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All comments by Phillip Martin
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Not sure where these numbers are coming from. As is often true, counting cases is easier and less error-prone than calculating percentages. Since there are seven spades outstanding, East is just as likely to have 4 spades as to have 3. And he is twice as likely to have a stiff club honor as to have QJ tight. So 4441 is twice as likely as 3442. If your calculations don't show that, you did something wrong.
Dec. 3
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Agreed. It's probably most people's biggest failing in card play. The problem is, the first thing you do when playing a hand is to make lots of assumptions to narrow down the number of deals you have to worry about. As the hand progresses and you learn more, it becomes easy to forget that some of your earlier conclusions were based on assumptions that it might be advisable to revisit. I can offer some pretty embarrassing examples from my own experience.
Dec. 3
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If you assume East has one of the two heart honors (I understand from Mike's comment above that turns out to be false–nonetheless I still think it's a reasonable assumption), then the finesse still works out to be about a 7 to 6 favorite. If we include the possibility that 1 is a tactical bid and East can have more than four hearts as he did, then that of course makes the finesse even better.
Dec. 3
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My assumption that East had the AK was a blind spot. I was assuming West had the A, which of course he needn't have.
Dec. 3
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The last multiplicand in each of your products should be 1, 1, 1, and 2–not 6, 6, 6, and 4. The club holding is known to be QJ or stiff honor.
Dec. 2
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It's pretty close, but I believe the finesse is a slight favorite. If we assume East has AK, he is a 15 to 6 favorite to have four hearts rather than three. He is also more likely to have four diamonds than to have three (the exact odds depending on what assumption you make about the 10). So the 4342 and 4432 shapes occur less likely than the other two shapes. They occur often enough that the finesse is not 2-to-1 as restricted choice would indicate, but it is still odds on.
Dec. 2
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I bid 2 the first time. This time, I think I'll pass.
Dec. 2
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The worst case holding for standard carding is 32 doubleton. You are just as likely to hold the two lowest cards as to hold the two highest cards, so readability does not make one method any better than the other. What does matter is the frequency with which you can't play the proper signal because doing so would cost a trick. In standard carding, you lose when your high card might be a winner; in upside-down carding, you lose when your low card might block the suit.
Dec. 2
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How does playing the jack help? The one thing partner will know for sure is you don't have a doubleton. It's either a singleton, which seems unlikely, or jack third. So the jack in fact guarantees partner will go wrong. Your only chance is to play the eight and hope partner concludes J8 doubleton is more likely than whatever else there is for him to play for.
Dec. 1
Phillip Martin edited this comment Dec. 1
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No, it's not count; it's attitude. With four (as with three), you should discourage. If you encourage, partner will assume you are ruffing the third round. So the eight could be from 8654.
Dec. 1
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They have eight clubs (since partner should have 3 for his double); we have eight or nine spades. So total tricks should be 16 or 17. I'm not bidding 3 over 3.
Nov. 27
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Thanks, Charles. Reformulating the question that way not only solves the time problem, it also avoids the issue of how you obtained the information. If we ask “from the set of all the two-children families who have a boy named Dylan, etc.,” it is clear that knowing the name affect the probability, whereas it might or might not (depending on how you obtained the information) as the problem was originally stated. Now I'm going to have to think about this type of rewording in the context of the Sleeping Beauty problem. Maybe I can finally get a handle on it.
Nov. 26
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I'm sure you could see an advantage to playing in diamonds if you worked at it. How about opposite x AJx KQJxxx KJx? Or, if partner is the aggressive sort, x Ax KQJxxx Kxxx?
Nov. 24
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1/2, 1/3, 13/27, ~1/2 (I'm ignoring the remark about P(Dylan) being 1/2. Easier just to say that the probability is small. The answer approaches 1/2 as the probability of the imposed condition approaches 0.)
Nov. 23
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Partner is supposed to be 3-suited with short spades and with clubs his longest or equal-longest suit. My doubleton spade suggests he doesn't have that. I suspect the auction didn't go as he had hoped. Still, whatever he has, I don't think he would double with a stiff diamond. So a cautious 2 seems unnecessary. I"ll go ahead and bid 2.
Nov. 21
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It's not just that it's an ace; it's the fact that it's a spade as well. Roger Stern once asserted that he would happily play rubber bridge against anyone if he were guaranteed to be dealt the space deuce on every hand.
Nov. 20
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Actually, there are 31 possible bids.
Nov. 20
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I disagree with both of you. It's not a take-out double, but doubling isn't silly.
Nov. 19
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At total points, giving the opponents an ace would be a handicap that would be almost impossible to overcome. So I choose the second option. Besides, it sounds like a good opportunity to psyche a cue-bid, then laugh at the opponents when they switch the king so that it's onside.
Nov. 19
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I think your partner should have found the spade lead. It's simply more likely to work than a heart. One way to think about it is this: It is likely declarer can find nine tricks without attacking spades. So the spade ace is not an entry. The only way it can be a working asset is if spades is your source of tricks. And it is unlikely you have enough extra strength that you can defeat this contract unless all your assets are working.
Nov. 18
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