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LHO leads ♦7, East playing the 4.
If you play a trump, LHO plays the 8 and RHO plays the 7.
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JEC #7, board 7 Tricky 6S
#1
- Andy Bowles
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Posted 2012-April-24, 01:01
LHO leads ♦7, East playing the 4.
If you play a trump, LHO plays the 8 and RHO plays the 7.
... that would still not be conclusive proof, before someone wants to explain that to me as well as if I was a 5 year-old. - gwnn
#2
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Posted 2012-April-24, 02:13
I think I got lucky on this one, am looking forward to the analysis.
"One of the painful things about our time is that those who feel certainty are stupid, and those with any imagination and understanding are filled with doubt and indecision"
-- Bertrand Russell
-- Bertrand Russell
#3
- World International Master without a clue
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Posted 2012-April-24, 05:05
when trumps don't break I don't see how to avoid club finese, so IMO only extra chance is to play LHO fpr ♠Jxxx and RHO for ♣K, play ♠10 to the ace and ♠Q
#4
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Posted 2012-April-24, 13:28
Do we have any knowledge of the opponent's leading style? For instance, from three small do they lead small, top, or 2nd high? What about from four small? From two small?
You are going to have to place the ♣ king with east, and from the ♦7 maybe the T-9-4 of diamonds at least (if they don't lead 2nd best). That places three potential cards in East's hand, compared to one in west (the ♦7. This isn't a lot to go on, but as fluffy said, with more cards (the club king) "know" (or we are screwed) in east's hand, play west to have 4♠ if either. The plan seems to be to win 4♠, 5♦ 3♣ for 12 tricks. if someone shows up with five diamonds, you will have to keep control in clubs after pulling trumps, and win 1♥, 4♦, 4♠, and 3♣.
My math is probably off, but with one more known card with East (the ♣K), assuming all else is equal, that one card being in east's hand makes a 4-1 spade split more likely to be with West (a 4:3 favorite if there is 4-1 split).
If we decide the ♦T9 are with east, as well as the ♣K) the odds that if the spades split 4-1 that west will have the four goes up. It's hard for me to decide, should i remove the low diamond spots from both hand, making it 1 know card for west (diamond x, to east diamond T9x, club K), or not. I always get confused about what cards can be used as "played" for these vacant space things. If we assume 12 vacant spaces for west and 10 for east, a 4-1 spade split is 2:1 favorite to be with West having four. If you can ignore the spot cards (7 and small diamond) then it is 13 vacant spaces for west, to 10 for east, which is just under 3:1 favorite.
All in all, since you have to place the ♣King with East, and probably the ♦T9 (it depends upon their leading agreements, and if they randomize their spot cards against slam) it is clearly correct to play west (if either) to be long in spades. You do note that they both played their highest spot spade to try to get you to play their partner for the long spade. Best to ignore such things against good opponents and play on the percentages.
You are going to have to place the ♣ king with east, and from the ♦7 maybe the T-9-4 of diamonds at least (if they don't lead 2nd best). That places three potential cards in East's hand, compared to one in west (the ♦7. This isn't a lot to go on, but as fluffy said, with more cards (the club king) "know" (or we are screwed) in east's hand, play west to have 4♠ if either. The plan seems to be to win 4♠, 5♦ 3♣ for 12 tricks. if someone shows up with five diamonds, you will have to keep control in clubs after pulling trumps, and win 1♥, 4♦, 4♠, and 3♣.
My math is probably off, but with one more known card with East (the ♣K), assuming all else is equal, that one card being in east's hand makes a 4-1 spade split more likely to be with West (a 4:3 favorite if there is 4-1 split).
If we decide the ♦T9 are with east, as well as the ♣K) the odds that if the spades split 4-1 that west will have the four goes up. It's hard for me to decide, should i remove the low diamond spots from both hand, making it 1 know card for west (diamond x, to east diamond T9x, club K), or not. I always get confused about what cards can be used as "played" for these vacant space things. If we assume 12 vacant spaces for west and 10 for east, a 4-1 spade split is 2:1 favorite to be with West having four. If you can ignore the spot cards (7 and small diamond) then it is 13 vacant spaces for west, to 10 for east, which is just under 3:1 favorite.
All in all, since you have to place the ♣King with East, and probably the ♦T9 (it depends upon their leading agreements, and if they randomize their spot cards against slam) it is clearly correct to play west (if either) to be long in spades. You do note that they both played their highest spot spade to try to get you to play their partner for the long spade. Best to ignore such things against good opponents and play on the percentages.
--Ben--
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