Contemporary Development With Functional Programming

The Schnapsen Log

July 8, 2012

How Much Help Can You Expect?

Martin Tompa

You are back at home, playing against your sister Emmi again. Your clever Uncle Hans has come to visit and has been watching over your shoulder. You take this as another sign that Hans likes you better than Emmi, which makes you happy. As he’s kibitzing, the following situation comes up at trick 2.

Unseen cards:
♣ K

Your cards:


Trump: J
Stock: 7 face-down cards
Game points: Emmi 4, You 5
Trick points: Emmi 0, You 15
On lead: You

You already have trump control, but nothing exciting to lead, so you decide to lead a passive ♣J. To your surprise, Emmi trumps with T. You draw J from the stock, and next Emmi shows the marriage and leads the queen. This doesn’t seem worth trumping, so you discard ♣Q. Emmi now closes the stock, leads A on which you must follow with J, and immediately shows you the marriage and announces, “That’s more than enough. I get 2 game points.” You are disgusted by this turn of events.

“What could I have done, Hans?” you plead, looking for sympathy.

“Well, dear,” Hans replies, “it is truly bad luck for you that your sister had both of those marriages. But I tell you what, let’s go back to trick 2 when you had the lead and see if you had any other good options.”

Hans takes the cards and reconstructs the situation at trick 2 for you. “You had trump control and all these clubs you could have run. Did you consider closing the stock yourself when you were on lead here and your sister hadn’t taken a single trick yet?” he inquires.

“How could I?” you demand, looking at your hand. “I only had 15 points in my tricks, and another, let’s see, 11, 21, 24, 27, 29, yes 29 points in my hand. That only comes to 44 points. What if Emmi threw a bunch of jacks and queens on my tricks? I’d never make it to 66.”

Hans nods. “Yes, you’re right. With bad luck, Emmi might contribute only 14 points to your tricks with these cards.” He pulls out two jacks, two queens, and a king from the remaining cards. “But with ordinary luck instead of bad luck, how many points do you think she’d contribute?”

Since both you and Emmi look blank and don’t reply, Hans continues. “We can actually figure it out. As you said, dear, you were looking at 44 points in your tricks and hand. You could also see that J on the table, that makes 46 points. How many points are there in the whole deck?”

“One hundred twenty,” you and Emmi say in unison.

“All right,” says Hans. “Since you can see 46 points, that leaves 74 points among the 12 cards you can’t see. On average, those 12 cards are then worth 74/12 points apiece. 74/12 is a little more than 6, right? Let’s call it 6. This means that, on average, Emmi’s five cards will have a total value of 5 times 6, or 30 points. Add that to your 44 points, and what do you get? On average, you will have 74 points in your tricks by the time the deal is over, as long as you close the stock and start by pulling trumps. That is so much more than you need to win. The chance that you wouldn’t get to 66 points is actually quite small. I’ll tell you what, let’s do an experiment. Take the 12 cards you hadn’t seen, shuffle them well, and deal out Emmi’s 5 cards. Add their points to your 44 and see if the total comes to at least 66. Then repeat that experiment several more times. I’ll bet that you get to 66 nearly every time.”

Once again, your Uncle Hans has shown what a smart man he is. Let’s generalize his formula, so that you can use it yourself at the table.

Suppose you have seen a total of T trick points (in all tricks, your own hand, and the face-up trump). Suppose there are c cards you haven’t yet seen (in your opponent’s hand and face-down in the stock). If you close the stock and can collect all five remaining tricks, the expected number of trick points your opponent will contribute to these five tricks is


(There are 120−T trick points distributed among the c unseen cards, so the average unseen card is worth (120−T)/c trick points.)

This notion of “expected number of trick points” is analogous to that of Expected Game Points that we’ve seen in earlier columns. It simply means the average number of trick points, averaged over all possible arrangements of the unseen cards between your opponent’s hand and the stock.

In practice, you’ll often find yourself in a situation where you want to close the stock, but can only take four of the last 5 tricks. In that case, your opponent will probably contribute about 10 fewer points, since he or she will probably have a ten or ace that you won’t be able to capture. So you can expect to capture about 5(120−T)/c − 10 trick points from your opponent.

Now it’s your turn to apply these ideas. Hans has sat down in Emmi’s place, and you find yourself in this position:

Unseen cards:

Your cards:

♣ T

Trump: J
Stock: 3 face-down cards
Game points: Hans 6, You 7
Trick points: Hans 19, You 13
On lead: You

Plan your play. When you think you have a good plan, you are welcome to read my analysis.

© 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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