The Schnapsen Log

Exit Cards, Stoppers, and Optimal Endplay (stoppers)

Martin Tompa

Let’s continue making Deal 2 more interesting by adding an exit card for East:

Now whichever player is on lead has an elimination play available: cash the singleton ace (removing the opponent’s exit card) and then exit with the singleton king, forcing the opponent to open up the spade suit. This means the outcome of Deal 3 (to West, as usual) is (2, 1). Notice that this outcome is very much the opposite of the outcome (0, 1) of Deal 1: in Deal 3, it is an advantage to be on lead. Let’s add up the values of the individual suits and see what we get for the value of the deal as a whole. The value (to West) of the newly introduced club suit is given by the exchange formula: it is 1 − ε₀. Therefore the value of Deal 3, which is the sum of the values of its three suits, is ½ + ε₀ + (1 − ε₀) = 1½ + ε₀ − ε₀.

At this point we must be very careful about arithmetic with these infinitesimals. We already know from the identity ε₀ + ε₀ = ε₀ that infinitesimal arithmetic is strange. It may not be that ε₀ − ε₀ equals 0. In fact, we already know that ε₀ − ε₀ cannot be equal to 0, for otherwise the value of Deal 3 would be 1½, which doesn’t correspond to its known outcome (2, 1). I will denote ε₀ − ε₀ by ⊛, which Wästlund calls a “fuzzy” value. ⊛ is “fuzzy” because it is incomparable to the number 0, neither less than, equal to, nor greater than 0. What we do know is that −ε₀ < 0 < ε₀ and −ε₀ < ⊛ < ε₀.

The value of Deal 3, then, is 1½ + ⊛. Since the outcome of Deal 3 is (2, 1), this leads us to a new rule: if a value is equidistant between two integers and fuzzy, you round in favor of the player on lead.

Next, let’s add some protection for West’s exit card:

Given the lead, West can cash both aces and exit with ♥J, forcing East to open up the spade suit and making 3 tricks in all. This play shows that ♥J in Deal 4 is playing the same role of exit card that ♥K in Deal 3 was playing. But ♥A acts as “protection” for that exit card: if East is on lead, East can no longer immediately eliminate West’s exit card by cashing a heart. East’s best play is indeed to lead a heart to knock out the protector ♥A, but West has time to cash ♣A and exit with ♥J, again forcing East to open up the spade suit and again making 3 tricks. Therefore the outcome of Deal 4 is (3, 3). Since ♥A stops East from eliminating West’s exit card, it will be called a stopper.

What we see from the discussion of Deal 4 is that an exit card with a stopper in its suit is better than an unprotected exit card. In order to assign a value to the heart suit in Deal 4, Wästlund introduces another infinitesimal ε₁, with the property that ε₀ < ε₁. The value of the heart suit in Deal 4 is 1 + ε₁. This means the value of Deal 4 as a whole is

½ + (1 + ε₁) + (1 − ε₀) = 2½ + ε₁ − ε₀.

Since ε₀ < ε₁, this value is closer to 3 than to 2, so rounding to the nearest integer always produces 3. This is corroborated by the known outcome (3, 3) of Deal 4.

Here is a table summarizing all the possible distributions of 2- and 4-card suits and their assigned values. Every second row of the table comes from the previous row by applying the exchange formula.

By adding the appropriate values from this table, we now know how to calculate the value of any multi-suit deal (of up to 5 cards to each player) and how to calculate, from that value, the deal’s outcome. This tells us the maximum number of tricks West can take.

All that’s left is to discuss a simple method for finding the sequence of plays for West that will yield that maximum number of tricks.

© 2012 Martin Tompa. All rights reserved.

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