Deal 75 |
♠ ? ♥ ? ♦ ? ♣ ? |
You are South and it is your bid. Decide what you would say, then click on BID . |
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♠ A J 9 3 ♥ K 8 2 ♦ A 8 ♣ Q J 5 3 |
Deal 75 |
♠ ? ♥ ? ♦ ? ♣ ? |
With 15 points and a balanced hand of course you open 1NT. Partner's response is 2♣. What do you bid? |
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♠ A J 9 3 ♥ K 8 2 ♦ A 8 ♣ Q J 5 3 |
Deal 75 |
♠ Q 10 4 ♥ A Q J 6 ♦ 7 5 4 ♣ K 10 8 |
Partner's 2♣ bid is Stayman asking you to name a 4-card Major. You bid 2♠, but that must not have been the Major he wanted so he bids 3NT. South plays 3NT. West leads the ♦2. East plays ♦Q. You hold up on this trick and East returns the ♦3. Make a Plan, then click NEXT . |
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♠ A J 9 3 ♥ K 8 2 ♦ A 8 ♣ Q J 5 3 |
Deal 75 |
♠ Q 10 4 ♥ A Q J 6 ♦ 7 5 4 ♣ K 10 8 |
South plays 3NT. West leads the ♦2. East plays ♦Q. You hold up on this trick and East returns the ♦3. Winners: ♠ 1 : ♥ 4 : ♦ 1 : ♣ 0 : Total = 6 You need to get three more winners and you have two very clear possibilities. If East holds the ♠K you can finesse him out of it and win an additional 3 tricks in the suit. The problem with this approach is that half the time West will hold the ♠K and you will go down. Or you can drive out the ♣A and definitely set up an additional 3 tricks in that suit. The problem with this approach is that if the defender's ♦s split 5-3 they will be able to win 4 ♦s and the ♣A. If you pay attention to the opening lead this becomes a simple decision. West led the ♦2. Assuming that was fourth down then he must have led from a 4-card suit. So it looks like the ♦s must be splitting 4-4 and it is safe for you to establish ♣s. So you drive out the ♣A. The defenders take their other 2 ♦s but you make the game losing only 3 ♦s and the ♣A. Click NEXT to see the full deal. |
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♠ A J 9 3 ♥ K 8 2 ♦ A 8 ♣ Q J 5 3 |
Deal 75 |
♠ Q 10 4 ♥ A Q J 6 ♦ 7 5 4 ♣ K 10 8 |
In this layout the ♠ finesse fails, as it would do half the time.
West's lead of the ♦2 should have made you pretty sure that he had led from a 4-card suit. If the ♦2 is fourth down, there cannot be a fifth down. Additionally, East's return of the ♦3 confirmed that conclusion. If East had started with only 3 ♦s, (say ♦ Q 6 3) his correct return at trick 2 would have been his next highest, the ♦6. With 4 cards in the suit his correct return is his original fourth down. Deal 76 |
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♠ K 7 2 ♥ 10 5 4 ♦ K J 9 2 ♣ 7 6 4 |
♠ 8 6 5 ♥ 9 7 3 ♦ Q 10 6 3 ♣ A 9 2 |
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♠ A J 9 3 ♥ K 8 2 ♦ A 8 ♣ Q J 5 3 |