The Schnapsen Log

Keep Calm (solution)

Martin Tompa

There is nothing you can do to prevent Rudi from winning a trick with the ♣Q he showed as part of the marriage at trick 1. If he is to fall short of 66 trick points, then, your only hope is to force him to trump ♦J with that ♣Q, which would give him a total of 65. This means you must assume that he does not hold ♣A and that he does not hold two diamonds. These assumptions make what we are constructing a desperation play, where you assume whatever you must in order to have a chance of success.

After winning his ♠T lead, then, you will want to tackle diamonds and knock out his ♣Q immediately, while you maintain control of the spade and heart suits. But this still leaves one question: does Rudi have one diamond or none? If he has one, you must start with ♦A, but if he has none, you must start with ♦J. How do you decide?

From the unseen cards at the end of trick 3, there are more 5-card hands with one diamond that Rudi could hold than 5-card hands with no diamond. But this isn’t the right way to think about the situation, because all those possible 5-card hands are not equally probable. After all, you have the additional information that Rudi closed the stock and led ♠T. He would not have made that play from some of the possible 5-card hands. So what can you infer from his actions?

Surely Rudi does not hold ♥KQ, or he would have declared that marriage and ended the game. Could he hold ♥TK? Let’s construct such a hand for Rudi and look at it from his point of view, in what I earlier called a role reversal.

Concealed cards:
♠ AKQ
♥ AQ
♣ A
♦ ATQJ

Rudi’s cards:
♠ T
♥ TKJ
♣ Q
♦ —

Could Rudi have a hand containing ♥TK such as this? Not likely, since from this hand he would lead ♥K rather than ♠T in order to establish his ♥T as a winner if you hold ♥A. Leading ♥K wins the deal unless Rudi’s opponent holds ♣A, ♥A, no ♥Q, and either ♠A or no spades at all.

We have eliminated two possible heart holdings Rudi could have, given his play. Can we infer more? Could Rudi hold ♠Q? Let’s again construct such a hand for Rudi and look at it from his point of view.

Concealed cards:
♠ AK
♥ AKQJ
♣ A
♦ AQJ

Rudi’s cards:
♠ TQ
♥ T
♣ Q
♦ T

It seems unlikely that Rudi would lead ♠T from such a hand. He is usually better off leading one of the red suits and hoping that you open up the spade suit, in case you hold ♠AK. The inference that Rudi does not hold ♠Q is not iron-clad; for instance, Rudi’s lead of ♠T would be reasonable from a hand such as this:

Rudi’s cards:
♠ TQ
♥ TQ
♣ Q
♦ —

Here Rudi could lead ♠T, hoping that you either don’t have ♠A or that you will open up the heart suit for him. Clearly, though, Rudi could just as well have chosen to lead ♥T from this hand, so the probability that he holds this hand should be weighted by ½.

Let’s summarize. Rudi does not have ♣A, does not have ♠Q with a few exceptions, and has none of the combinations ♦TQ, ♥KQ, and ♥TK. Given these inferences, the only hands Rudi could hold that contain no diamond are these:

1 ♠T ♥TQJ ♣Q ♦-
½ ♠TQ ♥TQ ♣Q ♦-
1 ♠TQ ♥TJ ♣Q ♦-

The first column shows the weight of the hand’s probability, weighted by ½ if the lead of ♠T is one of two equal choices.

In contrast, the list of hands Rudi could hold that contain one diamond is as follows:

½ ♠T ♥TQ ♣Q ♦T
1 ♠T ♥TQ ♣Q ♦Q
½ ♠T ♥TJ ♣Q ♦T
1 ♠T ♥TJ ♣Q ♦Q
½ ♠T ♥KJ ♣Q ♦T
1 ♠T ♥KJ ♣Q ♦Q
½ ♠T ♥QJ ♣Q ♦T
1 ♠T ♥QJ ♣Q ♦Q

These lists show that there are the equivalent of 2.5 hands Rudi might hold containing no diamond, and 6 hands Rudi might hold containing one diamond.

Since it is more likely Rudi holds one diamond than none, you should win his ♠T lead, then lead ♦A (holding your breath until he follows suit), and finally lead ♦J (holding your breath until he trumps with ♣Q). If Rudi doesn’t have ♣A up his sleeve, he will get no further than 65 trick points and you will win the game. Well inferred, maximizing your chances of defeating him on this deal!

© 2016 Martin Tompa. All rights reserved.

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