. I personally use this feature because there is someone of the same name as me at my school who seems incapable of giving his correct e-mail to his friends and coworkers. I have increased the list of people who my filters treat as spam (so that they go straight to the trash) and send them a fake bounce or sometimes a curt e-mail explaining that they should check to whom they are sending e-mails. This situation is different than normal spammers in that these people are usually good intentioned and will remove my address from their mass e-mails when they see it doesn't work.

Thanks for the suggestion, it's interesting. What do you play from dummy, the Jack? Won't this always lose three tricks when RHO has AQx (not an unlikely situation)?

I thought this hand was interesting. It occurred during a match between the University of Michigan and UCSF on BBO. [hv=pc=n&n=sa876hqj7dj54ct94&e=sq952hak63d73cq85&d=w&v=0&b=8&a=1n(15-17)p2cp2hp4hppp]266|200[/hv] Trick one was 6♠, 2, J, K Trick 2: 2♥, 7♥, A♥, 9♥ Trick 3: K♥, 8♦, 4♥, J♥ (standard signals) Trick 4: 3♥, 2♦, 5♥, Q♥ I underled my A because I had no lead that struck me as "obvious" and did not expect either declarer or dummy to have a singleton spade. Since partner played the Jack, I decided that he couldn't have the Ten, and that it could not be right to underlead my Ace again. Sadly, that was probably the only way to beat the hand. Partner had played the Jack from JT doubleton of spades. The whole hand. Furthermore on the tenth trick of the hand partner was endplayed. On the trick I was able to overtake, but it would have promoted a trick for declarer just as much as the endplay did. Is there a name for such an unfortunate situation?

I asked a friend who was teaching me a similar question once. He responded, "That's very easy to answer. In IMPs make sure you bid your games and slams; in matchpoints preserve your plus scores and take all your tricks; in a Board-a-Match all you have to do is play perfectly."

The opponents were beginners, but I'm not sure how my partner should have played the trump suit given the extra information from the bidding and lead. Thanks for the advice. [hv=pc=n&s=sk875hk5dt8753c73&n=sj943ha87dakj6cat&d=e&v=n&b=2&a=pp3cdp3sppp]266|200[/hv] Trick 1 was: 4♥, A, 9, 5. What now? Thanks to Justin for suggesting this rephrasing and representation of the problem.

I haven't played in several months, so I may be remembering the details of how it worked incorrectly.

It was a misclick, obviously.

At the top of my list is "2001: A Space Odyssey." I know it is very widely regarded, and I do enjoy several other Kubrick films, but I thought this one was a pretentious, and self-indulgent piece of "art."

In addition to cyberyeti's questions, I'm curious about what happens if the auction continues: 1♦-2♦-X-2♥ P-P-X-P P-3♦-X-P P-P If East "ethically" passes 2♥ and south doubles, then can East run to 3♦ and now reveal the misbid to his partner? Are we allowed to consider this when deciding if a correction needs to be made. It should be noted that I know virtually nothing of ACBL laws, and actually nothing of EBU laws...I'm just curious.

It's also amusing to note that if your spades were 3-3 rather than 4-2 (with the T still in West and Q in East) then they only get 3 spade tricks and a diamond. Tough break. Sometimes 4-2 spades is better as a stopper, sometimes 3-3.

I'll bid 4♥ if I expect partner will accept the "transfer" to 4 spades. Otherwise I'll just bid 4♠ myself. I'm not going to try for slam, or 5♦. I'd leave it in 3NT before considering going past the 4 level, I don't want to count on partner for too many aces and high cards.

Sorry, I wasn't quite able to discern one detail from this. Does your system only allow a bid of 4NT with the trump queen? It seems like if you don't have the trump queen you have to "answer" the blackwood bid whether it was asked or not. Is this correct?

Hi docshu, welcome to the forums. I hope you recover quickly. Allow me to address your comments for now and more experienced, and directly involved people will probably address them when they get to it. 1. This is one of the most often requested suggestions and, for reasons I do not fully appreciate, it is more difficult than I would imagine to implement. But I believe it is somewhere on a long list of things to do. 2. I agree that tricks sometimes close quickly, however as long as you haven't played to the next trick (either from your hand or dummy) then you can click on the trick (on the lower right of the screen) and see the previous trick again. 3. You can resign by clicking on the "Director!" button and choosing to resign. Not everyone chooses to do this, and so it is fairly usual (I've never seen it otherwise) that it takes 55 minutes to get the results. C'est la vie. Sincerely, Ben

I agree completely with the sentiments already stated (4♥ is fine, X is terrible, 5♥ is iffy but better than the double) but I wanted to say that I don't hate downgrading this 15 points when 3 of the points are QJ tight of spades. It's of course better than xx of spades, but not 3 points better.

Yes, it seemed like either 3♥ or 3♣ were reasonable second bids for opener, but it's not clear to me what 4♦ should be. I guess it's a splinter in support of clubs? Not clear to me that this is standard.

[hv=pc=n&s=sahaqjt43djt7caq8&n=sqt62hk2d4ckjt954]133|200|Dealer SOUTH[/hv] Thanks for the help.

Thank you for all the updates and details. From what I'm reading in the news (NYTimes) it seems that things are getting worse, but I can't fully understand it. When you get a chance, an update would be really helpful. Thanks again for the information.

FWIW I don't think any of my partnerships would find it after west opened a 15-17 NT. Maybe a strong club pair could get there.

Yes, there is a fancy explanation (both link to wikipedia). Given any base "b" you are correct in realizing that all that matters is the last digit. Another way of saying this is that what matters is the remainder of the number when you divide by "b". One can do arithmetic on the numbers from 0 to b-1 (the remainders). It's referred to as "clock arithmetic" sometimes or "modular arithmetic." If you add to numbers, you imagine "wrapping around." For example, if "b = 5" then "2 + 4 = 6 = 1" (since the remainder of 6 and 1 is the same when divided by 5). We can also multiply "3*4 = 12 = 2" (since 12 and 2 have the same remainder when divided by 5). You may see what this has to do with your problem. We want to know the first time that 2*2*2....*2 = 2 (with b = 10). There is a theorem that if "b" is prime then this number must always divide b-1. In general it will divide the Euler Totient function of "b." To define this we assume that "b" factors into products of powers of prime numbers as: b = p_1^(e_1)*p_2^(e_2)...*p_k^(e_k) Then the Euler Totient function of "b" is (p_1-1)*...*(p_k-1)*p_1^(e_1-1)*p_2^(e_2-1)...*p_k^(e_k-1) As an example with b = 10. It factors as 2*5, so the totient function is (2-1)*(5-1) = 4. This is why all the "periods" you see divide 4. I fear I may have given too much detail (and at once not enough), but I'm a bit too tired to edit this so please ask follow up questions if you want more details about parts of this.

Ah yes, of course...it's only helpful in cases that humans can clearly see are 100%, but those are not easy to program.

Continuing the "newbie" questions: How can the teams get different scores if it's a knockout. Only one of the teams can move on.

I think that is how it currently works. Out of curiosity, would the following improvement be possible? When there are few enough remaining cards, the computer simply exhaust all possible hands. This would certainly prevent trick 9 and 10 difficulties. I dunno how much slower the computer would play.

Here's the hand. I don't understand the explanation either.

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