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♠ 4 3 2 |
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♥ J 4 |
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♦ A Q 10 4 2 |
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♣ A 10 5 |
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♠ 6 5 |
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♠ 9 8 7 |
♥ K Q 10 9 7 6 5 |
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♥ 3 |
♦ 6 |
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♦ J 9 8 7 5 |
♣ Q 9 6 |
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♣ J 7 3 2 |
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♠ A K Q J 10 |
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♥ A 8 2 |
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♦ K 3 |
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♣ K 8 4 |
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100%? More like 1000%. Win the ♥A and play five rounds of spades, discarding a heart and a club from dummy, and continue with the ♦K and a diamond.
If West follows to the second diamond, and could conceivably have four diamonds (i.e., he had fewer than three spades), insert the ♦10 and claim. If it wins, you have at least 12 and almost certainly 13 tricks: five spades, five diamonds, two clubs and one heart. If it loses you have 12 tricks, (one less diamond trick).
If West shows out on the second diamond (or the first), and assuming East has correctly saved five diamonds (otherwise you can concede a diamond for 12 tricks), East’s last seven cards must be five diamonds and two clubs. Win the ♦A, play the ♣K A, stripping East of clubs, and exit dummy with a low diamond. East must win and return a
diamond into dummy’s Q–10.
You must avoid the trap of playing fewer than five rounds of spades even if spades divide 3–2 or 4–1. Say you play four rounds of spades and then the ♦K and a diamond only to find East with five diamonds. The lead is now in the dummy and you can’t conveniently force a club discard from East. Say you enter your hand with the ♣K and play your last spade, discarding a club from dummy. East discards a diamond and you have to lose two more tricks. If you discard a diamond rather than a
club from dummy, East also discards a diamond. Now the best you can do is play the ♣A 10 and hope East has to take the trick and lead a diamond. Don’t hold your breath.