The Schnapsen Log
A Tangled Web (solution)
Martin Tompa
Because Katharina was on lead the previous trick and didn’t show any marriage, there is a good chance that one of the club or diamond marriage cards is the last card in the stock. If Katharina already holds ♦AKQ, she will take whatever you lead with ♦A and show the royal marriage for the championship. But if ♦Q is still in the stock, you may have a chance. Let’s assume, then, that ♦Q is the lone face-down card. This analysis will be an example of role reversal, because we will have to look at the position from Katharina’s point of view several times.
If you weren’t the Schnapsen master you have become, you would do the obvious: declare your marriage to bring you to 61 points and lead ♠Q. Katharina must trump to avoid losing immediately, but with which trump? She will quickly go through the possible cards she might draw: If it is ♦T she will win however she chose to trump, due to her club marriage, and if it is ♣A or ♠J she will lose however she chose. But if it is ♦Q, trumping with ♦A will present her with the royal marriage and the tournament. Your obvious play forces her into the assumption that the mystery stock card is exactly the card you assumed it to be, and exactly the card you’d like to hide from her! That’s not good.
You won’t do any better leading ♦T: when she takes it with ♦A, she will be pleasantly surprised to pick up the royal marriage. You also won’t do any better leading ♣A: she can trump with ♦K, cash ♦A, and show the club marriage for 70 points.
It’s looking bleak, isn’t it? The only thing that can work is a very rare, deceptive play. You lead ♠Q without declaring the marriage! The idea is to convince Katharina that the lone face-down card in the stock must be ♠K. (Leading ♠J rather than declaring the marriage is similarly deceptive, but less convincing. In addition, as we will see, it turns out to be important that Katharina thinks ♠K is in the stock and not ♠Q.) Let’s see how the deception of leading ♠Q works.
The first thing Katharina will consider is whether she can win by ducking, discarding her ♣J. That would lead to this position:
Katharina: (20 points)
♠ —
♥ —
♣ KQ
♦ AKJYou: (46 points)
♠ KJ
♥ —
♣ A
♦ TQ
From here, she can foresee your elimination play: You will cash ♣A, bringing your total to 60, breaking her marriage, and removing an exit card. You will then lead ♠J to throw her in. After this sequence, she can’t stop you from making your ♦T.
Ducking your ♠Q lead doesn’t work, so she will have to trump it. When she trumps, what card will she draw? Obviously ♠K, which you certainly cannot be holding since you didn’t show the marriage. In that case, she will see that trumping with ♦K is the winning move. This is the position she visualizes:
Katharina: (27 points)
♠ K
♥ —
♣ KQJ
♦ AYou: (41 points)
♠ J
♥ —
♣ A
♦ TQJ
From here she can lay down her cards and claim the win: cash ♦A (capturing ♦J), cash ♠K (capturing ♠J), and show the marriage for exactly 66 trick points. (This is why she needed to be convinced that the last face-down card in the stock was ♠K rather than ♠Q, which would leave her 1 trick point short.)
Imagine Katharina’s shock when she trumps with ♦K and draws not ♠K, but ♦Q! It will take her a few moments to reorient herself, but when she does she will curse roundly, for she has just thrown away the royal marriage and the title. Here is the true final position:
Katharina: (27 points)
♠ —
♥ —
♣ KQJ
♦ AQYou: (41 points)
♠ KJ
♥ —
♣ A
♦ TJ
She is now stuck, because you cannot be stopped from winning both ♣A and ♦T, which will give you at least 67 trick points. If she leads a club (say, after declaring the marriage to bring her total to 47), you will win ♣A and force her to trump ♠J with ♦Q (52 trick points). This establishes your control of trumps. She can cash ♦A (65 trick points), but then must concede. If, instead of leading a club, she plays trumps, you will win ♦T that much sooner and cash ♣A.
The crowd erupts in wild cheers, and hails you as a Schnapsen grandmaster!
© 2014 Martin Tompa. All rights reserved.
