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The Schnapsen Log

August 8, 2012

Exit Cards, Stoppers, and Optimal Endplay (exits)

Martin Tompa

Let’s start with a simple and familiar example consisting of only one suit:

Deal 1
West (you):East (opponent):

(In the examples, I will always use AKQJ for the ranks in a 4-card suit. But you must recognize that the hands in Deal 1 could equivalently be TQ opposite AK, or TJ opposite AK, or TJ opposite AQ, or KJ opposite TQ. I say “equivalently”, because all we care about today is the number of tricks taken, not the trick points in those tricks.)

All regular readers of this column will recognize Deal 1 as a suit that neither player wants to lead. From our point of view as West (by convention, we will always measure a hand in terms of how many tricks our alter ego West can take), if West is on lead we will take 0 tricks, and if East is on lead we will take 1 trick. Wästlund denotes this by the pair of integers (0, 1) called the outcome, where the first integer is the number of tricks West will take if West is on lead, and the second integer is the number of tricks West will take if East is on lead.

In addition to assigning an outcome to every deal, Wästlund also assigns a value. Following our convention, this is the value of the deal to West. Deal 1 is assigned a value of ½, for reasons that will be clear in a moment. First I must explain how to convert a value (which is one number) into an outcome (which is a pair of numbers). To convert a value into the number of tricks West will take, you round the value to the nearest integer; as an example, a value just slightly greater than ½ would correspond to the outcome (1, 1). But if the value is a number equidistant between two consecutive integers, such as ½, you round in favor of the player not on lead. Hence, a value of ½ corresponds to the outcome (0, 1), which we know is the correct outcome of Deal 1, and ½ is the only value that maps to this outcome.

It is interesting to see that the value of ½ assigned to Deal 1 makes some intuitive sense for a completely different reason. Consider the following deal, in which the same suit distribution of Deal 1 appears twice:

Deal 1d

In Deal 1d, West takes 1 trick no matter who is on lead (assuming optimal play by both players, which we will always assume). If East is on lead, the best East can do is pick a suit and then play ace and queen of that suit in that order, giving West 1 trick. If West is on lead, East will take both tricks in the suit West leads, but must then give up a trick to the king of the other suit. Therefore, Deal 1d has value to West of 1 trick. We will always assign values to individual suits in such a way that the value of a multi-suit deal is the sum of the values of its individual suits. From this, it makes sense that the value of each individual suit in Deal 1d (and hence the value of Deal 1) should be ½.

If we exchange the two hands in Deal 1, we get

Deal 1e

The outcome of this deal is (1, 2): West gets 1 trick if on lead, and 2 tricks if East is on lead. The value of Deal 1e is 1½: round it to the nearest integer, in favor of the player not on lead (since 1½ is a number equidistant between 1 and 2). In general, if you have a deal with value v in which each player holds n cards, when you exchange the two hands the value (to West) becomes nv. Let’s call this the exchange formula.

Now let’s make Deal 1 more interesting by adding a second suit:

Deal 2

Even though we’ve given West a losing heart, the outcome of this deal is (1, 1), an improvement for West. If East is on lead, West must get a spade trick. If West is on lead, West leads K to throw East in, and must then get a spade trick. The benefit of that losing heart to West is as an exit card for the throw-in.

What value does Wästlund assign to the heart suit in this hand? This is where infinitesimals make their first appearance. The value assigned to this heart suit is such an infinitesimal ε0. As a (positive) infinitesimal, it satisfies the two conditions

  1. ε0 > 0 and
  2. ε0 + ε0 = ε0.

In order to explain these conditions and why they make sense for such exit cards, recall that the value of a multi-suit deal is just the sum of the values of its individual suits. In the case of Deal 2, the value (to West, as usual) is ½ + ε0. Now let’s convert this value into an outcome. Remember that the rule is to round the value to the nearest integer. Because of the first condition that ε0 > 0, the value ½ + ε0 is closer to 1 than to 0, so the outcome corresponding to this value is (1, 1). Remember what this means: West takes 1 trick whoever is on lead, which is exactly what we figured out above when we analyzed what would happen in Deal 2.

To explain that weird second condition ε0 + ε0 = ε0, we need to consider a deal in which West has 2 suits with exit cards:

Deal 2d
♣ K♣ A

Whereas the heart suit in Deal 2 added some value for West, it’s not hard to see that the club suit in Deal 2d adds no further value. When West is on lead, either exit card will do, and the other is then useless for the remainder of the deal. The value of Deal 2, ½ + ε0, and the value of Deal 2d, ½ + ε0 + ε0, should be equal. In other words, ε0 + ε0 = ε0.

Next we’ll see what happens when we add an exit card for East.

© 2012 Martin Tompa. All rights reserved.


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About the Author

Martin Tompa

Martin Tompa (

I am a Professor of Computer Science & Engineering at the University of Washington, where I teach discrete mathematics, probability and statistics, design and analysis of algorithms, and other related courses. I have always loved playing games. Games are great tools for learning to think logically and are a wonderful component of happy family or social life.

Read about Winning Schnapsen, the very first and definitive book on the winning strategy for this fascinating game.


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