The Schnapsen Log
Exit Cards, Stoppers, and Optimal Endplay (strategy)
Martin Tompa
It’s something to be able to calculate the maximum number of tricks you will be able to take, but it’s not so useful if you can’t also calculate the sequence of plays necessary to take all those tricks. Fortunately, Wästlund provides a simple recipe for this.
If you are on lead, the first task is to choose which suit to lead. This is done as follows:
- If you have a suit whose value contains a negative infinitesimal part, choose any such suit.
- Otherwise, if you have a suit whose value contains a positive infinitesimal part, choose any such suit.
- Otherwise, choose any suit.
The second task is to choose which card to lead from the chosen suit. This decision can be based purely on what cards the players are holding in that suit, ignoring all the other suits. Even so, in Wästlund’s general version of suits with arbitrarily many cards, this decision is complicated enough that he wrote an entirely separate paper solving this problem, entitled A Solution of Two-Person Single-Suit Whist. Fortunately for us, we only have to deal with 1- and 2-card suits in the hand. In that case, the solution is simple: you can always just lead the higher card from the chosen 2-card suit.
The last task is to choose with which card to follow when your opponent leads. Here again the solution is simple and natural: if you can take the trick, do so with the lowest card possible; otherwise, discard your lowest card.
If you’ve followed today’s discussion, you now know everything you need to play 5-card symmetric whist optimally. Let’s try it out on Deal 3, repeated here for convenience:
| West: | East: |
| ♠ KJ | ♠ AQ |
| ♥ K | ♥ A |
| ♣ A | ♣ K |
Suppose you are West in this deal and you are on lead. You compute the value of the spade suit as ½, the value of the heart suit as ε0, and the value of the club suit as 1 − ε0, for a total value of 1½ + ⊛. The suit-choice method above says to lead a suit whose value contains a negative infinitesimal, which is the club suit in this hand, so you lead ♣A. After this trick, the value of the deal increases to (½ + ε0) + 1 (that extra 1 is for the club trick you have already taken) and the position is reduced to Deal 2, which is repeated below for convenience. (If, in Deal 3, you mistakenly led ♥K instead of ♣A, the value of the deal would have decreased to 1½ − ε0 and you would only make 1 trick instead of the 2 that you should make.)
| West: | East: |
| ♠ KJ | ♠ AQ |
| ♥ K | ♥ A |
Continuing the example in Deal 2, since there isn’t any suit whose value contains a negative infinitesimal, the suit-choice method says to lead a suit whose value contains a positive infinitesimal. This is the heart suit in this hand, so you continue with ♥K. This actually decreases the value of the deal from (½ + ε0) + 1 to ½ + 1. But that’s okay, because you’ve transferred the lead to East, and this value gets rounded in favor of the player not on lead. The position is now reduced to Deal 1 with East on lead.
| West: | East: |
| ♠ KJ | ♠ AQ |
Whatever East leads, you will now make a spade trick in addition to the club trick you have already taken.
Notice that the elimination play (♣A) and throw-in (♥K) happen automatically with Wästlund’s procedure. No endplay planning is required at all on your part. It’s quite beautiful, in a way that only mathematics can be.
© 2012 Martin Tompa. All rights reserved.
