Contemporary Development With Functional Programming

The Schnapsen Log

July 8, 2012

How Much Help Can You Expect? (solution)

Martin Tompa

Because of your trump control and nontrump strength, closing the stock is definitely worth considering. The problem is the whereabouts of that nasty ♣A. If it is still in the stock, then you will collect all five tricks after closing. You have 45 trick points in your hand and 13 in your tricks, for a total of 58. Hans must contribute much more than the 8 points you need in order to reach 66. So this is the easy case.

But it is more likely that ♣A is in Hans’s hand: since there are 8 cards you haven’t seen and Hans is holding 5 of them, the probability is 5/8 that ♣A is among his 5. In this case, even if you run your other four tricks first, Hans will probably have no difficulty deciding to retain ♣A for last, given how many clubs are still outstanding. This means you will only add 35 trick points from your hand to the 13 already in your tricks, for a total of 48. How many trick points can you expect Hans to contribute to your four tricks? You have seen T = 19+13+45+2 = 79 trick points so far in all tricks, your hand, and J on the table. There are c = 8 cards you haven’t yet seen. This means Hans is expected to hold 5(120−79)/8 ≈ 25 trick points in hand. Assuming one of Hans’s cards is ♣A, you can expect to capture 25−11 = 14 trick points from Hans, for a total of 48+14 = 62. Not enough to close.

Actually, the reasoning in the previous paragraph is flawed. The reason is that we are assuming Hans holds ♣A, yet we ignored that fact when we computed the value of five average cards in his hand, averaging over arrangements of the cards including those where ♣A wasn’t in Hans’s hand. But we can correct the flaw using similar ideas. Just assume that ♣A is one of the cards you have seen, as though Hans showed it to you in his hand. In that case, you have seen T = 79+11 = 90 trick points, and there are only c = 7 cards you haven’t yet seen. This means that you expect Hans’s other 4 cards to total 4(120−90)/7 ≈ 17 trick points, not 14 trick points as we calculated in the previous paragraph. But this still only brings your expected total to 48+17 = 65 trick points, still not quite enough to close with much confidence.

Well then, if you’re not going to close the stock, you will have to choose a card to lead. You would like to keep Hans under 33 trick points. With your powerful trump suit, leading A seems best. You will still retain trump control and the lead, and your intention is to close the stock next trick if you draw something advantageous from the stock. The cards you could draw that make closing the stock an easy decision are K, A, ♣A, ♣K, and K, any of which gives you an extra trick. Let’s examine the position if you draw K, the weakest of these cards. Supposing that Hans discarded ♣J on your A, here is the position:

Unseen cards:

You: (26 points)

♣ T

If you now close the stock, your own spades and diamonds will add 28 trick points to your 26 already collected, for a total of 54. Hans must contribute at least 13 more points to these four tricks, for a total of 67.

© 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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