# The Schnapsen Log

## Exit Cards, Stoppers, and Optimal Endplay (exits)

#### Martin Tompa

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

West (you): | East (opponent): |

♠ KJ | ♠ AQ |

(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:

West: | East: |

♠ KJ | ♠ AQ |

♥ KJ | ♥ AQ |

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

West: | East: |

♠ AQ | ♠ KJ |

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 *n* − *v*. Let’s call this the *exchange formula*.

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

West: | East: |

♠ KJ | ♠ AQ |

♥ K | ♥ A |

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
ε_{0}. As a (positive) infinitesimal, it satisfies the two
conditions

- ε
_{0}> 0 and - ε
_{0}+ ε_{0}= ε_{0}.

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 + ε_{0}. 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 ε_{0} > 0, the value ½ +
ε_{0} 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 ε_{0} + ε_{0}
= ε_{0}, we need to consider a deal in which West has 2 suits
with exit cards:

West: | East: |

♠ KJ | ♠ AQ |

♥ K | ♥ A |

♣ 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 +
ε_{0}, and the value of Deal 2d, ½ + ε_{0} +
ε_{0}, should be equal. In other words, ε_{0} +
ε_{0} = ε_{0}.

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

© 2012 Martin Tompa. All rights reserved.